a) \(\sqrt{4+2x-x^2}=x-2\)

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

\(\Leftrightarrow4+2x-x^2=x^2-4x+4\)

\(\Leftrightarrow-x^2+6x=0\)

\(\Leftrightarrow x\left(6-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\6-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

hình như bài này sai đó! em mới học lớp 8 thôi !

lê thị thu huyền:

sai rồi đó em, nhưng mà nhờ em chị mới biết chị sai chỗ nào. Không hiểu đầu óc kiểu gì mà lại thấy 2x+4x=8x mới chết chứ !!!

a.

ĐKXĐ: \(x\ne-1\)

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

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

Đặt \(\sqrt{x^2+x+2}=t>0\)

\(\Rightarrow t^2-2\left(x+1\right)t+4x=0\)

\(\Leftrightarrow t\left(t-2x\right)-2\left(t-2x\right)=0\)

\(\Leftrightarrow\left(t-2\right)\left(t-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+2}=2\\\sqrt{x^2+x+2}=2x\left(x\ge0\right)\end{matrix}\right.\)

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

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

b.

ĐKXĐ: \(x\ge-1\)

\(x^2-5x+14-4\sqrt{x+1}=0\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(x+1-4\sqrt{x+1}+4\right)=0\)

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

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

\(\Leftrightarrow x=3\)

Lời giải:
ĐKXĐ: $x\geq -1$

PT $\Leftrightarrow (x^2-6x+9)+[(x+1)-4\sqrt{x+1}+4]=0$

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

Vì $(x-3)^2; (\sqrt{x+1}-2)^2\geq 0$ với mọi $x\geq -1$
Do đó để tổng của chúng $=0$ thì:

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

ĐKXĐ: \(x\ge-1\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(x+1-4\sqrt{x+1}+4\right)=0\)

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

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

\(a,PT\Leftrightarrow x\sqrt{3}=x+2\\ \Leftrightarrow3x^2=x^2+4x+4\\ \Leftrightarrow2x^2-4x-4=0\Leftrightarrow x^2-2x-2=0\\ \Delta=4+8=12\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2-2\sqrt{3}}{2}=1-\sqrt{3}\\x=\dfrac{2+2\sqrt{3}}{2}=1+\sqrt{3}\end{matrix}\right.\)

\(b,ĐK:x\ge\dfrac{2}{3}\\ PT\Leftrightarrow3x-2=7-4\sqrt{3}\\ \Leftrightarrow3x=9-4\sqrt{3}\\ \Leftrightarrow x=\dfrac{9-4\sqrt{3}}{3}\left(tm\right)\)

\(c,ĐK:x\ge-1\\ PT\Leftrightarrow\left(x+1-4\sqrt{x+1}+4\right)+\left(x^2-6x+9\right)=0\\ \Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x=3\end{matrix}\right.\Leftrightarrow x=3\left(tm\right)\)

ĐKXĐ: \(x\ge-1\)

pt \(\Leftrightarrow-x+4\sqrt{x+1}-5=x^2-5x+14-5\)

\(\Leftrightarrow-\left(x+1-4\sqrt{x+1}+4\right)=x^2-6x+9\)

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

Ta thấy \(VT\le0\forall x\ge-1;VP\ge0\forall x\)

nên pt \(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x+1}-2\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow x=3\left(TM\right)\)

Vậy ...

ĐK: x \(\ge\)-1 . pt \(\Leftrightarrow\) (x+1 -4\(\sqrt{x+1}\) +4) + \(^{x^2}\) -6x+9 =0 \(\Leftrightarrow\) (\(\sqrt{x+1}\) -2)^2 +(x-3 )^2 =0 Do (\(\sqrt{x+1}\)-2)^2 \(\ge\)0 ; (x-3)^2 \(\ge\)0 \(\Rightarrow\) \(\left\{{}\begin{matrix}\sqrt{x+1}=2\\x-3=0\end{matrix}\right.\)\(\Rightarrow\) x=3 (tm)

Đk:\(x\ge-1\)

\(pt\Leftrightarrow4\sqrt{x+1}-8=x^2-5x+6\)

\(\Leftrightarrow4\left(\sqrt{x+1}-2\right)=x^2-5x+6\)

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

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{4}{\sqrt{x+1}-2}-x+2\right)=0\)

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\) (thỏa mãn)

Ta có x2−5x+14≐(x−3)2+x+5≥x+5≥x+1+4≥4√x+1x2−5x+14≐(x−3)2+x+5≥x+5≥x+1+4≥4x+1
⇒VT≥VP⇒VT≥VP
Để VT=VP thì x=3.(dấu "=" xảy ra) 

ĐK: x + 1 ≥ 0 <=> x ≥ - 1

<=> 16( x + 1) = x⁴ + 25x² + 196 - 10x³ + 28x² - 140x

<=> 16x + 16 = x⁴ - 10x³ + 53x² - 140x +196

<=> x⁴ - 10x³ + 53x² - 156x + 180 = 0

<=> ( x - 3)²(x² - 4x + 20 ) = 0

<=> x = 3

\(\left(ĐKXĐ:x\ge-1\right)\)

\(4\sqrt{x-1}-x-5=x^2-6x+9\)

\(\Leftrightarrow\frac{16\left(x+1\right)-\left(x+5\right)^2}{4\sqrt{x+1}+x+5}=\left(x-3\right)^2\)

\(\Leftrightarrow\frac{16x+16-x^2-10x-25}{4\sqrt{x+1}+x+5}-\left(x-3\right)^2\)\(=0\)

\(\Leftrightarrow\frac{-\left(x-3\right)^2}{4\sqrt{x+1}+x+5}-\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)^2\left(-\frac{1}{4\sqrt{x+1}+x+5}-1\right)\)\(=0\)

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

\(\Leftrightarrow x=3\)