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) \[2,5\,\,.\,\,{\left( { - \frac{1}{2}} \right)^3}.\,\,0,4{\rm{ + 41}}\frac{2}{7}.\frac{3}{8} - 38\frac{2}{7}.\frac{3}{8}\]

\[ = (2,5\,.\,0,4)\,\,.\,\,\left( { - \frac{1}{8}} \right) + \frac{3}{8}\,\,\left( {41\frac{2}{7} - 38\frac{2}{7}} \right)\]

 \( = - \frac{1}{8} + \frac{3}{8}\,.\,3 = - \frac{1}{8} + \frac{9}{8} = 1\)

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

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

 \(x\, = \frac{{31}}{{12}}\,:2 = \frac{{31}}{{24}}\).

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: \[d'\parallel d''\]

\( \Rightarrow {\widehat D_1} = \widehat A = 61^\circ \) (hai góc so le trong)

\( \Rightarrow \widehat {{C_2}} = \widehat B = 100^\circ \) (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 \(\left\{ \begin{array}{l}Am\parallel Bx\\Bx\parallel Cy\end{array} \right. \Rightarrow Am\parallel Cy\) (ba đường thẳng song song)

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

\( \Rightarrow \) \[\widehat {mAC} = 40^\circ \]

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

\( \Rightarrow \)\(BA \bot CA\)\( \Rightarrow \) 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) \({2^{12}}\, = \,{\left( {{2^2}} \right)^6}\, = \,{4^6}\)

  \({4^{18}}\, = \,{\left( {{4^3}} \right)^6}\, = \,{64^6}\)