a/

\(\Leftrightarrow x^2-2x+4-4=0\\ \Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow x=0;x-2=0\)

\(\Leftrightarrow x=0;x=2\)

b/

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

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

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

\(\Rightarrow x=3\)

Ta có (\(^{x^{2^{ }}^{ }+3x}\)) (\(^{x^{2^{ }}+3x+4}\))

Đặt \(x^{2^{ }^{ }}+3x\) là a ta có

a.(a+4)=-4

4a+\(a^2\) -4=0

\(^{ }\left(a-2\right)^2\)=0

Suy ra a=2

hay \(x^{2^{ }^{ }^{ }}+3x=2\)

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

𝑥=−3±17√/2

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 \(S=\left\{5;-1\right\}\)

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

chỗ dấu suy ra thứ 2 e ko hiểu lắm ạ