\(a,ĐK:1\le x\le3\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\\\sqrt{3-x}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(PT\Leftrightarrow a+b-ab=1\Leftrightarrow a+b-ab-1=0\\ \Leftrightarrow\left(a-1\right)\left(1-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=1\\3-x=1\end{matrix}\right.\Leftrightarrow x=2\left(tm\right)\)

\(b,ĐK:0\le x\le9\\ PT\Leftrightarrow9+2\sqrt{x\left(9-x\right)}=-x^2+9x+9\\ \Leftrightarrow2\sqrt{-x^2+9x}-\left(-x^2+9x\right)=0\\ \Leftrightarrow\sqrt{-x^2+9x}\left(2-\sqrt{-x^2+9x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x^2+9x=0\\\sqrt{-x^2+9x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\\x^2-9x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=9\left(n\right)\\x=\dfrac{9+\sqrt{65}}{2}\left(n\right)\\x=\dfrac{9-\sqrt{65}}{2}\left(n\right)\end{matrix}\right.\)

1/ Đặt \(\sqrt{x^2+x+1}=a>0\)

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

\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=2\end{cases}}\)

2/ \(\sqrt{x+5}-\sqrt{x}=\sqrt{x-3}\)

\(\Leftrightarrow\sqrt{x+5}=\sqrt{x}+\sqrt{x-3}\)

\(\Leftrightarrow8-x=2\sqrt{x\left(x-3\right)}\)

\(\Leftrightarrow-3x^2-4x+64=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{16}{3}\\x=4\end{cases}}\)

PS: Điều kiện b tự làm rồi tự chọn nghiệm nhé

MÌnh đang học lớp 8 nên chỉ giải được câu 1 thôi :(

1) \(x^2+x+1-3\sqrt{x^2+x+1}+2.25-0.25=0\)

\(\left(x^2+x+1-1.5\right)^2=0.25\)

\(=>\left(x^2+x-1.5\right)^2=0.5^2\)

=> \(x^2+x-1.5=0.5\)                                     \(x^2+x-1.5=-0.5\)

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

\(x^2+x+\frac{1}{4}=\frac{9}{4}\\ \left(x+\frac{1}{2}\right)^2=\frac{9}{4}\)                                       Đến đây bạn tự làm nốt nhé mình lười quá Sorry

Lời giải:

1. ĐKXĐ: $x\geq \frac{-5+\sqrt{21}}{2}$

PT $\Leftrightarrow x^2+5x+1=x+1$

$\Leftrightarrow x^2+4x=0$

$\Leftrightarrow x(x+4)=0$

$\Rightarrow x=0$ hoặc $x=-4$

Kết hợp đkxđ suy ra $x=0$

2. ĐKXĐ: $x\leq 2$

PT $\Leftrightarrow x^2+2x+4=2-x$

$\Leftrightarrow x^2+3x+2=0$

$\Leftrightarrow (x+1)(x+2)=0$

$\Leftrightarrow x+1=0$ hoặc $x+2=0$

$\Leftrightarrow x=-1$ hoặc $x=-2$
3.

ĐKXĐ: $-2\leq x\leq 2$

PT $\Leftrightarrow \sqrt{2x+4}=\sqrt{2-x}$

$\Leftrightarrow 2x+4=2-x$

$\Leftrightarrow 3x=-2$

$\Leftrightarrow x=\frac{-2}{3}$ (tm)

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

\(\Leftrightarrow\left|x+5\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)

2) \(ĐK:x\ge2\)

\(\Leftrightarrow\sqrt{x-2}=2\)

\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

4) \(ĐK:x\ge0\)

\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)

\(a,\sqrt{2x+5}=\sqrt{1-x}\)

\(\Rightarrow2x+5=1-x\)

\(2x+x=1-5\)

\(3x=-4\Leftrightarrow x=\frac{-4}{3}\)

Vậy \(S=\left\{-\frac{4}{3}\right\}\)thuộc tập nghiệm của pt trên