ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow x-\sqrt{x^2-1}+x+\sqrt{x^2-1}+2\sqrt{x^2-\left(x^2-1\right)}=4\)

\(\Leftrightarrow2x+2=4\)

\(\Leftrightarrow x=1\)

Bạn tự xét ĐKXĐ nhé ^^

Ta có : \(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)

\(\Leftrightarrow\left(\sqrt{3x^2-5x+1}-\sqrt{3}\right)-\left(\sqrt{x^2-2}-\sqrt{2}\right)-\left[\sqrt{3\left(x^2-x-1\right)}-\sqrt{3}\right]+\left(\sqrt{x^2-3x+4}-\sqrt{2}\right)=0\)

\(\Leftrightarrow\frac{3x^2-5x+1-3}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x^2-2-2}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3x^2-3x-3-3}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x^2-3x+4-2}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(3x+1\right)}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{\left(x-2\right)\left(x+2\right)}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3\left(x-2\right)\left(x+1\right)}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{\left(x-2\right)\left(x-1\right)}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{3x+1}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x+2}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3x+3}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x-1}{\sqrt{x^2-3x+4}+\sqrt{2}}\right)=0\)Tới đây bạn tự làm tiếp ^^

Dài quá ^^

mn giúp mink 

mn giúp mink nha

=>(-4) x y -5= -9

=>(-4) x y= -4

=> y=1

a, A= \(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\left(\sqrt{x}\right)^2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{x}{\sqrt{x}+2}\right)\)

A=\(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)}+\frac{x}{\sqrt{x}+2}\right)\)

A=\(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\sqrt{x}+x}{\left(\sqrt{x}+2\right)}\right)\)

A=\(\frac{1}{x+2\sqrt{x}}\)

b, A >= \(\frac{1}{3\sqrt{x}}\)

=> \(\frac{1}{x+2\sqrt{x}}\) >= \(\frac{1}{3\sqrt{x}}\)

=> x <= -1 , x >= 4 (x khác 0)

Bài 4:

a: Ta có: \(-\left|x+1.1\right|\le0\forall x\)

\(\Leftrightarrow-\left|x+1.1\right|+1.5\le1.5\forall x\)

Dấu '=' xảy ra khi x=-1,1

b: Ta có: \(-4\left|x-2\right|\le0\forall x\)

\(\Leftrightarrow-4\left|x-2\right|+10\le10\forall x\)

Dấu '=' xảy ra khi x=2

a) Ta có: \(B=\left(\dfrac{x+3\sqrt{x}-3}{x-16}-\dfrac{1}{\sqrt{x}+4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-4}\)

\(=\left(\dfrac{x+3\sqrt{x}-3-\sqrt{x}+4}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-4}\)

\(=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\dfrac{\sqrt{x}-4}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+4}\)

Ta có: \(\left\{{}\begin{matrix}3\left|x-1\right|+2\left(x-y\right)=4\\4\left|x-1\right|-\left(x-y\right)=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}12\left|x-1\right|+8\left(x-y\right)=16\\12\left|x-1\right|-3\left(x-y\right)=27\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}11\left(x-y\right)=-11\\3\left|x-1\right|+2\left(x-y\right)=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\3\left|x-1\right|=4-2\left(x-y\right)=4-2\cdot\left(-1\right)=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\\left|x-1\right|=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-y=-1\\x-1=2\end{matrix}\right.\\\left\{{}\begin{matrix}x-y=-1\\x-1=-2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=x+1=3+1=4\\x=3\end{matrix}\right.\\\left\{{}\begin{matrix}y=x+1=-1+1=0\\x=-1\end{matrix}\right.\end{matrix}\right.\)

Vậy: \(\left(x,y\right)\in\left\{\left(3;4\right);\left(-1;0\right)\right\}\)