Phương trình chứa căn

Này hử .-. \(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)

Nhân 2 vế của đẳng thức với \(\sqrt{x^2+3}-x\) có:

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

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

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

Tương tự nhân 2 vế đẳng thức với \(\sqrt{y^2+3}-y\) cũng có:

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

Cộng theo vế 2 đẳng thức \(\left(1\right);\left(2\right)\) có:

\(2\left(x+y\right)=0\Leftrightarrow x+y=0\)

Và \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=0\)

\(10x^2-9x-8x\sqrt{2x^2-3x+1}+3=0\)

Đặt \(a=\sqrt{2x^2-3x+1}\ge0\) thì:

\(4x^2+3a^2-8ax=0\)

\(\Leftrightarrow\left(2x-a\right)\left(2x-3a\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}2x=\sqrt{2x^2-3x+1}\\2x=3\sqrt{2x^2-3x+1}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}4x^2=2x^2-3x+1\\4x^2=9\left(2x^2-3x+1\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x^2+3x-1=0\\\left(3-2x\right)\left(7x-3\right)=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{7}\\x=\dfrac{3}{2}\\x=\dfrac{\sqrt{17}}{4}-\dfrac{3}{4}\end{matrix}\right.\)

đkxđ:\(\left\{{}\begin{matrix}x\sqrt{x}-1\ne0\\\sqrt{x}\ge0\\2x+\sqrt{x}-1\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge1\\x\ge0\\x\ne\dfrac{1}{4}\end{matrix}\right.\)

\(\Rightarrow x\ge1\)

sory, x>1 bạn nhé

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

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

ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ge0\\x+3\ge0\\x-2\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge-3\\x\ge2\end{matrix}\right.\)\(\Leftrightarrow x\ge2\)

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

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

\(\Leftrightarrow2.\sqrt{\left(x+3\right).\left(x-2\right)}=-x\)

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

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

\(\Leftrightarrow4.\left(x+3\right).\left(x-2\right)=x^2\)

\(\Leftrightarrow\left(4x+12\right).\left(x-2\right)-x^2=0\)

\(\Leftrightarrow4x^2-8x+12x-24-x^2=0\)

\(\Leftrightarrow3x^2+4x-24=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+\sqrt{76}}{3}\left(\text{thỏa mãn}\right)\\x=\dfrac{-2-\sqrt{76}}{3}\left(\text{loại }\right)\end{matrix}\right.\)

Vậy ......

.

\(x^2-3x-\sqrt{x^2-3x+4}+2=0\) ĐK : \(x^2-3x+4\ge0\)

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

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

Đặt : \(\sqrt{x^2-3x+4}=t\) \(\left(t\ge0\right)\)

\(pt\Leftrightarrow t^2-2=t\)

\(\Leftrightarrow t^2-t-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(tm\right)\\t=-1\left(l\right)\end{matrix}\right.\)

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

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

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

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

Ta có: \(x^2-3x-\sqrt{x^2-3x+4}+2=0\)

\(x^2-3x+4-\sqrt{x^2-3x+4}-2=0\)

Đặt \(t=\sqrt{x^2-3x+4}\left(t\ge0\right)\)

Ta có: \(t^2-t-2=0\)

\(1+\left(-2\right)-\left(-1\right)=0\)

\(\Rightarrow\)pt có 2 nghiệm.

\(\left[{}\begin{matrix}t_1=-1\left(loại\right)\\t_2=2\left(nhận\right)\end{matrix}\right.\)

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Vậy nghiệm của pt là \(\left\{0;3\right\}\)