Tìm x, biết: A 6/7 - X = 12/28. B - 5/3 + 7/10x = 0,2. C 2x — Không quảng cáo

Phương pháp giải

a), b) Chuyển vế để tìm x.

c) Với ${A^2} = {B^2}$, ta chia hai trường hợp: TH1: A = B; TH2: A = - B.

a) $\frac{6}{7} - x = \frac{{12}}{{28}}$

$x = \frac{6}{7} - \frac{{12}}{{28}}$

$x = \frac{{24}}{{28}} - \frac{{12}}{{28}}$

$x = \frac{{12}}{{28}} = \frac{3}{7}$

Vậy $x = \frac{3}{7}$

b) $\frac{{ - 5}}{3} + \frac{7}{{10}}x = 0,2$

$\frac{7}{{10}}x = \frac{1}{5} + \frac{5}{3}$

$\frac{7}{{10}}x = \frac{3}{{15}} + \frac{{25}}{{15}} = \frac{{28}}{{15}}$

$x = \frac{{28}}{{15}}:\frac{7}{{10}}$

$x = \frac{8}{3}$

Vậy $x = \frac{8}{3}$

c) ${\left( {2x + 6} \right)^2} = \frac{{81}}{{25}}$

${\left( {2x + 6} \right)^2} = {\left( {\frac{9}{5}} \right)^2} = {\left( {\frac{{ - 9}}{5}} \right)^2}.$

* TH1: $2x + 6 = \frac{9}{5}$

$2x = \frac{9}{5} - 6$

$2x = \frac{{ - 21}}{5}$

$x = \frac{{ - 21}}{5}:2$

$x = \frac{{ - 21}}{{10}}.$

* TH2: $2x + 6 = \frac{{ - 9}}{5}$

$2x = \frac{{ - 9}}{5} - 6$

$2x = \frac{{ - 39}}{5}$

$x = \frac{{ - 39}}{5}:2$

$x = \frac{{ - 39}}{{10}}.$

Vậy $x \in \left\{ {\frac{{ - 39}}{{10}};\frac{{ - 21}}{{10}}} \right\}$.