\(a)ĐK:x\ge-1\\ \Leftrightarrow x+1=2\sqrt{x+1}\\ \Leftrightarrow x^2+2x+1=4x+4\\ \Leftrightarrow x^2+2x-4x+1-4=0\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{3;-1\right\}\)

\(b)ĐK:x\ge2\\ \Leftrightarrow2x-4=\sqrt{x-2}\\ \Leftrightarrow4x^2-16x+16=x-2\\ \Leftrightarrow4x^2-16x-x+16+2=0\\ \Leftrightarrow4x^2-17x+18=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{9}{4};2\right\}\)

\(c)ĐK:x\ge3\\ \Leftrightarrow2\sqrt{9\left(x-3\right)}-\dfrac{1}{5}\sqrt{25\left(x-3\right)}-\dfrac{1}{7}\sqrt{49\left(x-3\right)}=20\\ \Leftrightarrow2.3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\\ \Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\\ \Leftrightarrow x=25+3\\ \Leftrightarrow x=28\left(tm\right)\)

Vậy \(S=\left\{28\right\}\)

a. 5 - 3(x + 4) = -1

⇔ 5 - 3x - 12 = -1

⇔ 3x = -1 - 5 + 12

⇔ 3x = 6

⇔ x = 2

\(d,2x^2-3=5\)

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\)

\(e,x\left(2x-1\right)=0\)

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

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

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

a)\(=>3\left(x+4\right)=6=>x+4=2=>x=-2\)

b)\(=>x-1-x-2=0\)

\(=>-3=0\left(vl\right)\) => x ko tồn tại

a: x=9/2-3/7=57/14

b: =>x=7/5x5/7=1

c: =>x=11/3:3/8=11/3x8/3=88/9

a : x = \(\dfrac{9}{2}\) - \(\dfrac{3}{7}\) = \(\dfrac{57}{14}\)

b : = > x = \(\dfrac{7}{5}\) x\(\dfrac{5}{7}\) = 1

 c : = >  \(\dfrac{11}{3}\) : \(\dfrac{3}{8}\) = \(\dfrac{11}{3}\) x \(\dfrac{8}{3}\) = \(\dfrac{88}{9}\)

a. x² - 6x = -9

<=> x² - 6x + 9 = 0

<=> (x - 3)² = 0

<=> x - 3 = 0

<=> x = 3

b. 2(x + 3) - x² + 3x = 0

<=> 2(x + 3) - x(x + 3) = 0

<=> (2 - x)(x + 3) = 0

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

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

\(a.5x=-20\\ x=-20:5=-4\\ b.\dfrac{5}{6}-x=\dfrac{7}{4}\\ x=\dfrac{5}{6}-\dfrac{7}{4}=\dfrac{-11}{12}\)

\(5x=-20\Leftrightarrow x=-20:5\Leftrightarrow x=-4\)

\(\dfrac{5}{6}-x=\dfrac{7}{4}\Leftrightarrow x=\dfrac{5}{6}-\dfrac{7}{4}\Leftrightarrow x=-\dfrac{11}{12}\)

a: Ta có: \(4x\left(x-7\right)-4x^2=56\)

\(\Leftrightarrow4x^2-7x-4x^2=56\)

hay x=-8

b: Ta có: \(12x\left(3x-2\right)-\left(4-6x\right)=0\)

\(\Leftrightarrow36x^2-24x-4+6x=0\)

\(\Leftrightarrow36x^2-18x-4=0\)

\(\text{Δ}=\left(-18\right)^2-4\cdot36\cdot\left(-4\right)=900\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{18-30}{72}=\dfrac{-1}{6}\\x_2=\dfrac{18+30}{72}=\dfrac{2}{3}\end{matrix}\right.\)

c: Ta có: \(4\left(x-5\right)-\left(x-5\right)^2=0\)

\(\Leftrightarrow\left(x-5\right)\left(4-x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)

a.\(\left(-12\right)x-14=-2\)

\(\left(-12\right)x=-2+14\)

\(\left(-12\right)x=12\)

\(x=12:\left(-12\right)\)

\(x=-1\)

\(b,\left(-8\right)x=\left(-5\right)\left(-7\right)-3\)

\(\left(-8\right)x=35-3\)

\(\left(-8\right)x=32\)

\(x=32:\left(-8\right)\)

\(x=-4\)

\(c,\left(-9\right)x+3=\left(-2\right)\left(-7\right)+16\)

\(\left(-9\right)x+3=14+16\)

\(\left(-9\right)x+3=30\)

\(\left(-9\right)x=30-3\)

\(\left(-9\right)x=27\)

\(x=27:\left(-9\right)\)

\(x=-3\)

`a,(5-x)(x-1) < 0`

`<=>5-x<0` hoặc `x-1<0`

`<=>5 <x` hoặc `x<1`

Vậy `S={x|5<x;x<1}`

`b,(x-4)(x+1/2) >= 0`

`<=>TH1 : {(x-4>=0),(x+1/2 >=0):}<=>{(x>=4(TM)),(x>= -1/2(L)):}`

`<=>TH2 :{(x-4<=0),(x+1/2 <= 0):} <=>{(x<=4(L)),(x<=-1/2(TM)):}`

`=>x<= -1/2` hoặc `x>=4`

Vậy `S={x|x<= -1/2 ; x>=4}`

\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)

a) Ta có: \(\left(2x-8\right)\left(2x+10\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-8\ge0\\2x+10\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le-5\end{matrix}\right.\)

b) Ta có: \(\left(\left|x\right|+5\right)\left(x-3\right)< 0\)

nên x-3<0

hay x<3