Bài tập cuối chương \(1\) trang \(27\) sách giáo khoa toán lớp \(7\) tập \(1\) NXB Chân Trời Sáng Tạo
\(1.\) Thực hiện phép tính.
a) \(\displaystyle\frac{2}{5}+\displaystyle\frac{3}{5}:\left(-\displaystyle\frac{3}{2}\right)+\displaystyle\frac{1}{2};\)
b) \(2\displaystyle\frac{1}{3}+\left(-\displaystyle\frac{1}{3}\right)^2-\displaystyle\frac{3}{2};\)
c) \(\left(\displaystyle\frac{7}{8}-0,25\right):\left(\displaystyle\frac{5}{6}-0,75\right)^2;\)
d) \((-0,75)-\left[(-2)+\displaystyle\frac{3}{2}\right]:1,5+\left(\displaystyle\frac{-5}{4}\right).\)
Giải
a) \(\displaystyle\frac{2}{5}+\displaystyle\frac{3}{5}:\left(-\displaystyle\frac{3}{2}\right)+\displaystyle\frac{1}{2}=\displaystyle\frac{2}{5}+\displaystyle\frac{3}{5}.\left(\displaystyle\frac{-2}{3}\right)+\displaystyle\frac{1}{2}\)
\(=\displaystyle\frac{2}{5}+\left(\displaystyle\frac{-2}{5}\right)+\displaystyle\frac{1}{2}=0+\displaystyle\frac{1}{2}=\displaystyle\frac{1}{2}.\)
b) \(2\displaystyle\frac{1}{3}+\left(-\displaystyle\frac{1}{3}\right)^2-\displaystyle\frac{3}{2}=\displaystyle\frac{7}{3}+\displaystyle\frac{1}{9}-\displaystyle\frac{3}{2}\)
\(=\displaystyle\frac{42}{18}+\displaystyle\frac{2}{18}-\displaystyle\frac{27}{18}=\displaystyle\frac{17}{18}.\)
c) \(\left(\displaystyle\frac{7}{8}-0,25\right):\left(\displaystyle\frac{5}{6}-0,75\right)^2=\left(\displaystyle\frac{7}{8}-\displaystyle\frac{1}{4}\right):\left(\displaystyle\frac{5}{6}-\displaystyle\frac{3}{4}\right)^2\)
\(=\left(\displaystyle\frac{7}{8}-\displaystyle\frac{2}{8}\right):\left(\displaystyle\frac{10}{12}-\displaystyle\frac{9}{12}\right)^2=\displaystyle\frac{5}{8}:\left(\displaystyle\frac{1}{12}\right)^2=\displaystyle\frac{5}{8}.12^2=90.\)
d) \((-0,75)-\left[(-2)+\displaystyle\frac{3}{2}\right]:1,5+\left(\displaystyle\frac{-5}{4}\right)\)
\(=\displaystyle\frac{-3}{4}-\left(\displaystyle\frac{-4}{2}+\displaystyle\frac{3}{2}\right):\displaystyle\frac{3}{2}+\left(\displaystyle\frac{-5}{4}\right)\)
\(=\displaystyle\frac{-3}{4}-\left(\displaystyle\frac{-1}{2}\right).\displaystyle\frac{2}{3}+\left(\displaystyle\frac{-5}{4}\right)\)
\(=\displaystyle\frac{-3}{4}-\left(\displaystyle\frac{-1}{3}\right)+\left(\displaystyle\frac{-5}{4}\right)=\displaystyle\frac{-3}{4}+\left(\displaystyle\frac{-5}{4}\right)-\left(\displaystyle\frac{-1}{3}\right)\)
\(=(-2)-\left(\displaystyle\frac{-1}{3}\right)=(-2)+\displaystyle\frac{1}{3}=\displaystyle\frac{-6}{3}+\displaystyle\frac{1}{3}=\displaystyle\frac{-5}{3}.\)
\(\)
\(2.\) Thực hiện phép tính (bằng cách hợp lí nếu có thể).
a) \(\displaystyle\frac{5}{23}+\displaystyle\frac{7}{17}+0,25-\displaystyle\frac{5}{23}+\displaystyle\frac{10}{17};\)
b) \(\displaystyle\frac{3}{7}.2\displaystyle\frac{2}{3}-\displaystyle\frac{3}{7}.1\displaystyle\frac{1}{2};\)
c) \(13\displaystyle\frac{1}{4}:\left(\displaystyle\frac{-4}{7}\right)-17\displaystyle\frac{1}{4}:\left(-\displaystyle\frac{4}{7}\right);\)
d) \(\displaystyle\frac{100}{123}:\left(\displaystyle\frac{3}{4}+\displaystyle\frac{7}{12}\right)+\displaystyle\frac{23}{123}:\left(\displaystyle\frac{9}{5}-\displaystyle\frac{7}{15}\right).\)
Giải
a) \(\displaystyle\frac{5}{23}+\displaystyle\frac{7}{17}+0,25-\displaystyle\frac{5}{23}+\displaystyle\frac{10}{17}\)
\(=\left(\displaystyle\frac{5}{23}-\displaystyle\frac{5}{23}\right)+\left(\displaystyle\frac{7}{17}+\displaystyle\frac{10}{17}\right)+\displaystyle\frac{1}{4}\)
\(=0+1+\displaystyle\frac{1}{4}=\displaystyle\frac{4}{4}+\displaystyle\frac{1}{4}=\displaystyle\frac{5}{4}.\)
b)\(\displaystyle\frac{3}{7}.2\displaystyle\frac{2}{3}-\displaystyle\frac{3}{7}.1\displaystyle\frac{1}{2}=\displaystyle\frac{3}{7}.\displaystyle\frac{8}{3}-\displaystyle\frac{3}{7}.\displaystyle\frac{3}{2}\)
\(=\displaystyle\frac{3}{7}.\left(\displaystyle\frac{8}{3}-\displaystyle\frac{3}{2}\right)=\displaystyle\frac{3}{7}.\left(\displaystyle\frac{16}{6}-\displaystyle\frac{9}{6}\right)\)
\(=\displaystyle\frac{3}{7}.\displaystyle\frac{7}{6}=\displaystyle\frac{1}{2}.\)
c) \(13\displaystyle\frac{1}{4}:\left(-\displaystyle\frac{4}{7}\right)-17\displaystyle\frac{1}{4}:\left(-\displaystyle\frac{4}{7}\right)\)
\(=13\displaystyle\frac{1}{4}.\left(\displaystyle\frac{-7}{4}\right)-17\displaystyle\frac{1}{4}.\left(-\displaystyle\frac{7}{4}\right)\)
\(=\left(\displaystyle\frac{-7}{4}\right).\left(13\displaystyle\frac{1}{4}-17\displaystyle\frac{1}{4}\right)\)
\(=\left(\displaystyle\frac{-7}{4}\right).(-4)=7.\)
d) \(\displaystyle\frac{100}{123}:\left(\displaystyle\frac{3}{4}+\displaystyle\frac{7}{12}\right)+\displaystyle\frac{23}{123}:\left(\displaystyle\frac{9}{5}-\displaystyle\frac{7}{15}\right)\)
\(=\displaystyle\frac{100}{123}:\left(\displaystyle\frac{9}{12}+\displaystyle\frac{7}{12}\right)+\displaystyle\frac{23}{123}:\left(\displaystyle\frac{27}{15}-\displaystyle\frac{7}{15}\right)\)
\(=\displaystyle\frac{100}{123}:\displaystyle\frac{16}{12}+\displaystyle\frac{23}{123}:\displaystyle\frac{20}{15}\)
\(=\displaystyle\frac{100}{123}:\displaystyle\frac{4}{3}+\displaystyle\frac{23}{123}:\displaystyle\frac{4}{3}\)
\(=\displaystyle\frac{100}{123}.\displaystyle\frac{3}{4}+\displaystyle\frac{23}{123}.\displaystyle\frac{3}{4}\)
\(=\displaystyle\frac{3}{4}.\left(\displaystyle\frac{100}{123}+\displaystyle\frac{23}{123}\right)=\displaystyle\frac{3}{4}.1=\displaystyle\frac{3}{4}.\)
\(\)
\(3.\) Thực hiện phép tính.
\(a)\ \displaystyle\frac{5^{16}.27^7}{125^5.9^{11}}; \quad \qquad b)\ (-0,2)^2.5-\displaystyle\frac{2^{13}.27^3}{4^6.9^5}; \quad \qquad c)\ \displaystyle\frac{5^6+2^2.25^3+2^3.125^2}{26.5^6}.\)
Giải
a) \(\displaystyle\frac{5^{16}.27^7}{125^5.9^{11}}=\displaystyle\frac{5^{16}.(3^3)^7}{(5^3)^5.(3^2)^{11}}\)
\(=\displaystyle\frac{5^{16}.3^{3.7}}{5^{3.5}.3^{2.11}}=\displaystyle\frac{5^{16}.3^{21}}{5^{15}.3^{22}}=\displaystyle\frac{5}{3}.\)
b) \((-0,2)^2.5-\displaystyle\frac{2^{13}.27^3}{4^6.9^5}=\left(\displaystyle\frac{-1}{5}\right)^2.5-\displaystyle\frac{(2^{13}.(3^3)^3}{(2^2)^6.(3^2)^5}\)
\(=\displaystyle\frac{1}{5^2}.5-\displaystyle\frac{2^{13}.3^9}{2^{12}.3^{10}}=\displaystyle\frac{1}{5}-\displaystyle\frac{2}{3}=\displaystyle\frac{-7}{15}.\)
c) \(\displaystyle\frac{5^6+2^2.25^3+2^3.125^2}{26.5^6}=\displaystyle\frac{5^6+2^2.(5^2)^3+2^3.(5^3)^2}{26.5^6}\)
\(=\displaystyle\frac{5^6+2^2.5^{2.3}+2^3.5^{3.2}}{26.5^6}=\displaystyle\frac{5^6+2^2.5^6+2^3.5^6}{26.5^6}\)
\(=\displaystyle\frac{5^6(1+2^2+2^3)}{26.5^6}=\displaystyle\frac{5^6.13}{26.5^6}=\displaystyle\frac{1}{2}.\)
\(\)
\(4.\) Tính giá trị các biểu thức sau.
a) A=\(\left[(-0,5)-\displaystyle\frac{3}{5}\right]:(-3)+\displaystyle\frac{1}{3}-\left(-\displaystyle\frac{1}{6}\right):(-2);\)
b) B=\(\left(\displaystyle\frac{2}{25}-0,036\right):\displaystyle\frac{11}{50}-\left[\left(3\displaystyle\frac{1}{4}-2\displaystyle\frac{4}{9}\right)\right].\displaystyle\frac{9}{29}.\)
Giải
a) A=\(\left[(-0,5)-\displaystyle\frac{3}{5}\right]:(-3)+\displaystyle\frac{1}{3}-\left(-\displaystyle\frac{1}{6}\right):(-2)\)
A=\(\left(\displaystyle\frac{-1}{2}-\displaystyle\frac{3}{5}\right).\left(\displaystyle\frac{-1}{3}\right)+\displaystyle\frac{1}{3}-\left(\displaystyle\frac{-1}{6}\right).\left(\displaystyle\frac{-1}{2}\right)\)
A=\(\left(\displaystyle\frac{-5}{10}-\displaystyle\frac{6}{10}\right).\left(\displaystyle\frac{-1}{3}\right)+\displaystyle\frac{1}{3}-\displaystyle\frac{1}{12}\)
A=\(\left(\displaystyle\frac{-11}{10}\right).\left(\displaystyle\frac{-1}{3}\right)+\displaystyle\frac{1}{3}-\displaystyle\frac{1}{12}\)
A=\(\displaystyle\frac{11}{30}+\displaystyle\frac{1}{3}-\displaystyle\frac{1}{12}\)
A=\(\displaystyle\frac{22}{60}+\displaystyle\frac{20}{60}-\displaystyle\frac{5}{60}=\displaystyle\frac{37}{60}.\)
b) B=\(\left(\displaystyle\frac{2}{25}-0,036\right):\displaystyle\frac{11}{50}-\left[\left(3\displaystyle\frac{1}{4}-2\displaystyle\frac{4}{9}\right)\right].\displaystyle\frac{9}{29}\)
B=\(\left(\displaystyle\frac{2}{25}-\displaystyle\frac{36}{1000}\right).\displaystyle\frac{50}{11}-\left[\left(3+\displaystyle\frac{1}{4}\right)-\left(2+\displaystyle\frac{4}{9}\right)\right].\displaystyle\frac{9}{29}\)
B=\(\left(\displaystyle\frac{80}{1000}-\displaystyle\frac{36}{1000}\right).\displaystyle\frac{50}{11}-\left(3+\displaystyle\frac{1}{4}-2-\displaystyle\frac{4}{9}\right).\displaystyle\frac{9}{29}\)
B=\(\displaystyle\frac{11}{250}.\displaystyle\frac{50}{11}-\left(1+\displaystyle\frac{1}{4}-\displaystyle\frac{4}{9}\right).\displaystyle\frac{9}{29}\)
B=\(\displaystyle\frac{50}{250}-\left(\displaystyle\frac{36}{36}+\displaystyle\frac{9}{36}-\displaystyle\frac{16}{36}\right).\displaystyle\frac{9}{29}\)
B=\(\displaystyle\frac{1}{5}-\displaystyle\frac{29}{36}.\displaystyle\frac{9}{29}=\displaystyle\frac{1}{5}-\displaystyle\frac{1}{4}=\displaystyle\frac{4}{20}-\displaystyle\frac{5}{20}=\displaystyle\frac{-1}{20}.\)
\(\)
\(5.\) Tìm x, biết:
a) \(\displaystyle\frac{-3}{5}x=\displaystyle\frac{12}{25};\)
b) \(\displaystyle\frac{3}{5}x-\displaystyle\frac{3}{4}=-1\displaystyle\frac{1}{2};\)
c) \(\displaystyle\frac{2}{5}+\displaystyle\frac{3}{5}:x=0,5;\)
d) \(\displaystyle\frac{3}{4}-(x-\displaystyle\frac{1}{2})=1\displaystyle\frac{2}{3};\)
e) \(2\displaystyle\frac{2}{15}:\left(\displaystyle\frac{1}{3}-5x\right)=-2\displaystyle\frac{2}{5};\)
g) \(x^2+\displaystyle\frac{1}{9}=\displaystyle\frac{5}{3}:3.\)
Giải
a) \(\displaystyle\frac{-3}{5}x=\displaystyle\frac{12}{25}\)
\(x=\displaystyle\frac{12}{25}:\left(-\displaystyle\frac{3}{5}\right)\)
\(x=\displaystyle\frac{12}{25}.\displaystyle\frac{-5}{3}=\displaystyle\frac{-4}{5}.\)
b) \(\displaystyle\frac{3}{5}x-\displaystyle\frac{3}{4}=-1\displaystyle\frac{1}{2}\)
\(\displaystyle\frac{3}{5}x-\displaystyle\frac{3}{4}=\displaystyle\frac{-3}{2}\)
\(\displaystyle\frac{3}{5}x=\displaystyle\frac{-3}{2}+\displaystyle\frac{3}{4}\)
\(\displaystyle\frac{3}{5}x=\displaystyle\frac{-3}{4}\)
\(x=\displaystyle\frac{-3}{4}:\displaystyle\frac{3}{5}\)
\(x=\displaystyle\frac{-3}{4}.\displaystyle\frac{5}{3}=\displaystyle\frac{-5}{4}.\)
c) \(\displaystyle\frac{2}{5}+\displaystyle\frac{3}{5}:x=0,5\)
\(\displaystyle\frac{2}{5}+\displaystyle\frac{3}{5}:x=\displaystyle\frac{1}{2}\)
\(\displaystyle\frac{3}{5}:x=\displaystyle\frac{1}{2}-\displaystyle\frac{2}{5}\)
\(\displaystyle\frac{3}{5}:x=\displaystyle\frac{1}{10}\)
\(x=\displaystyle\frac{3}{5}:\displaystyle\frac{1}{10}\)
\(x=\displaystyle\frac{3}{5}.10=6.\)
d) \(\displaystyle\frac{3}{4}-(x-\displaystyle\frac{1}{2})=1\displaystyle\frac{2}{3}\)
\(\displaystyle\frac{3}{4}-(x-\displaystyle\frac{1}{2})=\displaystyle\frac{5}{3}\)
\(x-\displaystyle\frac{1}{2}=\displaystyle\frac{3}{4}-\displaystyle\frac{5}{3}\)
\(x-\displaystyle\frac{1}{2}=\displaystyle\frac{-11}{12}\)
\(x=\displaystyle\frac{-11}{12}+\displaystyle\frac{1}{2}=\displaystyle\frac{-5}{12}.\)
e) \(2\displaystyle\frac{2}{15}:\left(\displaystyle\frac{1}{3}-5x\right)=-2\displaystyle\frac{2}{5}\)
\(\displaystyle\frac{32}{15}:\left(\displaystyle\frac{1}{3}-5x\right)=\displaystyle\frac{-12}{5}\)
\(\displaystyle\frac{1}{3}-5x=\displaystyle\frac{32}{15}:\displaystyle\frac{-12}{5}\)
\(\displaystyle\frac{1}{3}-5x=\displaystyle\frac{-8}{9}\)
\(5x=\displaystyle\frac{1}{3}-\left(\displaystyle\frac{-8}{9}\right)\)
\(5x=\displaystyle\frac{11}{9}\)
\(x=\displaystyle\frac{11}{9}:5=\displaystyle\frac{11}{45}.\)
g) \(x^2+\displaystyle\frac{1}{9}=\displaystyle\frac{5}{3}:3\)
\(x^2=\displaystyle\frac{5}{3}.\displaystyle\frac{1}{3}-\displaystyle\frac{1}{9}\)
\(x^2=\displaystyle\frac{4}{9}\)
\(TH1:\ x^2=\left(\displaystyle\frac{2}{3}\right)^2\)
\(x=\displaystyle\frac{2}{3}.\)
\(TH2:\ x^2=\left(\displaystyle\frac{-2}{3}\right)^2\)
\(x=\displaystyle\frac{-2}{3}.\)
\(\)
\(6.\) a) Tính diện tích hình thang ABCD có các kích thước như hình sau:
b) Hình thoi MNPQ có diện tích bằng diện tích hình thang ABCD ở câu a, đường chéo MP = \(\displaystyle\frac{35}{4}\) m. Tính độ dài NQ.
Giải
a) Diện tích hình thang ABCD bằng:
\(S=\displaystyle\frac{AB+CD}{2}.AH=\displaystyle\frac{\displaystyle\frac{11}{3}+\displaystyle\frac{17}{2}}{2}.3=\displaystyle\frac{73}{4}\ (m^2).\)
b) Do diện tích hình thoi MNPQ bằng diện tích hình thang ABCD ở câu a nên diện tích hình thoi MNPQ bằng \(\displaystyle\frac{73}{4}\ m^2.\)
Ta có diện tích MNPQ bằng \(\displaystyle\frac{MP.NQ}{2}\) nên độ dài NQ bằng:
\(2.\displaystyle\frac{73}{4}:\displaystyle\frac{35}{4}=\displaystyle\frac{73}{2}.\displaystyle\frac{4}{35}=\displaystyle\frac{146}{35}\ (m).\)
Vậy độ dài NQ bằng \(\displaystyle\frac{146}{35}\ m\).
\(\)
Bài tập cuối chương 1 (Phần 2: bài 7 đến bài 11)
Xem thêm các bài giải khác tại: Giải bài tập SGK Toán Lớp 7 – NXB Chân Trời Sáng Tạo.
Thông tin liên hệ & mạng xã hội:
Website: https://bumbii.com/
Facebook: https://www.facebook.com/bumbiiapp
Pinterest: https://www.pinterest.com/bumbiitech
