a: ĐKXĐ: 2-|x-2|>=0

=>|x-2|<=2

=>-2<=x-2<=2

=>0<=x<=4

TH1: 0<=x<2

=>x-2<0

mà \(\sqrt{2-\left|x-2\right|}>=0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{2-\left|x-2\right|}\) >x-2 với mọi x thỏa mãn

=>NHận

=>0<=x<2

TH2: 2<=x<=4

\(\sqrt{2-\left|x-2\right|}>x-2\)

=>\(2-\left|x-2\right|\ge\left(x-2\right)^2\)

=>\(\left(x-2\right)^2\le2-\left(x-2\right)=2-x+2=4-x\)

=>\(x^2-4x+4\le4-x\)

=>\(x^2-3x\le0\)

=>x(x-3)<=0

=>0<=x<=3

=>2<=x<=3

Vậy: 0<=x<=3

b: \(x^2+3x+2\ge2\cdot\sqrt{x^2+3x+5}\)

=>\(x^2+3x+5-2\sqrt{x^2+3x+5}-3\ge0\)

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

=>\(\sqrt{x^2+3x+5}-3\ge0\)

=>\(\sqrt{x^2+3x+5}\ge3\)

=>\(x^2+3x+5\ge9\)

=>\(x^2+3x-4\ge0\)

=>(x+4)(x-1)>=0

=>x>=1 hoặc x<=-4

dẹp cách bình phương bùng nhùng ta cùng đến với liên hợp hại não :v

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

\(pt\Leftrightarrow\sqrt{x^2+x+2}-2=\frac{3x^2+3x+2}{3x+1}-2\)

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

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

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

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

Dễ thấy: \(\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{3x}{3x+1}>0\forall x\ge-\frac{1}{3}\)

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

cứ tin ở t đi thầy có hỏi thì cứ nói là t ko chịu trách nhiệm :v

ĐKXĐ: \(x\ne-\frac{1}{3}\)

Do \(3x^2+3x+2>0\Rightarrow\) để pt có nghiệm thì \(x>-\frac{1}{3}\)

Ta có: \(\left(3x+1\right)\sqrt{x^2+x+2}=3x^2+3x+2\)

\(\Leftrightarrow\left(3x+1\right)^2\left(x^2+x+2\right)=9x^4+6x^2\left(3x+2\right)+9x^2+12x+4\)

\(\Leftrightarrow\left(9x^2+6x+1\right)\left(x^2+x+2\right)=9x^4+18x^3+21x^2+12x+4\)

\(\Leftrightarrow9x^4+15x^3+25x^2+13x+2=9x^4+18x^3+21x^2+12x+4\)

\(\Leftrightarrow3x^3-4x^2-x+2=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(3x+2\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\left(n\right)\\x=-\frac{2}{3}\left(l\right)\end{cases}}\)

Vậy pt có nghiệm x = 1.

mn ơi giúp em vs ạ !!!

giúp e vs

a/ ĐKXĐ: ...

\(\Leftrightarrow2\left(x^2-5x-6\right)+\sqrt{x^2-5x-6}-3=0\)

Đặt \(\sqrt{x^2-5x-6}=a\ge0\)

\(2a^2+a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-5x-6}=1\Leftrightarrow x^2-5x-7=0\)

b/ ĐKXĐ: ...

\(\Leftrightarrow5\sqrt{3x^2-4x-2}-2\left(3x^2-4x-2\right)+3=0\)

Đặt \(\sqrt{3x^2-4x-2}=a\ge0\)

\(-2a^2+5a+3=0\) \(\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{1}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{3x^2-4x-2}=3\Leftrightarrow3x^2-4x-11=0\)

c/ \(\Leftrightarrow x^2+2x-6+\sqrt{2x^2+4x+3}=0\)

Đặt \(\sqrt{2x^2+4x+3}=a>0\Rightarrow x^2+2x=\frac{a^2-3}{2}\)

\(\frac{a^2-3}{2}-6+a=0\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x^2+4x+3}=3\Leftrightarrow2x^2+4x-6=0\)

d/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{3x-1}{x}}=a>0\)

\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\)

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

\(\Rightarrow a=1\Rightarrow\sqrt{\frac{3x-1}{x}}=1\Leftrightarrow3x-1=x\)

e/ĐKXĐ: ...

\(\Leftrightarrow2\sqrt{\frac{6x-1}{x}}=\frac{x}{6x-1}+1\)

Đặt \(\sqrt{\frac{6x-1}{x}}=a>0\)

\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)

\(\Rightarrow a=1\Rightarrow\sqrt{\frac{6x-1}{x}}=1\Rightarrow6x-1=x\)

f/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{x}{2x-1}}=a>0\)

\(\frac{1}{a}+1+a=3a^2\)

\(\Leftrightarrow3a^3-a^2-a-1=0\)

\(\Leftrightarrow\left(a-1\right)\left(3a^2+2a+1\right)=0\)

\(\Leftrightarrow a=1\Rightarrow\sqrt{\frac{x}{2x-1}}=1\Rightarrow x=2x-1\)