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: Ta có: \(\sqrt[3]{x+1}=-5\)

\(\Leftrightarrow x+1=-125\)

hay x=-126

b: Ta có: \(\sqrt[3]{x+1}-1=x\)

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

\(\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)^2-1\right]=0\)

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

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

a: =>3x=8

hay x=8/3

b: =>6-3x-x-2=0

=>-4x+4=0

hay x=1

\(a,2x+x-3=5\\ \Rightarrow3x=5+3\\ \Rightarrow3x=8\\ \Rightarrow x=\dfrac{8}{3}\\ b,3\left(2-x\right)-\left(x+2\right)=0\\ \Rightarrow6-3x-x-2=0\\ \Rightarrow4-4x=0\\ \Rightarrow4x=4\\ \Rightarrow x=1\)

a,2x+x-3=5

   3x-3=5

=>x=\(\dfrac{8}{3}\)

b.3(2-x)-(x+2)=0

  6-3x-x-2=0

  4-4x=0

 =>x=1

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.

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

b.

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

\(\dfrac{2x}{x-1}+\dfrac{4}{x^2+2x-3}=\dfrac{2x-5}{x+3}\)

\(\Leftrightarrow\dfrac{2x}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+3\right)}=\dfrac{2x-5}{x+3}\)

\(ĐK:x\ne1;-3\)

\(\Leftrightarrow\dfrac{2x\left(x+3\right)+4}{\left(x-1\right)\left(x+3\right)}=\dfrac{\left(2x-5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}\)

\(\Leftrightarrow2x\left(x+3\right)+4=\left(2x-5\right)\left(x-1\right)\)

\(\Leftrightarrow2x^2+6x+4=2x^2-2x-5x+5\)

\(\Leftrightarrow13x=1\)

\(\Leftrightarrow x=\dfrac{1}{13}\left(tm\right)\)

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/

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

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

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

\(\Rightarrow x=3\)

\(a,x^2-2.\left(x-2\right)=4\\ \Leftrightarrow x^2-2x+4-4=0\\ \Leftrightarrow x^2-2x=0\\ \Leftrightarrow x.\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ \Rightarrow S=\left\{0;2\right\}\\ b,x^2-9-2x\left(x-3\right)=0\\ \Leftrightarrow x^2-4x^2+6x-9=0\\ \Leftrightarrow-3x^2+6x-9=0\\ \Leftrightarrow x^2-2x+3=0\\ \Leftrightarrow\left(x^2-2x+1\right)+2=0\\ \Leftrightarrow\left(x-1\right)^2=-2\left(vô.lí\right)\\ \Rightarrow Pt.vô.nghiệm\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x+6>=0\\x-2>=0\end{matrix}\right.\Leftrightarrow x>=2\)

\(\sqrt{x+6}-\sqrt{x-2}=2\)

=>\(\left(\sqrt{x+6}-\sqrt{x-2}\right)^2=4\)

=>\(x+6+x-2-2\sqrt{\left(x+6\right)\left(x-2\right)}=4\)

=>\(2\sqrt{\left(x+6\right)\left(x-2\right)}=2x+4-4=2x\)

=>\(\sqrt{\left(x+6\right)\left(x-2\right)}=x\)

=>\(\left\{{}\begin{matrix}x>=0\\\left(x+6\right)\left(x-2\right)=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=2\\x^2+4x-12=x^2\end{matrix}\right.\)

=>x=3

b: ĐKXĐ: \(x-3>=0\)

=>x>=3

\(2\sqrt{x-3}-2x+3=0\)

=>\(\sqrt{4x-12}=2x-3\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\4x-12=4x^2-12x+9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=3\\4x^2-12x+9-4x+12=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=3\\4x^2-16x+21=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)