Chương 1 – Bài 3. Lũy thừa của một số hữu tỉ

Chương \(1\)- Bài \(3\). Lũy thừa của một số hữu tỉ trang \(14\) vở bài tập toán lớp \(7\) tập \(1\) NXB Chân Trời Sáng Tạo.

\(1.\) Viết các số sau dưới dạng lũy thừa với số mũ lớn hơn \(1\).

\(9; \displaystyle\frac{1}{8}; \displaystyle\frac{-1}{27}; \displaystyle\frac{81}{16}; \displaystyle\frac{8}{125}; 0,0625.\)

Giải

\(9 = 3^{2}; \hspace{3cm} \quad \displaystyle\frac{1}{8}=\left(\displaystyle\frac{1}{2}\right)^{3}; \hspace{2,5cm} \displaystyle\frac{-1}{27}=\left(\displaystyle\frac{-1}{3}\right)^{3};\)

\(\displaystyle\frac{81}{16}=\left(\displaystyle\frac{3}{2}\right)^{4}; \hspace{2cm} \displaystyle\frac{8}{125}=\left(\displaystyle\frac{2}{5}\right)^{3}; \hspace{2cm} 0,0625=(0,5)^{4}.\)

\(2.\) Tính:

a) \(\left(\displaystyle\frac{-1}{3}\right)^4;\left(\displaystyle\frac{-2}{3}\right)^3;\left(2\displaystyle\frac{1}{2}\right)^3;(0,2)^3;(-125,9)^0;(0,3)^4.\)

b) \(\left(\displaystyle\frac{-1}{2}\right)^2;\left(\displaystyle\frac{-1}{2}\right)^3;\left(\displaystyle\frac{-1}{2}\right)^4;\left(\displaystyle\frac{-1}{2}\right)^5.\)

Giải

a) \(\left(\displaystyle\frac{-1}{3}\right)^4=\displaystyle\frac{(-1)^4}{3^4}=\displaystyle\frac{1}{81};\)

\(\left(\displaystyle\frac{-2}{3}\right)^3=\displaystyle\frac{(-2)^3}{3^3}=\displaystyle\frac{-8}{27};\)

\(\left(2\displaystyle\frac{1}{2}\right)^3=\left(\displaystyle\frac{5}{2}\right)^3=\displaystyle\frac{5^3}{2^3}=\displaystyle\frac{125}{8};\)

\((-0,2)^3=\left(\displaystyle\frac{-1}{5}\right)^3=\displaystyle\frac{1^3}{5^3}=\displaystyle\frac{-1}{125};\)

\((-125,9)^0=1\)

\((0,3)^4=\left(\displaystyle\frac{3}{10}\right)^4=\displaystyle\frac{3^4}{10^4}=\displaystyle\frac{81}{10000}.\)

b) \(\left(\displaystyle\frac{-1}{2}\right)^2=\displaystyle\frac{(-1)^2}{2^2}=\displaystyle\frac{1}{4};\)

\(\left(\displaystyle\frac{-1}{2}\right)^3=\displaystyle\frac{(-1)^3}{2^3}=\displaystyle\frac{-1}{8};\)

\(\left(\displaystyle\frac{-1}{2}\right)^4=\displaystyle\frac{(-1)^4}{2^4}=\displaystyle\frac{1}{16};\)

\(\left(\displaystyle\frac{-1}{2}\right)^5=\displaystyle\frac{(-1)^5}{2^5}=\displaystyle\frac{-1}{32}.\)

\(3.\) Tính:

a) \(\left(-\displaystyle\frac{2}{3}\right)^3.\left(-\displaystyle\frac{2}{3}\right)^2; \hspace{2,5cm} b) (0,15)^7:(0,15)^5;\)

c) \(\left(\displaystyle\frac{3}{5}\right)^{15}:\left(-\displaystyle\frac{27}{125}\right)^5; \hspace{2cm} d) (\displaystyle\frac{1}{7})^4.\displaystyle\frac{1}{7}.49^3.\)

Giải

a) \(\left(-\displaystyle\frac{2}{3}\right)^3.\left(-\displaystyle\frac{2}{3}\right)^2=\left(-\displaystyle\frac{2}{3}\right)^5=\displaystyle\frac{-32}{243};\)

b) \((0,15)^7:(0,15)^5=(0,15)^2=0,0225;\)

c) \(\left(\displaystyle\frac{3}{5}\right)^{15}:\left(\displaystyle\frac{27}{125}\right)^5=\left(\displaystyle\frac{3}{5}\right)^15:\left[\left(\displaystyle\frac{3}{5}\right)^3\right]^5\)

\(=\left(\displaystyle\frac{3}{5}\right)^{15}:\left(\displaystyle\frac{3}{5}\right)^{3.5}=\left(\displaystyle\frac{3}{5}\right)^{15}:\left(\displaystyle\frac{3}{5}\right)^{15}=1;\)

d) \(\left(\displaystyle\frac{1}{7}\right)^4.\displaystyle\frac{1}{7}.49^3=\left(\displaystyle\frac{1}{7}\right)^4.\displaystyle\frac{1}{7}.(7^2)^3\)

\(=\left(\displaystyle\frac{1}{7}\right)^5.7^6=\displaystyle\frac{1}{7^5}.7^6=7.\)

\(4.\) Tìm x, biết:

a) \(x:\left(\displaystyle\frac{-1}{3}\right)^3=\displaystyle\frac{-1}{3}; \hspace{2,5cm} b)x.\left(\displaystyle\frac{-3}{7}\right)^5=\left(\displaystyle\frac{-3}{7}\right)^7;\)

c) \(\left(\displaystyle\frac{-2}{3}\right)^{12}:x=\left(\displaystyle\frac{-2}{3}\right)^9; \qquad d) \left(x+\displaystyle\frac{1}{3}\right)^2=\displaystyle\frac{1}{25}.\)

Giải

a) \(x:\left(\displaystyle\frac{-1}{3}\right)^3=\displaystyle\frac{-1}{3}\)

\(x=\displaystyle\frac{-1}{3}.\left(\displaystyle\frac{-1}{3}\right)^3\)

\(x=\left(\displaystyle\frac{-1}{3}\right)^4=\displaystyle\frac{1}{81}.\)

b) \(x.\left(\displaystyle\frac{-3}{7}\right)^5=\left(\displaystyle\frac{-3}{7}\right)^7\)

\(x=\left(\displaystyle\frac{-3}{7}\right)^7:\left(\displaystyle\frac{-3}{7}\right)^5\)

\(x=\left(\displaystyle\frac{-3}{7}\right)^2=\displaystyle\frac{9}{49}\)

c) \(\left(\displaystyle\frac{-2}{3}\right)^{12}:x=\left(\displaystyle\frac{-2}{3}\right)^9\)

\(x=\left(\displaystyle\frac{-2}{3}\right)^{12}:\left(\displaystyle\frac{-2}{3}\right)^9\)

\(x=\left(\displaystyle\frac{-2}{3}\right)^3=\displaystyle\frac{-8}{27}.\)

d) \(\left(x+\displaystyle\frac{1}{3}\right)^2=\displaystyle\frac{1}{25}\)

\(\left(x+\displaystyle\frac{1}{3}\right)^2=\left(\displaystyle\frac{1}{5}\right)^2\)

TH\(1\): \(x+\displaystyle\frac{1}{3}=\displaystyle\frac{1}{5}\)

\(x=\displaystyle\frac{1}{5}-\displaystyle\frac{1}{3}\)

\(x=\displaystyle\frac{3}{15}-\displaystyle\frac{5}{15}=\displaystyle\frac{-2}{15}.\)

TH\(2\): \(x+\displaystyle\frac{1}{3}=\displaystyle\frac{-1}{5}\)

\(x=\displaystyle\frac{-1}{5}-\displaystyle\frac{1}{3}\)

\(x=\displaystyle\frac{-3}{15}-\displaystyle\frac{5}{15}=\displaystyle\frac{-8}{15}.\)

\(5.\) Tính:

a) \(\left[\left(\displaystyle\frac{2}{5}\right)^6.\left(\displaystyle\frac{2}{5}\right)^5\right]:\left(\displaystyle\frac{2}{5}\right)^9;\)

b) \(\left[\left(\displaystyle\frac{3}{7}\right)^8:\left(\displaystyle\frac{3}{7}\right)^7\right]:\left(\displaystyle\frac{3}{7}\right);\)

c) \(\left[\left(\displaystyle\frac{2}{5}\right)^9.\left(\displaystyle\frac{2}{5}\right)^4\right]:\left[\left(\displaystyle\frac{2}{5}\right)^7.\left(\displaystyle\frac{2}{5}\right)^3\right].\)

Giải

a) \(\left[\left(\displaystyle\frac{2}{5}\right)^6.\left(\displaystyle\frac{2}{5}\right)^5\right]:\left(\displaystyle\frac{2}{5}\right)^9=\left(\displaystyle\frac{2}{5}\right)^{11}:\left(\displaystyle\frac{2}{5}\right)^9=\left(\displaystyle\frac{2}{5}\right)^2=\displaystyle\frac{4}{25};\)

b) \(\left[\left(\displaystyle\frac{3}{7}\right)^8:\left(\displaystyle\frac{3}{7}\right)^7\right]:\left(\displaystyle\frac{3}{7}\right)=\displaystyle\frac{3}{7}.\displaystyle\frac{3}{7}=\displaystyle\frac{9}{49};\)

c) \(\left[\left(\displaystyle\frac{2}{5}\right)^9.\left(\displaystyle\frac{2}{5}\right)^4\right]:\left[\left(\displaystyle\frac{2}{5}\right)^7.\left(\displaystyle\frac{2}{5}\right)^3\right]=\left(\displaystyle\frac{2}{5}\right)^{13}:\left(\displaystyle\frac{2}{5}\right)^{10}=\left(\displaystyle\frac{2}{5}\right)^3=\displaystyle\frac{8}{125}.\)

\(6.\) Tính:

a) \(\left(\displaystyle\frac{2}{5}-\displaystyle\frac{1}{3}\right)^2; \hspace{4,5cm} b)\left(1\displaystyle\frac{1}{2}-1,25\right)^3;\)

c) \(\left(\displaystyle\frac{1}{2}+\displaystyle\frac{1}{3}\right)^2:\left(1\displaystyle\frac{1}{2}\right)^2; \hspace{2cm} d)2:\left(\displaystyle\frac{1}{2}-\displaystyle\frac{2}{3}\right)^3.\)

Giải

a) \(\left(\displaystyle\frac{2}{5}-\displaystyle\frac{1}{3}\right)^2=\left(\displaystyle\frac{6}{15}-\displaystyle\frac{5}{15}\right)^2=\left(\displaystyle\frac{1}{15}\right)^2=\displaystyle\frac{1}{225};\)

b) \(\left(1\displaystyle\frac{1}{2}-1,25\right)^3=\left(\displaystyle\frac{3}{2}-\displaystyle\frac{5}{4}\right)^3=\left(\displaystyle\frac{6}{4}-\displaystyle\frac{5}{4}\right)^3=\left(\displaystyle\frac{1}{4}\right)^3=\displaystyle\frac{1}{64};\)

c) \(\left(\displaystyle\frac{1}{2}+\displaystyle\frac{1}{3}\right)^2:\left(1\displaystyle\frac{1}{2}\right)^2=\left(\displaystyle\frac{5}{6}\right)^2:\left(\displaystyle\frac{3}{2}\right)^2=\displaystyle\frac{25}{36}:\displaystyle\frac{9}{4}=\displaystyle\frac{25}{81};\)

d) \(2:\left(\displaystyle\frac{1}{2}-\displaystyle\frac{2}{3}\right)^3=2:\left(\displaystyle\frac{-1}{6}\right)^3=2:\left(\displaystyle\frac{-1}{216}\right)=-432.\)

\(7.\) Tính giá trị các biểu thức.

a) \(\displaystyle\frac{9^3.2^{10}}{16^2.81^2}; \hspace{2,5cm} b) \displaystyle\frac{(-3)^7.(-3)^8}{7.9^7};\)

c) \(\displaystyle\frac{(0,3)^6.(0,04)^3}{(0,09)^4.(0,2)^4} \qquad d) \displaystyle\frac{2^3+2^4+2^5+2^6}{15^2}.\)

Giải

a) \(\displaystyle\frac{9^3.2^{10}}{16^2.81^2}=\displaystyle\frac{(3^2)^3.2^{10}}{(2^4)^2.(3^4)^2}=\displaystyle\frac{3^6.2^{10}}{2^8.3^8}=\displaystyle\frac{2^2}{3^2}=\displaystyle\frac{4}{9};\)

b) \(\displaystyle\frac{(-3)^7.(-3)^8}{7.9^7}=\displaystyle\frac{(-3)^{15}}{7.(3^2)^7}=\displaystyle\frac{(-3)^{15}}{7.3^{14}}=\displaystyle\frac{-3}{7};\)

c) \(\displaystyle\frac{(0,3)^6.(0,04)^3}{(0,09)^4.(0,2)^4}=\displaystyle\frac{(0,3)^6.[(0,2)^2]^3}{[(0,3)^2]^4.(0,2)^4}\)

\(=\displaystyle\frac{(0,3)^6.(0,2)^6}{(0,3^8).(0,2)^4}=\displaystyle\frac{(0,2)^2}{(0,3)^2}=\displaystyle\frac{4}{9}\)

d) \(\displaystyle\frac{2^3+2^4+2^5+2^6}{15^2}=\displaystyle\frac{8+16+32+64}{225}=\displaystyle\frac{120}{225}=\displaystyle\frac{8}{15}.\)

\(8.\) Khối lượng một số hành tinh trong Hệ Mặt Trời:

Sao Thổ \(5,6846 . 10^{26}\) kg, Sao Mộc \(1,8986 . 10^{27}\) kg, Sao Thiên Vương \(8,6810 . 10^{25}\) kg, Sao Hải Vương \(10,243 . 10^{25}\) kg, Trái Đất \(5,9736 . 10^{24}\) kg.

a) Sắp xếp khối lượng các hành tinh trên theo thứ tự từ nhẹ đến nặng.

b) Trong các hành tinh trên, hành tinh nào nhẹ nhất, hành tinh nào nặng nhất?

(Theo: https://vi.wikipedia.org/wiki/Hệ_Mặt_Trời)

Giải

a) Ta có: \(5,6846 . 10^{26} = 568,46 . 10^{24}; 1,8986 . 10^{27} = 1 898,6 . 10^{24};\)

\(8,6810 . 10^{25} = 86,810 . 10^{24}; 10,243 . 10^{25} = 102,43 . 10^{24}.\)

Vì \(5,9736 < 86,810 < 102,43 < 568,46 < 1 898,6.\)

Khối lượng các hành tinh trên theo thứ tự từ nhẹ đến nặng: \(5,9736 . 10^{24} < 8,6810 . 10^{25} < 10,243 . 10^{25} < 5,6846 . 10^{26} < 1,8986 . 10^{27}.\)

b) Trong các hành tinh trên, Trái Đất là hành tinh nhẹ nhất, Sao Mộc là hành tinh nặng nhất.

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Xem thêm các bài giải khác tại: Giải Bài tập Toán Lớp 7 – NXB Chân Trời Sáng Tạo

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