1:

a: =>(|x|+4)(|x|-1)=0

=>|x|-1=0

=>x=1; x=-1

b: =>x^2-4>=0

=>x>=2 hoặc x<=-2

d: =>|2x+5|=2x-5

=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0

=>x=0(loại)

a, \(-2x\ge-5\Leftrightarrow x\le\dfrac{5}{2}\)

b, TH1 : \(\left\{{}\begin{matrix}x-1>0\\x+3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>1\\x>-3\end{matrix}\right.\Leftrightarrow x>1\)

TH2 : \(\left\{{}\begin{matrix}x-1< 0\\x+3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 1\\x< -3\end{matrix}\right.\Leftrightarrow x< -3\)

a, \(\left(x-5\right)\left(x-5+3\right)=0\Leftrightarrow x=5;x=2\)

b, \(-4x=\dfrac{274}{21}\Leftrightarrow x=-\dfrac{137}{42}\)

c, đk x khác - 2 ; 2 

\(x^2-3x+2-x^2-2x=6-7x\Leftrightarrow-5x+2=6-7x\)

\(\Leftrightarrow2x-4=0\Leftrightarrow x=2\left(ktm\right)\)

Vậy pt vô nghiệm 

a.

\(\dfrac{x+1}{x-1}>0\Rightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)

b.

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+2\right)}{x-9}< 0\Rightarrow\left[{}\begin{matrix}x< -2\\1< x< 9\end{matrix}\right.\)

Bài 1:

a. 

$(4x^2+4x+1)-x^2=0$

$\Leftrightarrow (2x+1)^2-x^2=0$

$\Leftrightarrow (2x+1-x)(2x+1+x)=0$

$\Leftrightarrow (x+1)(3x+1)=0$

$\Rightarrow x+1=0$ hoặc $3x+1=0$

$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$

b.

$x^2-2x+1=4$

$\Leftrightarrow (x-1)^2=2^2$

$\Leftrightarrow (x-1)^2-2^2=0$

$\Leftrightarrow (x-1-2)(x-1+2)=0$

$\Leftrightarrow (x-3)(x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-1$

c.

$x^2-5x+6=0$

$\Leftrightarrow (x^2-2x)-(3x-6)=0$

$\Leftrightarrow x(x-2)-3(x-2)=0$

$\Leftrightarrow (x-2)(x-3)=0$

$\Leftrightarrow x-2=0$ hoặc $x-3=0$

$\Leftrightarrow x=2$ hoặc $x=3$

2c.

ĐKXĐ: $x\neq 0$

PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$

$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$

$\Leftrightarrow x=-4$ (tm)

2d.

ĐKXĐ: $x\neq 2$

PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$

$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$

$\Rightarrow 3x-5=3-x$

$\Leftrightarrow 4x=8$

$\Leftrightarrow x=2$ (không tm) 

Vậy pt vô nghiệm.

2f.

ĐKXĐ: $x\neq \pm 2$

PT $\Leftrightarrow \frac{(x-2)^2-3(x+2)}{(x+2)(x-2)}=\frac{2(x-11)}{(x-2)(x+2)}$

$\Rightarrow (x-2)^2-3(x+2)=2(x-11)$

$\Leftrightarrow x^2-4x+4-3x-6=2x-22$

$\Leftrightarrow x^2-7x-2=2x-22$

$\Leftrightarrow x^2-9x+20=0$

$\Leftrightarrow (x-4)(x-5)=0$

$\Leftrightarrow x-4=0$ hoặc $x-5=0$

$\Leftrightarrow x=4$ hoặc $x=5$ (tm)

Bài 2:

\(A=\dfrac{2}{-x^2-2x-2}=\dfrac{-2\left(-x^2-2x-2\right)-2x^2-4x-2}{-x^2-2x-2}\) \(=-2+\dfrac{2\left(x+1\right)^2}{-x^2-2x-2}\ge-2\)

  Dấu bằng xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

  Vậy \(A_{Min}=-2\) khi \(x=-1\)

Bài 1:

a) Ta có: \(2x^2-6=0\)

\(\Leftrightarrow2x^2=6\)

\(\Leftrightarrow x^2=3\)

hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)

Vậy: \(S=\left\{\sqrt{3};-\sqrt{3}\right\}\)

a)2.(x+3)-(3+x).(1`+2x)=0\(\Leftrightarrow\)2x+6-3-6x-x-2x\(^2\)=0

\(\Leftrightarrow\)-2x\(^2\)-5x+3=0\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+3=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy PT đã cho có tập nghiệm S=\(\left\{-3;\dfrac{1}{2}\right\}\)

b)x\(^2\)-4x+4=9\(\Leftrightarrow\)x\(^2\)-4x+4-9=0\(\Leftrightarrow\)x\(^2\)-4x-5=0

\(\Leftrightarrow\left\{{}\begin{matrix}5-x=0\\1+x=0\end{matrix}\right.\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

Vậy PT đã cho có tập nghiệm S=\(\left\{-1;5\right\}\)

\(a,\Leftrightarrow\left(x+3\right)\left(2-1-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\1-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\-2x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(b,\Leftrightarrow\left(x-2\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

a) \(2\left(x+3\right)-\left(x+3\right)\left(1+2x\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2-1-2x\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(1-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\1-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

-Vậy \(S=\left\{-3;\dfrac{1}{2}\right\}\)

b) \(x^2-4x+4=9\)

\(\Leftrightarrow\left(x-2\right)^2-9=0\)

\(\Leftrightarrow\left(x-2-3\right)\left(x-2+3\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
-Vậy \(S=\left\{5;-1\right\}\)

a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)

`a,x(x+3)-(2x-1).(x+30)=0`
`<=>x^2+3x-(2x^2+59x-30)=0`
`<=>x^2+56x-30=0`
`<=>x^2+56x+28^2=28^2+30`
`<=>(x+28)^2=28^2+30`
`<=>x=+-sqrt{28^2+30}-28`
`b,x(x-3)-5(x-3)=0`
`<=>(x-3)(x-5)=0`
`<=>` $\left[ \begin{array}{l}x=3\\x=5\end{array} \right.$
`c)1/(x-1)+5/(x-2)=(3x)/((x-1)(x-2))`
`đk:x ne 1,2`
`pt<=>x-2+5(x-1)=3x`
`<=>x-2+5x-5=3x`
`<=>6x-7=3x`
`<=>3x=7`
`<=>x=7/3`
`d)(x-1)/(x+1)+(x+1)/(x-1)=(4-2x^2)/(x^2-1)`
`đk:x ne +-1`
`pt<=>(x-1)^2+(x+1)^2=4-2x^2`
`<=>2x^2+2=4-2x^2`
`<=>4x^2=2`
`<=>x^2=1/2`
`<=>x=+-sqrt{1/2}`