1) pt có 2 dấu bằng.......t bỏ =1 được hong?

ĐK: \(\left\{{}\begin{matrix}x-2\sqrt{x-1}\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}\le x\\x\ge1\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}2x-1\le x^2\\x\ge1\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x^2-2x+1\ge0\\x\ge1\end{matrix}\right.\Leftrightarrow}x\ge1}\)

\(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\Leftrightarrow x-2\sqrt{x-1}=x-1\Leftrightarrow4x-4=1\Leftrightarrow x=\dfrac{5}{4}\left(N\right)\)

Kl: x= 5/4

2) \(\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}=\sqrt{\left(a-2\right)+2\cdot2\cdot\sqrt{a-2}+4}+\sqrt{\left(a-2\right)-2\cdot2\cdot\sqrt{a-2}+4}=\sqrt{\left(a-2+2\right)^2}+\sqrt{\left(a-2-2\right)^2}=a+a-4=2a-4\)

chép lại cái đk, ghét nhất cái trò này của H24!! Viết đã đời cuối cùng công cốc !!

\(\left\{{}\begin{matrix}x-2\sqrt{x-1}\ge0\\x-1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}\le x\\x\ge1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x-4\le x^2\\x\ge1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2-4x+4\ge0\\x\ge1\end{matrix}\right.\)\(\Leftrightarrow x\ge1\)

a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Có 

Bài 1:

a) \(A=\sqrt{8}+\sqrt{18}-\sqrt{32}\)

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

\(=\sqrt{2}\)

b) \(B=\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

\(=\sqrt{4-4\sqrt{5}+5}-\sqrt{5}\)

\(=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)

\(=\left|2-\sqrt{5}\right|-\sqrt{5}\)

\(=\sqrt{5}-2-\sqrt{5}\)

\(=-2\)

Bài 2:

a) \(\left\{{}\begin{matrix}2x-3y=4\\x+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\x+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

Vậy phương trình có nghiệm là: \(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

b) ĐKXĐ: \(x\ne\pm2\)

Với \(x\ne\pm2\), ta có:

\(\dfrac{10}{x^2-4}+\dfrac{1}{2-x}=1\)

\(\Leftrightarrow\dfrac{10}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}=1\)

\(\Leftrightarrow\dfrac{10-x-2}{x^2-4}=1\)

\(\Leftrightarrow\dfrac{8-x}{x^2-4}=1\)

\(\Rightarrow x^2-4=8-x\)

\(\Leftrightarrow x^2+x-12=0\)

\(\Leftrightarrow x^2-3x+4x-12=0\)

\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)

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

Vậy phương trình có tập nghiệm là: S ={3; -4}