2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

đk: \(\hept{\begin{cases}x^2-2x+5\ge0\\4x+5\ge0\end{cases}}\Leftrightarrow x\ge\frac{-5}{4}\)

Ta có: \(x^3-2x^2-\sqrt{x^2-2x+5}=2\sqrt{4x+5}-5x-4\)

\(\Leftrightarrow3x^3-6x^2+15x+12-3\sqrt{x^2-2x+5}-6\sqrt{4x+5}=0\)

\(\Leftrightarrow3\left(x+1-\sqrt{x^2-2x+5}\right)+2\sqrt{4x+5}\left(\sqrt{4x+5}-3\right)+3x^3-6x^2+4x-1=0\)

\(\Leftrightarrow\frac{12\left(x-1\right)}{x+1+\sqrt{x^2-2x+5}}+\frac{8\left(x-1\right)\sqrt{4x+5}}{\sqrt{4x+5}+3}+\left(x-1\right)\left(3x^2-3x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{12}{x+1+\sqrt{x^2-2x+5}}+\frac{8\sqrt{4x+5}}{\sqrt{4x+5}+3}+3x^2-3x+1\right)=0\Leftrightarrow x=1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+5x+12}=a>0\\\sqrt{2x^2+3x+2}=b>0\end{matrix}\right.\) \(\Rightarrow x+5=\dfrac{a^2-b^2}{2}\)

Phương trình trở thành:

\(a+b=\dfrac{a^2-b^2}{2}\)

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

\(\Leftrightarrow a-b-2=0\) (do \(a+b>0\))

\(\Leftrightarrow a=b+2\)

\(\Leftrightarrow\sqrt{2x^2+5x+12}=\sqrt{2x^2+3x+2}+2\)

\(\Leftrightarrow2x^2+5x+12=2x^2+3x+6+4\sqrt{2x^2+3x+2}\)

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

\(\Leftrightarrow x^2+6x+9=4\left(2x^2+3x+2\right)\)

\(\Leftrightarrow7x^2+6x-1=0\)

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

cảm ơn Thầy nhiều ạ

a: Ta có: \(\sqrt{4x^2+4x+3}=8\)

\(\Leftrightarrow4x^2+4x+1+2-64=0\)

\(\Leftrightarrow4x^2+4x-61=0\)

\(\Delta=4^2-4\cdot4\cdot\left(-61\right)=992\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-4-4\sqrt{62}}{8}=\dfrac{-1-\sqrt{62}}{2}\\x_2=\dfrac{-4+4\sqrt{62}}{8}=\dfrac{-1+\sqrt{62}}{2}\end{matrix}\right.\)

VP bạn bình phương sao vế trái bạn không bình phương ạ! 

5: ĐKXĐ: \(\frac{x+3}{x-7}>0\)

=>x>7 hoặc x<-3

Ta có: \(\left(x-7\right)\cdot\sqrt{\frac{x+3}{x-7}}=x+4\)

=>\(\sqrt{\left(x+3\right)\left(x-7\right)}=x+4\)

=>\(\begin{cases}x+4\ge0\\ \left(x+3\right)\left(x-7\right)=\left(x+4\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-4\\ x^2-4x-21=x^2+8x+16\end{cases}\)

=>\(\begin{cases}x\ge-4\\ -12x=37\end{cases}\Rightarrow x=-\frac{37}{12}\) (nhận)

6: ĐKXĐ: x>=4

Ta có: \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)

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

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

=>2x-3=x-1

=>2x-x=-1+3

=>x=2(loại)

7: ĐKXĐ: x>=1

Ta có: \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)

=>\(\sqrt{x-1+2\cdot\sqrt{x-1}+1}+\sqrt{x-1-2\cdot\sqrt{x-1}\cdot1+1}=\frac{x+3}{2}\)

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

=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\frac{x+3}{2}\) (1)

TH1: \(\sqrt{x-1}-1\ge0\)

=>\(\sqrt{x-1}\ge1\)

=>x-1>=1

=>x>=2

(1) sẽ trở thành: \(\sqrt{x-1}+1+\sqrt{x-1}-1=\frac{x+3}{2}\)

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

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

=>\(16\left(x-1\right)=\left(x+3\right)^2\)

=>\(x^2+6x+9=16x-16\)

=>\(x^2-10x+25=0\)

=>\(\left(x-5\right)^2=0\)

=>x-5=0

=>x=5(nhận)

TH2: \(\sqrt{x-1}-1<0\)

=>\(\sqrt{x-1}<1\)

=>0<=x-1<1

=>1<=x<2

(1) sẽ trở thành: \(\sqrt{x-1}+1+1-\sqrt{x-1}=\frac{x+3}{2}\)

=>\(\frac{x+3}{2}=2\)

=>x+3=4

=>x=1(nhận)

\(ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow2x-2\sqrt{2x^2+5x-3}=1+x\sqrt{2x-1}-2x\sqrt{x+3}\\ \Leftrightarrow\left(2x-2\right)-\left(2\sqrt{2x^2+5x-3}-4\right)=\left(x\sqrt{2x-1}-x\right)-\left(2x\sqrt{x+3}-4x\right)-3x+3\\ \Leftrightarrow2\left(x-1\right)-\dfrac{2\left(2x^2+5x-7\right)}{\sqrt{2x^2+5x-3}+4}=\dfrac{x\left(2x-2\right)}{\sqrt{2x-1}+1}-\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+4x}-3\left(x-1\right)\\ \Leftrightarrow2\left(x-1\right)-\dfrac{2\left(x-1\right)\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x\left(x-1\right)}{\sqrt{2x-1}+1}+\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+4x}+3\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left[2-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x}{\sqrt{2x-1}+2}+\dfrac{2x}{\sqrt{x+3}+4x}+3\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\2-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x}{\sqrt{2x-1}+2}+\dfrac{2x}{\sqrt{x+3}+4x}+3=0\left(1\right)\end{matrix}\right.\)

Với \(x\ge\dfrac{1}{2}\Leftrightarrow-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}>-\dfrac{2\cdot8}{4}=-4\)

\(-\dfrac{2x}{\sqrt{2x-1}+2}>-\dfrac{1}{2};\dfrac{2x}{\sqrt{x+3}+4x}>0\)

Do đó \(\left(1\right)>2-4-\dfrac{1}{2}+3=\dfrac{1}{2}>0\) nên (1) vô nghiệm

Vậy PT có nghiệm duy nhất \(x=1\)