\(\sqrt{2x^2+16x+18}+\sqrt{x^2+1}=2x+4\left(1\right)\)

\(ĐK:x\in R\)

\(pt\left(1\right)\Leftrightarrow2x^2+16x+18+x^2+1+2\sqrt[]{(2x^2+16x+18)\left(x^2+1\right)}=4x^2+16x+16\)

\(\Leftrightarrow3+2\sqrt{(2x^2+16x+18)\left(x^2+1\right)}=x^2\)

\(\Leftrightarrow2\sqrt{(2x^2+16x+8)\left(x^2+1\right)}=x^2-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-3\ge0\\4\left(2x^2+16x+8\right)\left(x^2+1\right)=x^4-6x^2+9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{3}\le x\le\sqrt{3}\\4\left(2x^4+16x^3+10x^2+16x+8\right)=x^4-6x^2+9\end{matrix}\right.\)

\(\Leftrightarrow7x^4+64x^3+46x^2+64x+23=0\)

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

Sửa đề: \(\frac{x+2}{x^2+2x+4}+\frac{x-2}{x^2-2x+4}=\frac{32}{x\left(x^4+4x^2+16\right)}\)

ĐKXĐ: x<>0

TA có: \(x^4+4x^2+16\)

\(=x^4+8x^2+16-4x^2\)

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

Ta có: \(\frac{x+2}{x^2+2x+4}+\frac{x-2}{x^2-2x+4}=\frac{32}{x\left(x^4+4x^2+16\right)}\)

=>\(\frac{\left(x+2\right)\left(x^2-2x+4\right)+\left(x-2\right)\left(x^2+2x+4\right)}{\left(x^2-2x+4\right)\left(x^2+2x+4\right)}=\frac{32}{x\left(x^2-2x+4\right)\left(x^2+2x+4\right)}\)

=>\(\frac{x^3+8+x^3-8}{\left(x^2-2x+4\right)\cdot\left(x^2+2x+4\right)}=\frac{32}{x\left(x^2-2x+4\right)\left(x^2+2x+4\right)}\)

=>\(\frac{2x^3}{\left(x^2-2x+4\right)\left(x^2+2x+4\right)}=\frac{32}{x\left(x^2-2x+4\right)\left(x^2+2x+4\right)}\)

=>\(2x^4=32\)

=>\(x^4=16\)

=>x=2(nhận) hoặc x=-2(nhận)

a: ĐKXĐ: x∈R

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

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

=>|x-1|=(x-1)(x+1)(1)

TH1: x>=1

(1) sẽ trở thành (x-1)(x+1)=(x-1)

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

=>x(x-1)=0

=>x=1(nhận) hoặc x=0(loại)

TH2: x<1

(1) sẽ trở thành: (x-1)(x+1)=-(x-1)

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

=>x=1(loại) hoặc x=-2(nhận)

b: ĐKXĐ: x∈R

\(\sqrt{x^2+x+\frac14}=x\)

=>\(\begin{cases}x\ge0\\ x^2+x+\frac14=x^2\end{cases}\Rightarrow\begin{cases}x\ge0\\ x+\frac14=0\end{cases}\)

=>x∈∅

c: \(\sqrt{x^4-8x^2+16}=2-x\)

=>\(\sqrt{\left(x^2-4\right)^2}=2-x\)

=>\(\begin{cases}2-x\ge0\\ \left(x^2-4\right)^2=\left(2-x\right)^2\end{cases}\Rightarrow\begin{cases}x\le2\\ \left(x-2\right)^2\cdot\left\lbrack\left(x+2\right)^2-1\right\rbrack=0\end{cases}\)

=>x<=2 và \(\left(x-2\right)^2\cdot\left(x+1\right)\left(x+3\right)\) =0

=>x∈{2;-1;-3}

=>(2x-3)(2x+3)(x-4)-(2x-3)(x-4)(x+4)=0

=>(2x-3)(x-4)(2x+3-x-4)=0

=>(2x-3)(x-4)(x-1)=0

=>\(x\in\left\{1;4;\dfrac{3}{2}\right\}\)

khó quá

khó gì fan gao bạc

\(2\sqrt{2x+4}+4\sqrt{2-x}=\sqrt{9x^2+16}\)

\(\Leftrightarrow\left(2\sqrt{2x+4}+4\sqrt{2-x}\right)^2=\left(\sqrt{9x^2+16}\right)^2\)

\(\Leftrightarrow4\left(2x+4\right)+16\left(2-x\right)+16\sqrt{2x+4}\sqrt{2-x}=9x^2+16\)

\(\Leftrightarrow4.2\left(4-x^2\right)+16\sqrt{2\left(4-x^2\right)}=x^2+8x\)

Đặt \(\sqrt{2\left(4-x^2\right)}=a\)

\(\Rightarrow4a^2+16a=x^2+8x\)

\(\Leftrightarrow\left(2a-x\right)\left(2a+x+8\right)=0\)

Làm nốt