Giải SBT Toán 7 Bài 3. Lũy thừa của một số hữu tỉ có đáp án

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

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

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

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

b) (0,15)⁷ : (0,15)⁵ = (0,15)^{7 – 5}

= (0,15)² = 0,0225;

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

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

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

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

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

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

$x={\left(\frac{-3}{7}\right)}^2$

$x=\frac{9}{49}$

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

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

$x={\left(\frac{-2}{3}\right)}^3$

$x=\frac{-8}{27}$

Trường hợp 1:   $x+\frac{1}{3}=\frac{1}{5}$

$x=\frac{1}{5}-\frac{1}{3}$

$x=\frac{3}{15}-\frac{5}{15}$

$x=\frac{-2}{15}$

Trường hợp 2: $x+\frac{1}{3}=\frac{-1}{5}$

$x=\frac{-1}{5}-\frac{1}{3}$

$x=\frac{-3}{15}-\frac{5}{15}$

$x=\frac{-8}{15}$.

Vậy $x=\{\frac{-2}{15};\frac{-8}{15}\}$

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

$=\frac{3}{7}.\frac{3}{7}=\frac{9}{49}$

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

$={\left(\frac{2}{5}\right)}^{13}:{\left(\frac{2}{5}\right)}^{10}={\left(\frac{2}{5}\right)}^3=\frac{8}{125}$

b) ${(1\frac{1}{2}-1,25)}^3={(\frac{3}{2}-\frac{5}{4})}^3$

$={(\frac{3}{2}-\frac{5}{4})}^3={\left(\frac{1}{4}\right)}^3=\frac{1}{64}$

c) ${(\frac{1}{2}+\frac{1}{3})}^2:{\left(1\frac{1}{2}\right)}^2={\left(\frac{5}{6}\right)}^2:{\left(\frac{3}{2}\right)}^2$

$=\frac{5^2}{6^2}:\frac{3^2}{2^2}=\frac{5^2}{6^2}.\frac{2^2}{3^2}$

$=\frac{5^2{.2}^2}{2^2{.3}^2{.3}^2}=\frac{5^2}{3^4}=\frac{25}{81}$

d) $2:{(\frac{1}{2}-\frac{2}{3})}^3=2:{\left(\frac{-1}{6}\right)}^3$

$=2:{\left(\frac{-1}{6}\right)}^3=2:\left(\frac{-1}{6^3}\right)$

b) $\frac{{(-3)}^7.{(-3)}^8}{7.9^7}=\frac{{(-3)}^{15}}{7.{\left[{(-3)}^2\right]}^7}$

$=\frac{{(-3)}^{15}}{7.{(-3)}^{14}}=\frac{-3}{7}$

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

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

$={\left(\frac{0,2}{0,3}\right)}^2={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$

d) $\frac{2^3+2^4+2^5+2^6}{{15}^2}$

$=\frac{8+16+32+64}{{15}^2}$

$=\frac{120}{225}=\frac{8}{15}$

Lời giải:

Ta có: 5,6846 . 10²⁶ = 568,46 . 10²⁴; 1,8986 . 10²⁷ = 1898,6 . 10²⁴;

8,6810 . 10²⁵ = 86,810 . 10²⁴; 10,243 . 10²⁵ = 102,43 . 10²⁴.

Vì 5,9736 < 86,810 < 102,43 < 568,46 < 1898,6.

Nên 5,9736.10²⁴ < 86,810.10²⁴ < 102,43. 10²⁴ < 568,46.10²⁴ < 1898,6.10²⁴.

Do đó 5,9736.10²⁴ < 8,6810 . 10²⁵ < 10,243 . 10²⁵ < 5,6846 . 10²⁶ < 1,8986 . 10²⁷.

a) Khối lượng các hành tinh được sắp xếp theo thứ tự từ nhẹ đến nặng là: 5,9736.10²⁴ kg; 8,6810 . 10²⁵ kg; 10,243 . 10²⁵ kg; 5,6846 . 10²⁶ kg; 1,8986 . 10²⁷ kg.