5(+x)-4=24

8

\(ĐK:-\dfrac{1}{3}\le x\le2\\ PT\Leftrightarrow\left(\sqrt{3x+1}-2\right)-x+1-\sqrt{2-x}\left(\sqrt{2-x}-1\right)=0\\ \Leftrightarrow\dfrac{3\left(x-1\right)}{\sqrt{3x+1}+2}-\left(x-1\right)-\dfrac{\sqrt{2-x}\left(1-x\right)}{\sqrt{2-x}+1}=0\\ \Leftrightarrow\left(x-1\right)\left(\dfrac{3}{\sqrt{3x+1}+2}+\dfrac{\sqrt{2-x}}{\sqrt{2-x}+1}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\\dfrac{3}{\sqrt{3x+1}+2}+\dfrac{\sqrt{2-x}}{\sqrt{2-x}+1}-1=0\end{matrix}\right.\)

Với \(x\ge-\dfrac{1}{3}\) thì \(\dfrac{3}{\sqrt{3x+1}+2}+\dfrac{\sqrt{2-x}}{\sqrt{2-x}+1}-1>0\)

Vậy pt có nghiệm duy nhất \(x=1\)

ĐKXĐ: \(-\dfrac{1}{3}\le x\le2\)

\(\sqrt{3x+1}=3-\sqrt{2-x}\) (do \(-\dfrac{1}{3}\le x\le2\Rightarrow3-\sqrt{2-x}\ge3-\sqrt{2+\dfrac{1}{3}}>0\))

\(\Leftrightarrow3x+1=9+2-x-6\sqrt{3-x}\)

\(\Leftrightarrow3\sqrt{2-x}=5-2x\)

\(\Leftrightarrow9\left(2-x\right)=\left(5-2x\right)^2\)

\(\Leftrightarrow4x^2-11x+7=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{4}\end{matrix}\right.\) (thỏa mãn)

ví dụ x âm thì sao căn x² bằng x được em?

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

\(\Leftrightarrow\sqrt{3}x-x-\sqrt{3}x-1=0\)

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

\(\Leftrightarrow-x=1\)

\(\Leftrightarrow x=-1\)

ĐKXĐ: \(x\ge1\)

\(\left(\sqrt{x-1}-1\right)+\left(\sqrt{x+7}-3\right)+\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\dfrac{x-2}{\sqrt{x-1}+1}+\dfrac{x-2}{\sqrt{x+7}+3}+\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{\sqrt{x-1}+1}+\dfrac{1}{\sqrt{x+7}+3}+x-1\right)=0\)

\(\Leftrightarrow x-2=0\)

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

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

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

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

\(\Leftrightarrow\left(x-\sqrt{2}\right)[\left(x-\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2]=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x=\sqrt{2}\\x=\sqrt{2}+\sqrt{3}\\x=\sqrt{2}-\sqrt{3}\end{cases}}\)

\(x=\sqrt{2}-\sqrt{3}\) nữa nhé!

Lời giải:

ĐKXĐ: $x\geq -1$

Đặt $\sqrt{x+1}=a(a\geq 0)$ thì PT trở thành:

$x^3-3x(x+1)+2\sqrt{(x+1)^3}=0$

$\Leftrightarrow x^3-3xa^2+2a^3=0$

$\Leftrightarrow (x^3-xa^2)-(2xa^2-2a^3)=0$

$\Leftrightarrow x(x-a)(x+a)-2a^2(x-a)=0$

$\Leftrightarrow (x-a)(x^2+ax-2a^2)=0$

$\Leftrightarrow (x-a)[(x+a)(x-a)+a(x-a)]=0$

$\Leftrightarrow (x-a)^2(x+2a)=0$

Nếu $x-a=0$

$\Rightarrow x^2=a^2\Leftrightarrow x^2=x+1$

$\Rightarrow x=\frac{1\pm \sqrt{5}}{2}$. Vì $x=a\geq 0$ nên $x=\frac{1+\sqrt{5}}{2}$

Nếu $x+2a=0$

$\Rightarrow x^2=4a^2\Leftrightarrow x^2=4(x+1)$

$\Rightarrow x=2\pm 2\sqrt{2}$. Mà $x=-2a\leq 0$ nên $x=2-2\sqrt{2}$

Vậy..........

ĐKXĐ: ...

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

Đặt \(\left\{{}\begin{matrix}x=a\\\sqrt{x+1}=b\end{matrix}\right.\)

\(\Rightarrow a^3-3ab^2+2b^3=0\)

\(\Leftrightarrow\left(a+2b\right)\left(a-b\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}2b=-a\\a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=-x\left(x\le0\right)\\x=\sqrt{x+1}\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-4=0\\x^2-x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2-2\sqrt{2}\\x=\frac{1+\sqrt{5}}{2}\end{matrix}\right.\)