Đề thi giữa kì 1 Toán 7 Kết nối tri thức (2022-2023) có đáp án - Đề 6

 a) $\frac{7}{8}.\left(\frac{2}{12}+\frac{4}{10}\right)=\frac{7}{8}\left(\frac{10}{60}+\frac{24}{60}\right)$

 $$=\frac{7}{8}.\frac{17}{30}$$=$\frac{119}{240}$

 b) $1\frac{3}{7}:\left(-\frac{4}{5}\right)-3\frac{3}{7}:\left(-\frac{4}{5}\right)=\left(\frac{10}{7}-\frac{24}{7}\right):\left(-\frac{4}{5}\right)$                            

 $=\left(\frac{10}{7}-\frac{24}{7}\right).\frac{-5}{4}$$$=\frac{-5}{4}.(-2)=\frac{5}{2}$$  

c) $\frac{9^2.{27}^3}{{81}^4}=\frac{{(3^2)}^2.{(3^3)}^3}{{(3^4)}^4}=\frac{3^4.3^9}{3^{16}}$

           $$=\frac{3^{13}}{3^{16}}=\frac{1}{27}$$

 d) $$2,5.{\left(-\frac{1}{2}\right)}^3.0,4+41\frac{2}{7}.\frac{3}{8}-38\frac{2}{7}.\frac{3}{8}$$

$$=(2,5.0,4).\left(-\frac{1}{8}\right)+\frac{3}{8}\left(41\frac{2}{7}-38\frac{2}{7}\right)$$

 $=-\frac{1}{8}+\frac{3}{8}.3=-\frac{1}{8}+\frac{9}{8}=1$

a) $2x-\frac{5}{4}=\frac{20}{15}$

$2x=\frac{4}{3}+\frac{5}{4}=\frac{31}{12}$

 $x=\frac{31}{12}:2=\frac{31}{24}$.

Vậy $x=\frac{31}{24}$.

 b) ${\left(x+\frac{1}{5}\right)}^3=\left(\frac{-1}{8}\right)$

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

 $x+\frac{1}{5}=\frac{-1}{2}$

$x=\frac{-1}{2}-\frac{1}{5}=\frac{-7}{10}$.

Vậy $x=\frac{-7}{10}$

 c) $\frac{-2}{3}x+1,6=\frac{3}{2}$

$\frac{-2}{3}x=\frac{3}{2}+\frac{16}{10}$

$\frac{-2}{3}x=\frac{3}{2}+\frac{8}{5}=\frac{31}{10}$

$x=\frac{-93}{20}$

Vậy $x=\frac{-93}{20}$

 d) $\frac{243}{3^{x+1}}=3^x$

\[243 = {3^x}\,.\,{3^x}{^ + ^1}\]

$$3^5={3^x}^{+x+1}$$

$$3^5={3^2}^{x+1}$$

$$2x+1=5$$

$$x=2$$.

Vậy $$x=2$$.

1. Ta có: $$d^{\prime }\parallel d^{\prime \prime }$$

$\Rightarrow {\hat{D}}_1=\hat{A}=61^{\circ }$ (hai góc so le trong)

$\Rightarrow \widehat{C_2}=\hat{B}=100^{\circ }$ (hai góc đồng vị)

Ta thấy : $\widehat{B_4}=\widehat{C_2}=100^{\circ }$ (hai góc so le trong)

Vì $\widehat{C_2}+\widehat{C_3}=180^{\circ }$(hai góc kề bù)

$\Rightarrow 100^{\circ }+\widehat{C_3}=180^{\circ }$$\Rightarrow \widehat{C_3}=80^{\circ }$

Kẻ tia $$Am\parallel Bx\Rightarrow \widehat{ABx}+\widehat{mAB}=180^{\circ }$$ (hai góc trong cùng phía)

$$\Rightarrow \widehat{mAB}=50^{\circ }$$

Ta có $\{\begin{array}{l}Am\parallel Bx\\ Bx\parallel Cy\end{array}\Rightarrow Am\parallel Cy$ (ba đường thẳng song song)

$$\Rightarrow \widehat{yCA}+\widehat{CAm}=180^{\circ }$$ (hai góc trong cùng phía)

$\Rightarrow$ $$\widehat{mAC}=40^{\circ }$$

Ta có $$\widehat{CAB}=\widehat{CAm}+\widehat{BAm}=90^{\circ }$$

$\Rightarrow$$BA\perp CA$$\Rightarrow$ đpcm

a) $2^{225}={\left(2^3\right)}^{75}=8^{75}$

 $3^{150}={\left(3^2\right)}^{75}=9^{75}$

 Vì $$8^{75}<9^{75}$$.

 Nên $$2^{225}<3^{150}$$

 b) $2^{12}={\left(2^2\right)}^6=4^6$

  $4^{18}={\left(4^3\right)}^6=64^6$