Phương trình chứa căn

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\\sqrt{x^2-4}=x-3\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\x^2-4=x^2-6x+9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\x=\dfrac{13}{6}< 3\left(loại\right)\end{matrix}\right.\)

Vậy pt có nghiệm duy nhất \(x=-3\)

\(\Leftrightarrow x^2-4x+13-\sqrt{x^2-4x+13}-6=0\)

Đặt \(\sqrt{x^2-4x+13}=t>0\)

\(\Rightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2< 0\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-4x+13}=3\)

\(\Leftrightarrow x^2-4x+13=9\)

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

ĐK: \(x\ge\dfrac{1}{2}\)

\(pt\Leftrightarrow\sqrt{x}-1+\sqrt{2x-1}-1+x^2+x-2=0\)

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

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

Vì \(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\sqrt{2x-1}+1}+x+2>0\) nên \(x-1=0\Leftrightarrow x=1\left(tm\right)\)

ĐK: \(x\ge\dfrac{1}{3}\)

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

\(\Leftrightarrow16x^2+24x-32=8\left(4x-3\right)\sqrt{3x-1}\)

\(\Leftrightarrow\left(4x-3\right)^2+16\left(3x-1\right)-8\left(4x-3\right)\sqrt{3x-1}=25\)

\(\Leftrightarrow\left(4x-3-4\sqrt{3x-1}\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-3-4\sqrt{3x-1}=5\\4x-3-4\sqrt{3x-1}=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x-1}=x-2\\2\sqrt{3x-1}=2x+1\end{matrix}\right.\)

TH1: \(\sqrt{3x-1}=x-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-1=\left(x-2\right)^2\\x-2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-7x+6=0\\x\ge2\end{matrix}\right.\)

\(\Leftrightarrow x=6\left(tm\right)\)

TH2: \(2\sqrt{3x-1}=2x+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}4\left(3x-1\right)=\left(2x+1\right)^2\\2x+1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-8x+5\\x\ge-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\) vô nghiệm

Vậy \(x=6\)

ĐKXĐ: \(\left[{}\begin{matrix}x\le-1\\x\ge\dfrac{3}{5}\end{matrix}\right.\)

\(\left(x+1\right)\left(45x^2-62x+25\right)=4\sqrt{\left(x+1\right)\left(5x-3\right)\left(5x-3\right)^2}\)

- Với \(x=-1\) là 1 nghiệm

- Với \(x< -1\Rightarrow\left\{{}\begin{matrix}VT< 0\\VP>0\end{matrix}\right.\) pt vô nghiệm

Với \(x\ge\dfrac{3}{5}\) ta có:

\(45x^3-17x^2-37x+25=4\sqrt{\left(x+1\right)\left(5x-3\right)\left(5x-3\right)^2}\)

\(\Leftrightarrow45x^3-17x^2-37x+25\le2\left[\left(x+1\right)\left(5x-3\right)+\left(5x-3\right)^2\right]\)

\(\Leftrightarrow45x^3-77x^2+19x+13\le0\)

\(\Leftrightarrow\left(x-1\right)^2\left(45x+13\right)\le0\)

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{3x+1}-\sqrt{2x+3}=\dfrac{x-2}{4}\)

\(\Leftrightarrow\dfrac{x-2}{\sqrt{3x+1}+\sqrt{2x+3}}=\dfrac{x-2}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\dfrac{1}{\sqrt{3x+1}+\sqrt{2x+3}}=\dfrac{1}{4}\left(1\right)\end{matrix}\right.\)

Xét (1) \(\Leftrightarrow\sqrt{3x+1}+\sqrt{2x+3}=4\)

\(\Leftrightarrow5x+4+2\sqrt{6x^2+11x+3}=16\)

\(\Leftrightarrow2\sqrt{6x^2+11x+3}=12-5x\) (\(x\le\dfrac{12}{5}\))

\(\Leftrightarrow4\left(6x^2+11x+3\right)=\left(12-5x\right)^2\)

\(\Leftrightarrow x^2-164x+132=0\Rightarrow\left[{}\begin{matrix}x=82-8\sqrt{103}\\x=82+8\sqrt{103}>\dfrac{12}{5}\left(loại\right)\end{matrix}\right.\)

\(\sqrt{x+3}+\sqrt{2x-1}=4-x\)(1)

ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow\sqrt{x+3}-2+\sqrt{2x-1}-1+x-1=0\)

\(\Leftrightarrow\dfrac{x-1}{\sqrt{x+3}+2}+\dfrac{2\left(x-1\right)}{\sqrt{2x-1}+1}+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{x+3}+2}+\dfrac{2}{\sqrt{2x-1}+1}+1\right)=0\)

\(\Leftrightarrow x-1=0\)( vì \(\dfrac{1}{\sqrt{x+3}+2}+\dfrac{2}{\sqrt{2x-1}+1}+1\)>0)

\(\Leftrightarrow x=1\)(thỏa mãn)

Vậy phương trình có nghiệm là x=1

\(x+\sqrt{5-x^2}+x\sqrt{5-x^2}=5\)

ĐKXĐ: \(-\sqrt{5}\le x\le\sqrt{5}\)

Đặt \(\sqrt{5-x^2}=t\)(\(t\)\(\ge0\))

Ta có: \(\left\{{}\begin{matrix}x+t+xt=5\\x^2+t^2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\left(x+t\right)+2xt=10\\\left(x+t\right)^2-2xt=5\end{matrix}\right.\)

\(\Rightarrow\left(x+t\right)^2+2\left(x+t\right)-15=0\Leftrightarrow\left(x+t+5\right)\left(x+t-3\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=-5-x\\t=3-x\end{matrix}\right.\)(giải hai trường hợp rồi kết luận nghiệm)