`a)` `|x-5|-7=2x`
`<=>|x-5|=2x+7`
$⇔\left\{\begin{matrix}2x+7\ge 0\\ \left[\begin{array}{l}x-5=2x+7\\x-5=-2x-7\end{array}\right. \end{matrix}\right.$ $⇔\left\{\begin{matrix}2x\ge -7\\>\left[\begin{array}{l}x-2x=5+7\\x+2x=5-7\end{array}\right. \end{matrix}\right.$ $⇔\left\{\begin{matrix}x\ge \dfrac{-7}{2}\\\left[\begin{array}{l}-x=12\\3x=-2\end{array}\right. \end{matrix}\right.$ $⇔\left\{\begin{matrix}x\ge \dfrac{-7}{2}\\\left[\begin{array}{l}x=-12\ (loại)\\x=\dfrac{-2}{3}\ (nhận)\end{array}\right. \end{matrix}\right.$
Vậy `x={-2}/3`
$\\$
`b)` `3x-1+|3x+1|=-2`
`<=>|3x+1|=-3x+1-2`
`<=>|3x+1|=-(3x+1)`
`<=>3x+1\le 0`
`<=>3x\le -1`
`<=>x\le {-1}/3`
Vậy `x\le {-1}/3`
$\\$
`c)` `|2x-3|+2x-3=0`
`<=>|2x-3|=-(2x-3)`
`<=>2x-3\le 0`
`<=>2x\le 3`
`<=>x\le 3/ 2`
Vậy `x\le 3/ 2`
$\\$
`d)` `|x^2-5x|-x=4x`
`<=>|x^2-5x|=x+4x=5x`
$⇔\left\{\begin{matrix}5x\ge 0\\ \left[\begin{array}{l}x^2-5x=5x\\x^2-5x=-5x\end{array}\right. \end{matrix}\right.$ $⇔\left\{\begin{matrix}x\ge 0\\ \left[\begin{array}{l}x^2-10x=0\\x^2=0\end{array}\right. \end{matrix}\right.$ $⇔\left\{\begin{matrix}x\ge 0\\ \left[\begin{array}{l}x(x-10)=0\\x=0\ \end{array}\right. \end{matrix}\right.$ $⇔\left\{\begin{matrix}x\ge 0\\ \left[\begin{array}{l}x=0\\x-10=0\end{array}\right. \end{matrix}\right.$
$⇔\left\{\begin{matrix}x\ge 0\\ \left[\begin{array}{l}x=0\ (nhận)\\x=10\ (nhận)\end{array}\right. \end{matrix}\right.$
Vậy `x=0;x=10`
$\\$
`e)` `|x-3|+|5-x|=12` $(1)$
+) $TH1: x<3$
`(1)<=>3-x+5-x-12=0`
`<=>-2x-4=0`
`<=>-2x=4`
`<=>x=4:(-2)=-2` (nhận)
+) $TH2: 3\le x<5$
`(1)<=>x-3+5-x=12`
`<=>0x=10` (vô lý)
+) $TH3: x\ge 5$
`(1)<=>x-3+x-5-12=0`
`<=>2x-20=0`
`<=>2x=20`
`<=>x=20:2=10` (nhận)
Vậy `x=-2; x=10`
$\\$
`f)` `|x+2|+|x+5|=3` $(2)$
+) $TH1: x<-5$
`(2)<=>-x-2-x-5-3=0`
`<=>-2x-10=0`
`<=>-2x=10`
`<=>x=10:(-2)=-5` (loại)
+) $TH2: -5\le x<-2$
`(2)<=>-x-2+x+5-3=0`
`<=>0x=0` (đúng)
+) $TH3: x\ge -2$
`(1)<=>x+2+x+5-3=0`
`<=>2x+4=0`
`<=>2x=-4`
`<=>x=-4:2=-2` (nhận)
Vậy `-5\le x\le -2`
$\\$
`g)` `|x+1|+|x+2|+|x+3|=12` $(3)$
+) $TH1: x<-3$
`(3)<=>-x-1-x-2-x-3=12`
`<=>-3x-6-12=0`
`<=>-3x=18`
`<=>x=18:(-3)=-6` (nhận)
+) $TH2: -3\le x<-2$
`(3)<=>-x-1-x-2+x+3-12=0`
`<=>-x=12`
`<=>x=-12` (loại)
+) $TH3: -2\le x<-1$
`(3)<=>-x-1+x+2+x+3-12=0`
`<=>x-8=0`
`<=>x=8` (loại)
+) $TH4: x\ge -1$
`(3)<=>x+1+x+2+x+3-12=0`
`<=>3x=6`
`<=>x=6:3=2` (nhận)
Vậy `x=-6;x=2`
$\\$
$h)$ `|x-y-1|+|x^2-1|=0` $(4)$
`\qquad |x-y-1|\ge 0 ` với mọi $x;y$
`\qquad |x^2-1|\ge 0` với mọi $x$
$(4)⇔\begin{cases}x-y-1=0\\x^2-1=0\end{cases}$ $⇔\begin{cases}y=x-1\\x^2=1\end{cases}$ $⇔\left[\begin{array}{l}\begin{cases}x=1\\y=1-1=0\end{cases}\\ \begin{cases}x=-1\\y=-1-1=-2\end{cases}\end{array}\right.$
Vậy `x=1;y=0` hoặc `x=-1;y=-2`
$\\$
$i)$ `|x^2-y|+|y^2+y|=0` $(5)$
`\qquad |x^2-y|\ge 0 ` với mọi $x;y$
`\qquad |y^2+y|\ge 0` với mọi $y$
$(5)⇔\begin{cases}x^2-y=0\\y^2+y=0\end{cases}$ $⇔\begin{cases}x^2=y\\y(y+1)=0\end{cases}$ $⇔\left[\begin{array}{l}\begin{cases}y=0\\x^2=0\end{cases}\\ \begin{cases}y+1=0\\x^2=y\end{cases}\end{array}\right.$ $⇔\left[\begin{array}{l}\begin{cases}y=0\\x=0\end{cases}\\ \begin{cases}y=-1\\x^2=-1<0\end{cases}\ (loại)\end{array}\right.$
Vậy `x=0;y=0`
