Đáp án:
Bài 1:
$|-2x-\dfrac{3}{4}|-\dfrac{4}{5}=\dfrac{3}{2}$
$|-2x-\dfrac{3}{4}|=\dfrac{3}{2}+\dfrac{4}{5}$
$|-2x-\dfrac{3}{4}|=\dfrac{23}{10}$
TH1: $-2x-\dfrac{3}{4}=\dfrac{23}{10}$
$-2x=\dfrac{23}{10}+\dfrac{3}{4}$
$-2x=\dfrac{61}{20}$
$x=-\dfrac{61}{40}$
TH2: $-2x-\dfrac{3}{4}=-\dfrac{23}{10}$
$-2x=-\dfrac{23}{10}+\dfrac{3}{4}$
$-2x=-\dfrac{31}{20}$
$x=\dfrac{31}{40}$
Bài 2:
a) $\left(-\dfrac{4}{3}\right)^2-\dfrac{7}{3}=\dfrac{16}{9}-\dfrac{7}{3}=\dfrac{5}{9}$
b) $|2x-1|=5$
TH1: $2x-1=5$
$2x=5+1$
$2x=6$
$x=3$
TH2: $2x-1=-5$
$2x=-5+1$
$2x=-4$
$x=-2$
c) Ta có: $\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}$
$\Rightarrow \dfrac{4(3x-2y)}{4.4}=\dfrac{3(2z-4x)}{3.3}=\dfrac{2(4y-3z)}{2.2}$
$\Rightarrow \dfrac{12x-8y}{416}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}$
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
$\Rightarrow \dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=0$
$\Rightarrow \dfrac{12x-8y}{16}=0\Rightarrow 12x=8y$
$\Rightarrow \dfrac{6z-12x}{9}=0\Rightarrow 6z=12x$
$\Rightarrow \dfrac{8y-6z}{4}=0 \Rightarrow 8y=6z$
$\Rightarrow 12x=8y=6z$
$\Rightarrow \dfrac{12x}{24}=\dfrac{8y}{24}=\dfrac{6z}{24}$
$\Rightarrow \dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}$
$\Rightarrow đpcm$
