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Bài tập tuần Toán lớp 7 Tuần 5 có đáp án chi tiết

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Bài tập tuần Toán lớp 7 Tuần 5 có đáp án

Bài 1: Tính

a) ${(- 0, 4)}^{2} - {(- 0, 4)}^{3} . (- 3)$

b) ${(1 \frac{3}{4})}^{3} - {(1 \frac{3}{4})}^{2} + {(- 1, 031)}^{0}$

c) ${(\frac{2}{3})}^{3} - 4. {(- 1 \frac{3}{4})}^{2} + {(- \frac{2}{3})}^{3}$

d) ${(- 0, 5)}^{5} : {(- 0, 5)}^{3} - {(\frac{17}{2})}^{7} : {(\frac{17}{2})}^{6}$

e) ${{(- 2, 7)}^{4}]}^{5} - {{(- 2, 7)}^{2}]}^{10}$

f) $(8^{14} : 4^{12}) : (16^{6} : 8^{2})$

Bài 2: Tìm x, biết:

a) ${(\frac{- 5}{9})}^{10} : x = {(\frac{- 5}{9})}^{8}$

b) $x : {(\frac{- 5}{9})}^{8} = {(\frac{- 9}{5})}^{8}$

c) $x^{3} = - 8$

d) $4 {(x + 5)}^{3} = - 27$

e) ${(2 x - 3)}^{3} = - 6$

f) ${(2 x - 3)}^{2} = 25$

Bài 3: So sánh:

a) $5^{300} v à 3^{500}$

b) $2^{24} v à 3^{16}$

c) ${(- 16)}^{11} v à {(- 32)}^{9}$

d) ${(2^{2})}^{3} v à {2^{2}}^{3}$

e) ${2^{9}}^{1} v à {2^{2}}^{3}$

f) $4^{30} v à 3 . 24^{10}$

Bài 4: Chứng minh rằng:

a) $7^{6} + 7^{5} - 7^{4} ⋮ 55$

b) $81^{7} - 27^{9} + 3^{29} ⋮ 33$

c) $8^{12} - 2^{33} - 2^{30} ⋮ 55$

d) $10^{9} + 10^{8} + 10^{7} ⋮ 555$

Bài 5: Chứng minh $\hat{D A x} = \hat{B C N}$ theo nhiều cách.

PHẦN HƯỚNG DẪN GIẢI

Bài 1:

a) ${(- 0, 4)}^{2} - {(- 0, 4)}^{3} . (- 3) = {(- \frac{4}{10})}^{2} - {(- \frac{4}{10})}^{3} . (- 3)$

$= \frac{4}{25} - \frac{8}{125} .3 = \frac{- 4}{125}$

b) ${(1 \frac{3}{4})}^{3} - {(1 \frac{3}{4})}^{2} + {(- 1, 031)}^{0}$

$= {(1 \frac{3}{4})}^{2} (1 \frac{3}{4} - 1) + 1$

$= {(\frac{7}{4})}^{2} (\frac{7}{4} - 1) + 1 = \frac{49}{16} . \frac{3}{4} + 1 = \frac{211}{64}$

c) ${(\frac{2}{3})}^{3} - 4. {(- 1 \frac{3}{4})}^{2} + {(- \frac{2}{3})}^{3}$

$= {(\frac{2}{3})}^{3} - {(\frac{2}{3})}^{3} - 4 {(- \frac{7}{4})}^{2}$

$= - 4. \frac{49}{16} = - \frac{49}{4}$

d) ${(- 0, 5)}^{5} : {(- 0, 5)}^{3} - {(\frac{17}{2})}^{7} : {(\frac{17}{2})}^{6}$

$= {(- 0, 5)}^{2} - \frac{17}{2} = \frac{1}{4} - \frac{17}{2} = - \frac{33}{4}$

e) ${{(- 2, 7)}^{4}]}^{5} - {{(- 2, 7)}^{2}]}^{10}$

$= {(2, 7)}^{20} - {(2, 7)}^{20} = 0$

f) $(8^{14} : 4^{12}) : (16^{6} : 8^{2})$

$= {(2^{3})}^{14} : {(2^{2})}^{12}] : {(2^{4})}^{6} : {(2^{3})}^{2}]$

$= (2^{42} : 2^{24}) : (2^{24} : 2^{6}) = 2^{18} : 2^{18} = 1$

Bài 2:

a) (đk: $x \neq 0$ )

${(\frac{- 5}{9})}^{10} : x = {(\frac{- 5}{9})}^{8} \Leftrightarrow x = {(\frac{- 5}{9})}^{10} : {(\frac{- 5}{9})}^{8} \Leftrightarrow x = {(\frac{- 5}{9})}^{2} \Leftrightarrow x = \frac{25}{81}$

b) $x : {(\frac{- 5}{9})}^{8} = {(\frac{- 9}{5})}^{8}$

$\Leftrightarrow x = {(\frac{- 9}{5})}^{8} . {(\frac{- 5}{9})}^{8} \Leftrightarrow x = 1$

c) $x^{3} = - 8$

$\Leftrightarrow x^{3} = {(- 2)}^{3} \Leftrightarrow x = - 2$

d) ${(x + 5)}^{3} = - 27$

$\Leftrightarrow {(x + 5)}^{3} = {(- 3)}^{3}$

$\Rightarrow x + 5 = - 3 \Leftrightarrow x = - 8$

e) ${(2 x - 3)}^{3} = - 64$

$\Leftrightarrow {(2 x - 3)}^{3} = {(- 4)}^{3}$

$\Leftrightarrow 2 x - 3 = - 4 \Leftrightarrow 2 x = - 1 \Leftrightarrow x = - \frac{1}{2}$

f) ${(2 x - 3)}^{2} = 25$

$\Leftrightarrow {(2 x - 3)}^{2} = 5^{2}$

$\Leftrightarrow 2 x - 3 = 5 \Leftrightarrow 2 x = 8 \Leftrightarrow x = 4$

Bài 3:

a) $5^{300}$ và $3^{500}$

Ta có: $5^{300} = {(5^{3})}^{100} = 125^{100};$ $3^{500} = {(3^{5})}^{100} = 243^{100} .$

Mà $125 < 243 \Rightarrow 125^{100} < 243^{100} .$

Vậy $5^{300} < 3^{500} .$

b) $2^{24}$ và $3^{16}$

Ta có: $2^{24} = {(2^{3})}^{8} = 8^{5}; 3^{16} = {(3^{2})}^{8} = 9^{5} .$

Mà $8 < 9 \Rightarrow 8^{3} < 9^{3} .$

Vậy $2^{24} < 3^{16} .$

c) ${(- 16)}^{11} v à {(- 32)}^{9}$

Ta có: ${(- 16)}^{11} = {(- 2^{4})}^{11} = {(- 2)}^{4};$ ${(- 32)}^{9} = {(- 2^{5})}^{9} = {(- 2)}^{45}$

Mà ${(- 2)}^{44} > {(- 2)}^{45} .$

Vậy ${(- 16)}^{11} > {(- 32)}^{9} .$

d) ${(2^{2})}^{3} v à 2^{2^{3}}$

Ta có : ${(2^{2})}^{3} = 2^{6} = 64 v à 2^{2^{3}} = 2^{8} = 256.$

Mà $64 < 256$

Vậy ${(2^{2})}^{3} < 2^{2^{3}}$

e) $2^{9^{1}} v à 2^{2^{3}}$

Ta có: $2^{9^{1}} = 2^{9} v à 2^{2^{3}} = 2^{8}$

Mà $2^{9} > 2^{8}$

Vậy $2^{9^{1}} > 2^{2^{3}} .$

f) $4^{30} v à {3.24}^{10}$

Ta có:

$4^{30} = 2^{30} {.2}^{30} = 2^{30} \cdot {(2^{2})}^{15} = 2^{30} \cdot 4^{15} = 2^{30} \cdot 4^{11} \cdot 4^{4} {3.24}^{10} = 3. {({3.2}^{3})}^{10} = {3.3}^{10} {.2}^{30} = 3^{11} \cdot 2^{30}$