Bộ 10 đề thi cuối kì 1 Toán 7 Cánh diều có đáp án - Đề 10

Tìm x , biết: a) 3/7 + 4/ 7 x = 1/ 3 ; b) − 4 /x = x /− 9 ;

10/14

(1,5 điểm) Tìm \(x\), biết:

a) \(\frac{3}{7} + \frac{4}{7}x = \frac{1}{3}\);    

b) \(\frac{{ - 4}}{x} = \frac{x}{{ - 9}}\);                                     

c) \(\frac{5}{{11}} + \frac{2}{{11}}:\left| {\frac{3}{7} - x} \right| = \frac{4}{5}\).

0/3000 ký tự
Giải thích

a) \(\frac{3}{7} + \frac{4}{7}x = \frac{1}{3}\)

     \(\frac{4}{7}x = \frac{1}{3} - \frac{3}{7}\)

     \(\frac{4}{7}x = \frac{{ - 2}}{{21}}\)

      \(x = \frac{{ - 2}}{{21}}:\frac{4}{7}\)

      \(x = \frac{{ - 1}}{6}\)

Vậy \(x = - \frac{1}{6}\).

b) \(\frac{{ - 4}}{x} = \frac{x}{{ - 9}}\)

      \({x^2} = \left( { - 4} \right).\left( { - 9} \right)\)

      \({x^2} = 36\)

Suy ra \(x = 6\) hoặc \(x = - 6\)

Vậy \(x \in \left\{ {6; - 6} \right\}\).

c) \(\frac{5}{{11}} + \frac{2}{{11}}:\left| {\frac{3}{7} - x} \right| = \frac{4}{5}\)

            \(\frac{2}{{11}}:\left| {\frac{3}{7} - x} \right| = \frac{4}{5} - \frac{5}{{11}}\)

            \(\frac{2}{{11}}:\left| {\frac{3}{7} - x} \right| = \frac{{19}}{{55}}\)

                  \(\left| {\frac{3}{7} - x} \right| = \frac{2}{{11}}:\frac{{19}}{{55}}\)

                  \(\left| {\frac{3}{7} - x} \right| = \frac{{10}}{{19}}\)

Trường hợp 1:

\(\frac{3}{7} - x = \frac{{10}}{{19}}\)

      \(x = \frac{3}{7} - \frac{{10}}{{19}}\)

      \(x = - \frac{{13}}{{133}}\)

Vậy \(x \in \left\{ { - \frac{{13}}{{133}};\,\frac{{127}}{{133}}} \right\}\).

Trường hợp 2:

\(\frac{3}{7} - x = - \frac{{10}}{{19}}\)

      \(x = \frac{3}{7} + \frac{{10}}{{19}}\)

      \(x = \frac{{127}}{{133}}\)