\(a.x^2-4x+4=0\)

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

=>x=2

b) \(2x^2-x=0\)

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

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

c) \(x^2-5x+6=0\)

\(x^2-2x-3x+6=0\)

\(\left(x-2\right)\left(x-3\right)=0\)

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

d) \(x^2+y^2=0\)

Vì \(x^2,y^2\ge0\forall x,y\)

=>x=y=0

e) \(x^2+6x+10=0\)

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

Vì \(\left(x+3\right)^2\ge0\forall x\)

=> VT>0 \(\forall x\)

=> phương trình vô nghiệm

a) \(x^2-4x+4=0\)

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

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

b) \(2x^2-x=0\)

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

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

c) \(x^2-5x+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\) \(\left(a+b+c=0\right)\)

d) \(x^2+y^2=0\)

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

e) \(x^2+6x+10=0\)

\(\Leftrightarrow x^2+6x+9+1=0\)

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

mà \(\left(x+3\right)^2+1\ge1>0,\forall x\in R\)

Nên phương trình (1) vô nghiệm

a.

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

b.

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

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) (x - 7)(2x + 8) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\2x=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)

Vậy: S = {7; -4}

b) Tương tự câu a

c)  (x - 1)(2x + 7)(x² + 2) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\\x^2+2=0\end{matrix}\right.\)

Mà: x² + 2 > 0 với mọi x

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{7}{2}\right\}\)

d) (2x - 1)(x + 8)(x - 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=-8\\x=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{1}{2};-8;5\right\}\)

a/ Pt \(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)

Vậy \(S=\left\{7;-4\right\}\)

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

c/ pt \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\) (\(x^2+2>0\forall x\))\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)

d/ pt \(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)

a)(x-7)(2x+8)=0

⇔x-7=0 hoặc 2x+8=0

1.x-7=0⇔x=7

2.2x+8=0⇔2x=-8⇔x=-4

phương trình có 1 nghiệm x=7 và x=-4

b)(3x+1)(5x-2)=0

⇔3x+1=0 hoặc 5x-2=0

1.3x+1=0⇔3x=-1⇔x=-1/3

2.5x-2=0⇔5x=2⇔x=5/2

phương trình có 2 nghiệm x=-1/3 và x=5/2

\(a,\left(đk:x\ge0\right)\) 

\(x=0\Rightarrow\sqrt{0+3}+0=0\left(vô-nghiệm\right)\)

\(x>0\)

\(\)\(\sqrt{x+3}+\dfrac{4x}{\sqrt{x+3}}=4\sqrt{x}\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}+\dfrac{4\sqrt{x}}{\sqrt{x+3}}=4\)

\(VT\ge2\sqrt{\dfrac{\sqrt{x+3}}{\sqrt{x}}.\dfrac{4\sqrt{x}}{\sqrt{x+3}}}=4\)

\(dấu"="xảy-ra\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}=\dfrac{4\sqrt{x}}{\sqrt{x+3}}\Leftrightarrow x+3=4x\Leftrightarrow x=1\left(tm\right)\)

\(b.2x^4-5x^3+6x^2-5x+2=0\Leftrightarrow\left(x-1\right)^2\left(2x^2-2x+2\right)\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2-2x+2=0\left(vô-nghiệm\right)\end{matrix}\right.\)

a) ĐKXĐ : \(x\ge0\)

PT <=> \(x+3-4\sqrt{x}\sqrt{x+3}+4x=0\)

<=> \(\left(\sqrt{x+3}-2\sqrt{x}\right)^2=0\)

<=> \(\sqrt{x+3}=2\sqrt{x}\)

<=> \(x+3=4x\)

<=> x = 1

Vậy x = 1 là nghiệm phương trình

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

a) Ta có \(a = 3 > 0\) và tam thức bậc hai \(f\left( x \right) = 3{x^2} - 2x + 4\) có \(\Delta ' = {1^2} - 3.4 =  - 11 < 0\)

=> \(f\left( x \right) = 3{x^2} - 2x + 4\) vô nghiệm.

=> \(3{x^2} - 2x + 4 > 0\forall x \in \mathbb{R}\)

b) Ta có: \(a =  - 1 < 0\) và \(\Delta ' = {3^2} - \left( { - 1} \right).\left( { - 9} \right) = 0\)

=> \(f\left( x \right) =  - {x^2} + 6x - 9\) có nghiệm duy nhất \(x = 3\).

=> \( - {x^2} + 6x - 9 < 0\forall x \in \mathbb{R}\backslash \left\{ 3 \right\}\)

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

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