\(A=\sqrt{x-1-2\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}\)

\(A=\sqrt{x-2-2\sqrt{x-2}+1}+\sqrt{x-2-6\sqrt{x-2}+9}\)

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

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

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

\(A\ge\left|\sqrt{x-2}-1+3-\sqrt{x-2}\right|=\left|2\right|=2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(\sqrt{x-2}-1\right)\left(3-\sqrt{x-2}\right)\ge0\)

TH1 : \(\hept{\begin{cases}\sqrt{x-2}-1\ge0\\3-\sqrt{x-2}\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le11\end{cases}\Leftrightarrow}3\le x\le11}\)

TH2 : \(\hept{\begin{cases}\sqrt{x-2}-1\le0\\3-\sqrt{x-2}\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le3\\x\ge11\end{cases}}}\) ( loại ) 

Vậy GTNN của \(A\) là \(2\) khi \(3\le x\le11\)

Chúc bạn học tốt ~ 

Xét hạng tử: \(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\)

Thay \(xy+yz+zx=1\); ta có:

\(x\sqrt{\frac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{x^2+xy+yz+zx}}=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)^2\left(x+z\right)}{\left(x+y\right)\left(x+z\right)}}\)

\(=x\sqrt{\left(y+z\right)^2}=xy+xz\)

Tượng tự: \(y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}=xy+yz;z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}=xz+yz\)

Do đó: \(A=2\left(xy+yz+zx\right)=2.1=2\)

ĐS:...

TÍNH NHA M.N  

a, \(\sqrt{8}+\sqrt{18}-\sqrt{\frac{1}{2}}=2\sqrt{2}+3\sqrt{2}-\frac{1}{2}\sqrt{2}\)

\(=\frac{9}{2}\sqrt{2}\)

b, \(\frac{3-\sqrt{3}}{\sqrt{3}}+\frac{2\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)

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

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

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

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

c,  PT xác định với mọi x nha!

\(\sqrt{x^2-2x+1}=3\) \(\Rightarrow x^2-2x+1=9\)

\(\Leftrightarrow x^2-2x-8=0\)

\(\Leftrightarrow\left(x^2-4x\right)+\left(2x-8\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)

Vậy...

bạn tự kl

biết làm rồi

Bt rồi đăng làm j nữa??

chiu thoi em lop 8