\(4x^2-5x-4\sqrt{x-1}-2=0\left(x\ge1\right)\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x\in\varnothing\end{matrix}\right.\)

Vậy với x = 2 thì thỏa mãn pt 

câu a bài 1:(2x+1)(3x-2)=(5x-8)(2x+1)

<=>(2x+1)(3x-2)-(5x-8)(2x+1)=0

<=>(2x+1)(3x-2-5x+8)=0

<=>(2x+1)(6-2x)=0

bước sau tự làm nốt nha !

câu b:gợi ý: tách 4x^2-1thành (2x-1)(2x+1) rồi làm như câu a

Bài 2: 

a: \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)

\(\Leftrightarrow x-3=5\left(2x-3\right)=10x-15\)

=>-9x=-12

hay x=4/3

b: \(\Leftrightarrow x\left(x+2\right)-x+2=2\)

=>x²+2x-x+2=2

=>x²+x=0

=>x=0(loại) hoặc x=-1(nhận)

c: \(\dfrac{x+1}{x-2}+\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow x^2+3x+2+x^2-3x+2=2x^2+4\)

=>4=4(luôn đúng)

Vậy: S={x|x<>2; x<>-2}

tìm ra đáp án chưa

Đc rồi chỉ mình với

Bài 3: 

a: \(\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)

\(\Leftrightarrow x^2+2x-15=x^2-1-8=x^2-9\)

=>2x=6

hay x=3(loại)

b: \(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

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

=>2x=3

hay x=3/2

\(\text{a) }\left(x^2+x\right)^2+4\left(x^2+x\right)=12\\ \Leftrightarrow\text{Đặt }x^2+x=y\\ \Leftrightarrow y^2+4y=12\\ \Leftrightarrow y^2+6y-2y-12=0\\ \Leftrightarrow\left(y^2+6y\right)-\left(2y+12\right)=0\\ \Leftrightarrow y\left(y+6\right)-2\left(y+6\right)=0\\ \Leftrightarrow\left(y+6\right)\left(y-2\right)=0\\ \Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\\ \Leftrightarrow\left(x^2+x+\dfrac{1}{4}+\dfrac{23}{4}\right)\left(x^2+2x-x-2\right)=0\\ \Leftrightarrow\left[\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{23}{4}\right]\left[\left(x^2+2x\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\right]\left[x\left(x+2\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\left(Vì\text{ }\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ne0\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\\ \text{Vậy }S=\left\{1;-2\right\}\\ \)

\(\text{b) }6x^4-5x^3-38x^2-5x+6=0\\ \Leftrightarrow x^2\left(6x^2-5x-38-\dfrac{5}{x}+\dfrac{6}{x^2}\right)=0\\ \Leftrightarrow x^2\left[\left(6x^2+12+\dfrac{6}{x^2}\right)-\left(5x+\dfrac{5}{x}\right)-50\right]=0\\ \Leftrightarrow x^2\left[6\left(x^2+2+\dfrac{1}{x^2}\right)-5\left(x+\dfrac{1}{x}\right)-50\right]=0\\ \Leftrightarrow x^2\left[6\left(x+\dfrac{1}{x}\right)^2-5\left(x+\dfrac{1}{x}\right)-50\right]=0\\ \text{Đặt }x+\dfrac{1}{x}=y\\ \Leftrightarrow x^2\left(6y^2-5y-50\right)=0\\ \Leftrightarrow x^2\left(6y^2-20y+15y-50\right)=0\\ \Leftrightarrow x^2\left[\left(6y^2-20y\right)+\left(15y-50\right)\right]=0\\ \Leftrightarrow x^2\left[2y\left(3y-10\right)+5\left(3y-10\right)\right]=0\\ \Leftrightarrow x^2\left(2y+5\right)\left(3y-10\right)=0\\ \Leftrightarrow x^2\left(2x+\dfrac{2}{x}+5\right)\left(3x+\dfrac{3}{x}-10\right)=0\\ \Leftrightarrow\left(2x^2+2+5x\right)\left(3x^2+3-10x\right)=0\\ \Leftrightarrow\left(2x^2+4x+x+2\right)\left(3x^2-9x-x+3\right)=0\\ \Leftrightarrow\left[\left(2x^2+4x\right)+\left(x+2\right)\right]\left[\left(3x^2-9x\right)-\left(x-3\right)\right]=0\\ \Leftrightarrow\left[2x\left(x+2\right)+\left(x+2\right)\right]\left[3x\left(x-3\right)-\left(x-3\right)\right]=0\\ \Leftrightarrow\left(2x+1\right)\left(x+2\right)\left(3x-1\right)\left(x-3\right)=0\\ \)

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

ĐKXĐ: ...

\(4x^2+\frac{1}{x^2}-4\left(2x+\frac{1}{x}\right)+7=0\)

Đặt \(2x+\frac{1}{x}=a\Rightarrow a^2=4x^2+\frac{1}{x^2}+4\Rightarrow4x^2+\frac{1}{x^2}=a^2-4\)

\(a^2-4-4a+7=0\)

\(\Leftrightarrow a^2-4a+3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=3\end{matrix}\right.\)

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

\(\frac{2}{x^2+1}+\frac{4}{x^2+3}+\frac{6}{x^2+5}=3+\frac{x^2-1}{x^2+6}\)

\(\Leftrightarrow\frac{x^2-1}{x^2+6}+1-\frac{2}{x^2+1}+1-\frac{4}{x^2+3}+1-\frac{6}{x^2+5}=0\)

\(\Leftrightarrow\frac{x^2-1}{x^2+6}+\frac{x^2-1}{x^2+1}+\frac{x^2-1}{x^2+3}+\frac{x^2-1}{x^2+5}=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(\frac{1}{x^2+6}+\frac{1}{x^2+1}+\frac{1}{x^2+3}+\frac{1}{x^2+5}\right)=0\)

\(\Rightarrow x=\pm1\)