a: \(\left(x+5\right)^2>=0\forall x\)

\(\left(2y-8\right)^2>=0\forall y\)

Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)

b: \(\left(x+3\right)\left(2y-1\right)=5\)

=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)

a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)

c: Ta có: \(\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)

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

b. 

PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$

$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$

$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$

$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$

$\Leftrightarrow (x-3)^2(2x+1)^2=0$

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

$\Leftrightarrow x-3=0$ hoặc $2x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$

d.

$x^2-2x=24$

$\Leftrightarrow x^2-2x-24=0$

$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$

$\Leftrightarrow x=-4$ hoặc $x=6$

a)Ta có:

\(x^2-7x+12=0\)

\(\Leftrightarrow x^2-3x-4x+12=0\)

\(\Leftrightarrow x\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)

b) Ta có:

\(x\left(x-4\right)-3\left(4-x\right)=0\)

\(\Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)

- Gửi lẻ câu hỏi ra nha bạn 2 3 câu 1 lần thôi .

a) (x-3)²-4=0

⇒ (x-3)²=4

⇒ hoặc x-3=2⇒x=5

hoặc x-3=-2⇒x=1

c) (x+4)²-(x+1)(x-1)=16

⇒ x²+8x+16-x²+1=16

⇒ 8x+17=16

⇒ 8x=-1

⇒ x=-1/8

\((2x-1)^2+(x+3)^2-5(x+7)(x-7)=0\)

\(< =>4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\\ < =>4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ < =>2x+255=0\\ < =>2x=-255=>x=\dfrac{-255}{2}\)

Vậy \(x=\dfrac{-255}{2}\)

\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Rightarrow2x+255=0\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)

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) 8x-3=0

⇔8x=3

⇔x=\(\dfrac{3}{8}\)

Vậy...

b)  -5x+7=-3x-9

⇔-5x+3x=-9-7

⇔-2x=-16

⇔x=8

Vậy...

e)

\(\dfrac{1}{x-2}+4=\dfrac{x+3}{x-2}\)

⇔\(\dfrac{1}{x-2}-\dfrac{x+3}{x-2}=4\)

⇔\(\dfrac{-x-2}{x-2}=4\)

⇔\(x+2=4x-8\)

⇔\(-3x=-10\)

⇔\(x=\dfrac{10}{3}\)

a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)

=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)

=>\(\left|2x+1\right|=\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

c: (2x-3)²=36

=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

d: \(7^{x+2}+2\cdot7^x=357\)

=>\(7^x\cdot49+7^x\cdot2=357\)

=>\(7^x=7\)

=>x=1

a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

\(---\)

b) \(\left|2x+1\right| +\dfrac{2}{3}=2\)

\( \Rightarrow\left|2x+1\right|=2-\dfrac{2}{3}\)

\(\Rightarrow\left|2x+1\right|=\dfrac{4}{3}\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{4}{3}\\2x+1=-\dfrac{4}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)

\(---\)

c) \(\left(2x-3\right)^2=36\)

\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(---\)

d) \(7^{x+2}+2\cdot7^x=357\)

\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)

\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)

\(\Rightarrow7^x\cdot\left(49+2\right)=357\)

\(\Rightarrow7^x\cdot51=357\)

\(\Rightarrow7^x=357:51\)

\(\Rightarrow7^x=7\)

\(\Rightarrow x=1\)