1) \(x^2+2x+1=\left(x+2\right)\sqrt[]{x^2+1}\left(1\right)\)

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

\(\Leftrightarrow\left(x^2+2x+1\right)^2=\left(x\sqrt[]{x^2+1}+2\sqrt[]{x^2+1}\right)^2\)

\(\Leftrightarrow x^4+4x^2+1+4x^3+2x^2+4x=x^2\left(x^2+1\right)+4\left(x^2+1\right)+4x\left(x^2+1\right)\)

\(\Leftrightarrow x^4+4x^3+6x^2+4x+1=x^4+x^2+4x^2+4+4x^3+4\)

\(\Leftrightarrow x^4+4x^3+6x^2+4x+1=x^4+4x^3+5x^2+4x+4\)

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow x=\pm\sqrt[]{3}\left(Tm.x\ge-2\right)\)

Vậy nghiệm của phương trình \(\left(1\right)\) là \(x=\pm\sqrt[]{3}\)

2) \(P=\sqrt[]{x^2-2x+13}+4\sqrt[]{x-3}\)

Ta có : 

\(\sqrt[]{x^2-2x+13}=\sqrt[]{x^2-2x+1+12}=\sqrt[]{\left(x-1\right)^2+12}\ge\sqrt[]{12}=2\sqrt[]{3},\forall x\in R\)

\(4\sqrt[]{x-3}\ge0,\forall x\ge3\)

\(\Rightarrow P=\sqrt[]{x^2-2x+13}+4\sqrt[]{x-3}\ge\sqrt[]{4+12}+0=4\left(khi.x=3\right),\forall x\ge3\)

Vậy \(Min\left(P\right)=4\left(tại.x=3\right)\)

????  

xin lỗi nha ! 

mình mới học lớp 3 

mà bài này khó nắm 

ko bt thì ko nhắn nha

a.A=\(\sqrt{12}-\sqrt{27}+\sqrt{4+2\sqrt{3}}\)\(=2\sqrt{3}-3\sqrt{3}+\sqrt{\left(\sqrt{3}+1\right)^2}\)               \(=-\sqrt{3}+\sqrt{3}+1\)                                                                                  =1                                                                                                        b. \(x^2-2x-4=0\)                                                                                   Δ= \(\left(-2\right)^2-4\times1\times-4=20>0\)                                             \(\Rightarrow\)   phương trình có 2 nghiệm pb                                                \(x1=\dfrac{2+\sqrt{20}}{2}=1+\sqrt{5}\)     \(x2=\dfrac{2-\sqrt{20}}{2}=1-\sqrt{5}\)                      c. \(\left\{{}\begin{matrix}2x-y=5\\x+3y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=5\\2x+6y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-7y=7\\2x-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\2x+1=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)

\(M=6\sqrt{2x}-\sqrt{2x}+2\sqrt{2x}-4\sqrt{2x}=3\sqrt{2x}\)

\(1,P=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(dkxd:x\ge0,x\ne9\right)\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-3\sqrt{x}-2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-1-2\sqrt{x}+6}\)

\(=\dfrac{-x-3\sqrt{x}}{\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}}{-\sqrt{x}+5}\)

\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}}{5-\sqrt{x}}\)

\(=-\dfrac{x}{5-\sqrt{x}}\)

\(2,x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\)

\(=2+\sqrt{3}+2-\sqrt{3}=4\)

\(x=4\Rightarrow P=-\dfrac{4}{5-\sqrt{4}}=\dfrac{-4}{5-2}=-\dfrac{4}{3}\)

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

\(\Leftrightarrow B=\left(2^4-2.2^3-2^2+2.2-1\right)^{2020}=\left(-1\right)^{2020}=1\)

1/ \(\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}\)

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

\(=\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|\)

\(=\sqrt{2x-1}+1+1-\sqrt{2x-1}\)

\(=2\)

2/ ĐKXĐ: \(a^2-1\ge0\Rightarrow a^2\ge1\Rightarrow\left[{}\begin{matrix}a\ge1\\a\le-1\end{matrix}\right.\)

3/ \(4\left|x\right|-\sqrt{\left(5x-1\right)^2}=4\left|x\right|-\left|5x-1\right|\)

\(=4x-\left(5x-1\right)=1-x\)

4/ \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}< \sqrt{7}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x< 7\end{matrix}\right.\) \(\Rightarrow0\le x< 7\)

5/ \(M=\sqrt{3-2\sqrt{2.3}+2}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

\(=\left|\sqrt{3}-\sqrt{2}\right|=\sqrt{3}-\sqrt{2}\)

6/ \(\left|x\right|-\sqrt{\left(x-1\right)^2}=\left|x\right|-\left|x-1\right|=x-\left(x-1\right)=1\)

1.

\(\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}\)

\(=\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}\)

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

\(=\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|\)

\(=\sqrt{2x-1}+1+1-\sqrt{2x-1}=2\)

2.

\(\sqrt{a^2-1}\text{ xác định }\Leftrightarrow a^2-1\ge0\)

\(\Leftrightarrow\left(a-1\right)\left(a+1\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a-1\ge0\\a+1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}a-1\le0\\a+1\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a\ge1\\a\le-1\end{matrix}\right.\)

3.

\(4\left|x\right|-\sqrt{1+25x^2-10x}\)

\(=4\left|x\right|-\sqrt{\left(5x-1\right)^2}\)

\(=4\left|x\right|-\left|5x-1\right|\)

\(=4x-5x+1=1-x\)

4.

ĐKXĐ: \(x\ge0\)

\(-\sqrt{x}>-\sqrt{7}\)

\(\Leftrightarrow\sqrt{x}< \sqrt{7}\)

\(\Leftrightarrow\text{ }x< 7\)

Vậy bât phương trình có nghiệm \(0\le x< 7\)

5.

\(\sqrt{5-2\sqrt{6}}=\sqrt{2-2\sqrt{2}.\sqrt{3}+3}\)

\(=\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)

\(=\sqrt{3}-\sqrt{2}\)

6.

\(\left|x\right|-\sqrt{1-2x+x^2}\)

\(=\left|x\right|-\sqrt{\left(1-x\right)^2}\)

\(=\left|x\right|-\left|x-1\right|\)

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