b: \(\left(x^2-3x+2\right)\left(x^2-12x+32\right)\le4x^2\)

=>(x-1)(x-2)(x-4)(x-8)<=\(4x^2\)

=>\(\left(x^2-9x+8\right)\left(x^2-6x+8\right)\le4x^2\)

=>\(\left(x^2+8\right)^2-15x\left(x^2+8\right)+54x^2-4x^2\le0\)

=>\(\left(x^2+8\right)^2-15x\left(x^2+8\right)+50x^2\le0\)

=>\(\left(x^2-5x+8\right)\left(x^2-10x+8\right)\le0\)

mà \(x^2-5x+8=x^2-5x+\frac{25}{4}+\frac74=\left(x-\frac52\right)^2+\frac74>0\forall x\)

nên \(x^2-10x+8\le0\)

=>\(x^2-10x+25-17\le0\)

=>\(\left(x-5\right)^2\le17\)

=>\(-\sqrt{17}\le x-5\le\sqrt{17}\)

=>\(-\sqrt{17}+5\) <=x<=\(\sqrt{17}+5\)

Câu 1: ĐKXĐ: ...

\(\Leftrightarrow4x\left(3x-1\right)+x-1=4x\sqrt{3x+1}\)

\(\Leftrightarrow12x^2-3x-1-4x\sqrt{3x+1}=0\)

\(\Leftrightarrow16x^2-\left(4x^2+4x\sqrt{3x+1}+3x+1\right)=0\)

\(\Leftrightarrow16x^2-\left(2x+\sqrt{3x+1}\right)^2=0\)

\(\Leftrightarrow\left(2x-\sqrt{3x+1}\right)\left(6x+\sqrt{3x+1}\right)=0\)

\(\Leftrightarrow...\)

Câu 2:

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

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

\(\Leftrightarrow y\left(y^2+3xy+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=0\Rightarrow...\\y^2+3xy+2=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow3xy=-y^2-2\Rightarrow x=\frac{-y^2-2}{3y}\)

\(\Rightarrow\left(\frac{y^2+2}{3y}\right)^2-1=3\left(1-y^2\right)\)

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

\(\Leftrightarrow\frac{\left(y-1\right)\left(y-2\right)\left(y+1\right)\left(y+2\right)}{9y^2}=3\left(1-y^2\right)\)

\(\Leftrightarrow\frac{\left(y^2-1\right)\left(y^2-4\right)}{9y^2}=3\left(1-y^2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\\frac{y^2-4}{9y^2}=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\28y^2=4\end{matrix}\right.\)

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

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

\(\Leftrightarrow\frac{12x^2-4x+x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{12x^2-3x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{\left(12x^2-3x-1\right)^2}{16x^2}=3x+1\)

\(\Leftrightarrow\left(12x^2-3x-1\right)^2=16x^2\left(3x+1\right)\)

\(\Leftrightarrow144x^4-120x^3-31x^2+6x+1=0\)

\(\Leftrightarrow144x^4-144x^3+24x^3-24x^2-7x^2+7x-x+1=0\)

\(\Leftrightarrow144x^3\left(x-1\right)+24x^2\left(x-1\right)+7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(144x^3+24x^2+7x-1\right)=0\)

Tìm được mỗi nghiệm thôi à :v

a) \(\left|3x+1\right|=\left|x+1\right|\)

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

c) \(\sqrt{9x^2-12x+4}=\sqrt{x^2}\)

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

\(\Leftrightarrow\left|3x-2\right|=\left|x\right|\)

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

d) \(\sqrt{x^2+4x+4}=\sqrt{4x^2-12x+9}\)

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

\(\Leftrightarrow\left|x+2\right|=\left|2x-3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-3\\x+2=-2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)

e) \(\left|x^2-1\right|+\left|x+1\right|=0\)

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

\(\Leftrightarrow x=-1\)

f) \(\sqrt{x^2-8x+16}+\left|x+2\right|=0\)

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

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

⇒ vô nghiệm

mong mọi người giải giúp em vs 

ĐKXĐ: -3<=x<=2

Ta có: \(4(x + 1)(\sqrt{x + 3} + \sqrt{2 - x}) = -x^2 + 12x + 13\)

=>\(4(x+1)(\sqrt{x + 3}+\sqrt{2 - x})+\left(x^2-12x-13\right)=0\)

=>\(4(x+1)(\sqrt{x + 3}+\sqrt{2 - x})+\left(x-13\right)\left(x+1\right)=0\)

=>\((x + 1) \left[ 4(\sqrt{x + 3} + \sqrt{2 - x}) - (13 - x) \right] = 0\)

TH1: x+1=0

=>x=-1(nhận)

TH2: \(4\left(\sqrt{x+3}+\sqrt{2-x}\right)-\left(13-x\right)=0\) (1)

Đặt \(t=\sqrt{x+3}+\sqrt{2-x}\)

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

=>\(t^2=5+2\cdot\sqrt{-x^2-x+6}\)

(1) sẽ trở thành 4t=13-x

=>x=13-4t

\(-x^2-x+6=-(13-4t)^2-(13-4t)+6\)

\(=-16t^2+108t-176\)

Do đó, ta sẽ có phương trình sau:

\(t^2 = 5 + 2\sqrt{-16t^2 + 108t - 176}\)

=>\(t^2 - 5 = 2\sqrt{-16t^2 + 108t - 176}\)

=>\((t^2 - 5)^2 = 4(-16t^2 + 108t - 176)\) và \(t\ge\sqrt5\)

=>\(t^4 - 10t^2 + 25 = -64t^2 + 432t - 704\) và \(t\ge\sqrt5\)

=>\(t^4+54t^2-432t+729=0\) và \(t\ge\sqrt5\)

=>\((t-3)^2(t^2+6t+81)=0\) và \(t\ge\sqrt5\)

=>t=3

=>x=13-4*3=1(nhận)