Trắc nghiệm bài tập theo tuần Toán 7-Tuần 5 có đáp án

(đk: $x\ne 0$) ${\left(\frac{-5}{9}\right)}^{10}:x={\left(\frac{-5}{9}\right)}^8\iff x={\left(\frac{-5}{9}\right)}^{10}:{\left(\frac{-5}{9}\right)}^8\iff x={\left(\frac{-5}{9}\right)}^2\iff x=\frac{25}{81}$  (t/m)

$x:{\left(\frac{-5}{9}\right)}^8={\left(\frac{-9}{5}\right)}^8\iff x={\left(\frac{-9}{5}\right)}^8.{\left(\frac{-5}{9}\right)}^8\iff x=1$ 

$x^3=-8\iff x^3={\left(-2\right)}^3\iff x=-2$ 

$5^{300}$  và $3^{500}$ 

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

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

$2^{24}$ và  $3^{16}$ 

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

Mà $8<9\Rightarrow 8^3<9^3.$ 

Vậy $2^{24}<3^{16}.$ 

${\left(-16\right)}^{11}$ và  ${\left(-32\right)}^9$

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

Mà ${\left(-2\right)}^{44}>{\left(-2\right)}^{45}.$

Vậy ${\left(-16\right)}^{11}>{\left(-32\right)}^9.$ 

${\left(2^2\right)}^3$ và $2^{2^3}$ 

Ta có : ${\left(2^2\right)}^3=2^6=64$ và $2^{2^3}=2^8=256.$ 

Mà $64<256$ 

Vậy ${\left(2^2\right)}^3<2^{2^3}$ 

$2^{9^1}$ và $2^{2^3}$ 

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

Mà $2^9>2^8$ 

Vậy $2^{9^1}>2^{2^3}.$ 

$4^{30}$ và ${3.24}^{10}$ 

Ta có: $4^{30}=2^{30}{.2}^{30}=2^{30}\cdot {\left(2^2\right)}^{15}=2^{30}\cdot 4^{15}=2^{30}\cdot 4^{11}\cdot 4^4;{3.24}^{10}=3.{\left({3.2}^3\right)}^{10}={3.3}^{10}{.2}^{30}=3^{11}\cdot 2^{30}$

Mà $4^{11}{.4}^4>3^{11}$  nên $4^{30}>{3.24}^{10}$

$\frac{3}{1^2\cdot 2^2}+\frac{5}{2^2\cdot 3^2}+\frac{7}{3^2\cdot 4^2}+\dots +\frac{19}{9^2\cdot {10}^2}$ và 1

Ta có:

Vậy $\frac{3}{1^2\cdot 2^2}+\frac{5}{2^2\cdot 3^2}+\frac{7}{3^2\cdot 4^2}+\dots +\frac{19}{9^2\cdot {10}^2}<1$

${81}^7-{27}^9+3^{29}⋮33$ 

Ta có: ${81}^7-{27}^9+3^{29}={\left(3^4\right)}^7-{\left(3^3\right)}^9+3^{29}=3^{28}-3^{27}+3^{29}=3^{26}\cdot \left(3^2-2+3^3\right)=3^{26}.33⋮33.$ 

Vậy ${81}^7-{27}^9+3^{29}⋮33$ 

$8^{12}-2^{33}-2^{30}⋮55$ 

Ta có $8^{12}-2^{33}-2^{30}={\left(2^3\right)}^{12}-2^{33}-2^{30}=2^{36}-2^{33}-2^{30}=2^{30}.\left(2^6-2^3-1\right)=2^{30}.55⋮55$

Vậy $8^{12}-2^{33}-2^{30}⋮55$ 

${10}^9+{10}^8+{10}^7⋮555$ 

Ta có ${10}^9+{10}^8+{10}^7={10}^6\cdot \left({10}^3+{10}^2+10\right)={10}^6.1110={10}^6\mathrm{.555.2}⋮555$ 

Vậy ${10}^9+{10}^8+{10}^7⋮555$ 

Ta có $Mx\text{ // }Ny$ vì cùng vuông góc với $MN.$ 

Vẽ $Dz\text{ // }Mx\text{ // }Ny.$ 

Ta có: $\widehat{BCN}+\widehat{DCy}=90°$; $\widehat{DCy}=\widehat{zDC}$;

Suy ra: $\widehat{BCN}+\widehat{zDC}=90°$ (1)

Lại có: $\widehat{zDC}+\widehat{zDA}=90°$; $\widehat{zDC}+\widehat{zDA}=90°$.

Suy ra: $\widehat{zDC}+\widehat{DAx}=90°$ (2)

Từ (1) và (2) suy ra đpcm.

Cách 2: Vẽ $Bt\text{ // }Mx\text{ // }Ny$

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