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

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

lx

lỗi r bn

lx

a: =>(x^2+x)^2-2(x^2+x)+(x^2+x)-2=0

=>(x^2+x-2)(x^2+x+1)=0

=>(x+2)(x-1)=0

=>x=-2 hoặc x=1

b: ĐKXĐ: x<>4; x<>1

PT =>\(\dfrac{x+3+3x-12}{x-4}=\dfrac{6}{1-x}\)

=>(4x-9)(1-x)=6(x-4)

=>4x-4x^2-9+9x=6x-24

=>-4x^2+13x-9-6x+24=0

=>-4x^2+7x+15=0

=>x=3(nhận) hoặc x=-5/4(nhận)

\(x=7\)

ĐKXĐ:\(x\ge-9\)

\(x-\sqrt{x+9}=3\\ \Leftrightarrow\sqrt{x+9}=x-3\left(x\ge3\right)\\ \Leftrightarrow x+9=x^2-6x+9\\ \Leftrightarrow x^2-7x=0\\ \Leftrightarrow x\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy x=7

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:

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

Bài 1:
a. ĐKXĐ: $x\geq \frac{2}{5}$

PT $\Leftrightarrow 5x-2=7^2=49$

$\Leftrightarrow 5x=51$

$\Leftrightarrow x=\frac{51}{5}=10,2$

b. ĐKXĐ: $x\geq 3$

PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{25(x-3)}=24$

$\Leftrightarrow 3\sqrt{x-3}+5\sqrt{x-3}=24$

$\Leftrightarrow 8\sqrt{x-3}=24$

$\Leftrightarrow \sqrt{x-3}=3$

$\Leftrightarrow x-3=9$

$\Leftrightarrow x=12$ (tm)

Bài 1:

c. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow x^2-5x+6-2(\sqrt{x-2}-1)=0$

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

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

$x-3=0$ hoặc $x-2=\frac{2}{\sqrt{x-2}+1}$

Nếu $x-3=0$

$\Leftrightarrow x=3$ (tm) 

Nếu $x-2=\frac{2}{\sqrt{x-2}+1}$

$\Leftrightarrow a^2=\frac{2}{a+1}$ (đặt $\sqrt{x-2}=a$)

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

$\Leftrightarrow a^2(a-1)+2a(a-1)+2(a-1)=0$

$\Leftrightarrow (a-1)(a^2+2a+2)=0$

Hiển nhiên $a^2+2a+2=(a+1)^2+1>0$ với mọi $a$ nên $a-1=0$

$\Leftrightarrow a=1\Leftrightarrow \sqrt{x-2}=1\Leftrightarrow x=3$ (tm)

Vậy pt có nghiệm duy nhất $x=3$.

Bài 2:

ĐKXĐ: $x\geq 0; x\neq 4$

\(A=\frac{\sqrt{x}(\sqrt{x}-2)-\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}+2)\sqrt{x}-2)}.\frac{\sqrt{x}+2}{2}\\ =\frac{-4\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}+2}{2}\\ =\frac{-2\sqrt{x}}{\sqrt{x}-2}=\frac{2\sqrt{x}}{2-\sqrt{x}}\)

a. ĐKXĐ: \(-1\le x\le1\)

Đặt \(\sqrt{1+x}+\sqrt{1-x}=t>0\)

\(\Rightarrow t^2=2+2\sqrt{1-t^2}\)

Pt trở thành:

\(t.t^2=8\Leftrightarrow t^3=8\Leftrightarrow t=2\)

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

\(\Leftrightarrow2+2\sqrt{1-x^2}=2\)

\(\Leftrightarrow1-x^2=0\Rightarrow x=\pm1\)

b.

ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)

\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\)

Pt trở thành:

\(t=t^2-4-16\Leftrightarrow...\)