\(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(đk:x\ge3\right)\)

\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

\(ĐK:x\le-3;x\ge3\\ PT\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

ĐKXĐ: \(x^2-4x+1\ge0\)

\(2x+2+2\sqrt{x^2-4x+1}=6\sqrt{x}\)

\(\Leftrightarrow2x+2-5\sqrt{x}+2\sqrt{x^2-4x+1}-\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{4x^2-17x+4}{2x+2+5\sqrt{x}}+\dfrac{4x^2-17x+4}{2\sqrt{x^2-4x+1}+\sqrt{x}}=0\)

\(\Leftrightarrow\left(4x^2-17x+4\right)\left(\dfrac{1}{2x+2+5\sqrt{x}}+\dfrac{1}{2\sqrt{x^2-4x+1}+\sqrt{x}}\right)=0\)

\(\Leftrightarrow4x^2-17x+4=0\)

\(\Leftrightarrow...\)

a,\(\sqrt{1-x}=\sqrt[3]{27}\left(đk:x\le1\right)\Leftrightarrow\sqrt{1-x}=3\)

\(< =>\sqrt{1-x}^2=9< =>1-x=9< =>x=-8\)tm

b,\(\sqrt{x^2-10x+25}=x+1\)

\(< =>\sqrt{\left(x-5\right)^2}=x+1\)

\(< =>|x-5|=x+1\)

\(< =>\orbr{\begin{cases}-x+5=x+1\left(x< 5\right)\\x-5=x+1\left(x\ge5\right)\end{cases}}\)

\(< =>\orbr{\begin{cases}2x=4< =>x=2\left(tm\right)\\-5-1=0\left(vo-li\right)\end{cases}}\)

c, Đặt \(\sqrt{x}=t\left(t\ge0\right)\)khi đó pt tương đương

\(t^2+t-6=0< =>t^2-2t+3t-6=0\)

<\(< =>t\left(t-2\right)+3\left(t-2\right)=0< =>\left(t+3\right)\left(t-2\right)=0\)

\(< =>\orbr{\begin{cases}t+3=0\\t-2=0\end{cases}}< =>\orbr{\begin{cases}t=-3\left(ktm\right)\\t=2\left(tm\right)\end{cases}}\)

khi đó ta được \(\sqrt{x}=t< =>x=4\)

a) \(\sqrt{1-x}=\sqrt[3]{27}\)

\(\Leftrightarrow\sqrt{1-x}=3\)

\(\Leftrightarrow1-x=9\)

\(\Rightarrow x=-8\)

b) \(\sqrt{x^2-10x+25}=x+1\)

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

\(\Leftrightarrow\left|x-5\right|=x+1\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=x+1\\x-5=-x-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}0=6\left(vl\right)\\2x=4\end{cases}}\Rightarrow x=2\)

c) \(x+\sqrt{x}-6=0\)

\(\Leftrightarrow\left(x+3\sqrt{x}\right)-\left(2\sqrt{x}+6\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\\sqrt{x}+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=-3\left(vl\right)\end{cases}}\Rightarrow x=4\)

ĐKXĐ:  \(x\ge1\)

\(\Rightarrow\left(\sqrt{x-1}+\sqrt{2x+1}\right)^2=1\Leftrightarrow x-1+2x+1+2\sqrt{\left(x-1\right)\left(2x+1\right)}=1\Leftrightarrow3x+2\sqrt{2x^2-x-1}=1\) \(\Leftrightarrow2\sqrt{2x^2-x-1}=1-3x\Rightarrow\left(2\sqrt{2x^2-x-1}\right)^2=\left(1-3x\right)^2\Leftrightarrow8x^2-4x-4=9x^2-6x+1\) \(\Leftrightarrow x^2-2x+5=0\Leftrightarrow\left(x-1\right)^2+4=0\Leftrightarrow\left(x-1\right)^2=-4\) vô lí vì VT\(\ge0\) mà VP<0 \(\Rightarrow\) ko có x Vậy...

cần gấp thì mình làm cho 

\(\sqrt{x^2+2x+1}=\sqrt{x+1}\left(đk:x\ge1\right)\)

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

\(< =>x+1=\sqrt{x+1}\)

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

\(< =>\sqrt{x+1}=1< =>x=0\left(ktm\right)\)

ĐKXĐ : \(x\ge-1\)

Bình phương 2 vế , ta có :

\(x^2+2x+1=x+1\)

\(\Leftrightarrow x^2+2x+1-x-1=0\)

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

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

Vậy ...............................

 mình nhầm  

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

\(< =>x^2+2x+1=x+1\)

\(< =>x^2+x=0< =>\orbr{\begin{cases}x=0\\x=-1\end{cases}\left(tmđk\right)}\)

ĐKXĐ: \(2x+3\ge0\Leftrightarrow x\ge-\frac{3}{2}\\ \)

Ta có: \(x+\sqrt{2x+3}=0\Leftrightarrow\sqrt{2x+3}=-x\)

\(\Leftrightarrow\hept{\begin{cases}-x\ge0\\\left(\sqrt{2x+3}\right)^2=\left(-x\right)^2\end{cases}\Leftrightarrow\hept{\begin{cases}x\le0\\2x+3=x^2\end{cases}\Leftrightarrow}\hept{\begin{cases}-\frac{3}{2}\le x\le0\\x^2-2x-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}-\frac{3}{2}\le x\le0\\\left(x-3\right).\left(x+1\right)=0\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}-\frac{3}{2}\le x\le0\\\orbr{\begin{cases}x=3\\x=-1\end{cases}}\end{cases}\Leftrightarrow x=-1}\)\(\Leftrightarrow\hept{\begin{cases}-\frac{3}{2}\le x\le0\\x=3,x=-1\end{cases}\Leftrightarrow x=-1}\)

Vậy x=-1

\(\Leftrightarrow\hept{\begin{cases}-\frac{3}{2}\le x\le0\\\orbr{\begin{cases}x=3\\x=-1\end{cases}}\end{cases}\Leftrightarrow x=-1}\)

cảm ơn bạn nhiều nhiều ^-^

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

\(2x+3=x^2\)

\(x^2-2x-3=0\)\(\)

\(\left(x+1\right)\left(x-3\right)=0\)