Bài 1
a) $x-\frac{3}{10}=\frac{7}{15}.\frac{3}{5}\Rightarrow x-\frac{3}{10}=\frac{7}{25}\Rightarrow x=\frac{7}{25}+\frac{3}{10}\Rightarrow x=\frac{29}{50}$
b) ${{\left( \frac{2}{5} \right)}^{x}}+\frac{98}{125}=1\Leftrightarrow {{\left( \frac{2}{5} \right)}^{x}}=1-\frac{98}{125}\Leftrightarrow {{\left( \frac{2}{5} \right)}^{x}}=\frac{27}{135}\Leftrightarrow {{\left( \frac{2}{5} \right)}^{x}}={{\left( \frac{3}{5} \right)}^{3}}\Leftrightarrow x=3$
c) $17-\left| \frac{2}{3}-4x \right|=9\Rightarrow \left| \frac{2}{3}-4x \right|=17-9\Rightarrow \left| \frac{2}{3}-4x \right|=8$
Trường hợp 1: \[\frac{2}{3}-4x=8\Rightarrow 4x=\frac{2}{3}-8\Rightarrow 4x=\frac{-22}{3}\Rightarrow x=\frac{-11}{6}\]
Trường hợp 2: \[\frac{2}{3}-4x=-8\Rightarrow 4x=\frac{2}{3}+8\Rightarrow 4x=\frac{26}{3}\Rightarrow x=\frac{13}{6}\]

Bài 2
$A=3,7+\left| 4,3-x \right|$
Ta có: $\left| 4,3-x \right|\ge 0\forall x\Rightarrow 3,7+\left| 4,3-x \right|\ge 3,7$
Dấu “=” xảy ra khi $4,3-x=0\Rightarrow x=4,3$
Vậy giá trị nhỏ nhất của A là 3,7 khi x = 4,3

Bài 3
a) Vì Ax // By nên $\widehat{CDB}=\widehat{CAx}=50{}^\circ $(2 góc so le trong)
b) Xét ∆CBD có: $\widehat{BCD}+\widehat{CBD}+\widehat{CDB}=180{}^\circ \Rightarrow \widehat{BCD}+40{}^\circ +50{}^\circ =180{}^\circ \Rightarrow \widehat{BCD}=90{}^\circ $
Suy ra $BC\bot AC\Rightarrow \widehat{ACB}=90{}^\circ $