\(\frac{1}{3+\sqrt{7}}+\frac{1}{3-\sqrt{7}}=\frac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}+\frac{3+\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}=\frac{3-\sqrt{7}}{2}+\frac{3+\sqrt{7}}{2}=\frac{6}{2}=3\)

CHÚC BẠN HỌC TỐT 

khó vậy bạn có đăng bài nào lớp 3456 ko mih làm cho nhưng bài dễ mih làm cho 

Sửa lại đề nha , đề đúng nè :

\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{x}{x-1}\right):\)\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{x}{x+2\sqrt{x}+1}\right)\)

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

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

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

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

\(\)

ĐK \(x\ge2\)

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

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

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)+2x\sqrt{x-2}=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-2}=0\\\left(x-1\right)\sqrt{x-2}+2x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\\left(x+1\right)^2\left(x+2\right)=4x^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x^3+5x+2=0\left(loại\right)\end{cases}}\)

Vậy x=2 là nghiệm của pt

Ta có:

\(a+b+c\ge abc\) (gt)

mà \(a^2+b^2+c^2\ge a+b+c\forall a,b,c\ge0\) 

\(\Rightarrow a^2+b^2+c^2\ge abc\left(đpcm\right)\)

nếu sd bổ đề thì ít nhất bạn cx cần nói sơ qua về nó hoặc cm nó ạ

\(P=\frac{x\sqrt{x}-8}{x+2\sqrt{x}+4}+3\left(1-\sqrt{x}\right).\)

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

\(=\frac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{x+2\sqrt{x}+4}+3-3\sqrt{x}\)

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

\(Q=\frac{2P}{1-P}=\frac{2\left(-2\sqrt{x}+1\right)}{1-\left(-2\sqrt{x}+1\right)}\)

\(=\frac{-4\sqrt{x}+2}{1+2\sqrt{x}-1}=\frac{-2\sqrt{x}+1}{\sqrt{x}}\)

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

\(Q\in Z\Leftrightarrow-2+\frac{1}{\sqrt{x}}\in Z\Rightarrow\frac{1}{\sqrt{x}}\in Z\)

\(\Rightarrow1\)\(⋮\)\(\sqrt{x}\)\(\Rightarrow\sqrt{x}\inƯ_1\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=1\\\sqrt{x}=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x\in\varnothing\end{cases}}}\)

Vậy \(Q\in Z\Leftrightarrow x=1\)