\(\frac{\left(x-2\right)^2}{3}-\frac{2x-1}{4}=4-\frac{\left(2x-3\right)^2}{6}.\)

\(\Rightarrow\frac{4\left(x-2\right)^2}{12}-\frac{3\left(2x-1\right)^2}{12}=\frac{48}{12}-\frac{2\left(2x-3\right)^2}{12}\)

\(\Rightarrow4\left(x^2-4x+4\right)-3\left(4x^2-4x+1\right)=48-2\left(4x^2-12x+9\right)\)

\(\Rightarrow4x^2-16x+16-12x^2+12x-3=48-8x^2+24x-18\)

\(\Rightarrow-16x+12x+16-3=24x+48-18\)

\(\Rightarrow28x=-17\Leftrightarrow x=-\frac{17}{28}\)

nhung sao lai binh phuong len vay

\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)  ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)

\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)

\(\Leftrightarrow x-3=10x-15\)

\(\Leftrightarrow x-10x=3-15\)

\(\Leftrightarrow-9x=-12\)

\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)

KL :....

\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)   ĐKXĐ : \(x\ne0;2\)

\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)

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

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

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

KL ::

\(c,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)    ĐKXĐ : \(x\ne\pm2\)

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

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

\(\Leftrightarrow0x=0\)

KL : PT vô số nghiệm 

\(\hept{\begin{cases}\frac{2x-3y}{2y-5}=\frac{3x+1}{3y-4}\left(1\right)\\2\left(x-3\right)-3\left(y+2\right)=-16\left(2\right)\end{cases}}\)

Nhân chéo và chuyển vế phương trình (1) và nhân phân phối, chuyển vế phương trình (2), ta được:

\(\hept{\begin{cases}7x-11y=-17\\2x-3y=-4\end{cases}}\)

<=>\(\hept{\begin{cases}x=7\\y=6\end{cases}}\)

\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\) \(ĐK:x\ne-1;x\ne-3\)

\(\Leftrightarrow\frac{4x}{x^2+4x+3}-\frac{x^2+4x+3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)}{2\left(x+3\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)\left(x+3\right)}\right]\)

\(\Leftrightarrow\frac{4x-x^2-4x-3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)-x-3}{2\left(x+3\right)\left(x+1\right)}\right]\)

\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=6\left[\frac{2x+2-x-3}{2\left(x^2+4x+3\right)}\right]\)

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\left(loại\right)\end{cases}}\) 

Vậy x = 0 

\(ĐK:x\ne\frac{-1}{2};x\ne\frac{-3}{2}\)

\(\frac{3}{2x+1}=\frac{6}{2x+3}+\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow\frac{3}{2x+1}-\frac{6}{2x+3}=\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow\frac{3\left(2x+3\right)-6\left(2x+1\right)}{\left(2x+1\right)\left(2x+3\right)}=\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow\frac{6x+9-12x-6}{4x^2+8x+3}=\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow-6x+3=8\)

\(\Leftrightarrow x=-\frac{5}{6}\)

Vậy ... 

1. ĐKXĐ: ...

\(\Leftrightarrow tan\left(x+\frac{\pi}{3}\right)=\frac{1}{tan\left(2x-\frac{\pi}{4}\right)}\)

\(\Leftrightarrow tan\left(x+\frac{\pi}{3}\right)=cot\left(2x-\frac{\pi}{4}\right)\)

\(\Leftrightarrow tan\left(x+\frac{\pi}{3}\right)=tan\left(\frac{3\pi}{4}-2x\right)\)

\(\Leftrightarrow x+\frac{\pi}{3}=\frac{3\pi}{4}-2x+k\pi\)

\(\Rightarrow x=\frac{5\pi}{36}+\frac{k\pi}{3}\)

2.

ĐKXĐ: ...

\(\Leftrightarrow tan\left(x+1\right)=\frac{1}{cot\left(2x+3\right)}\)

\(\Leftrightarrow tan\left(x+1\right)=tan\left(2x+3\right)\)

\(\Leftrightarrow2x+3=x+1+k\pi\)

\(\Rightarrow x=-2+k\pi\)

3.

ĐKXĐ: ...

\(\Leftrightarrow tan^22x+\left(\frac{1}{cos^22x}+1\right)=8\)

\(\Leftrightarrow tan^22x+tan^22x=8\)

\(\Leftrightarrow tan^22x=4\)

\(\Rightarrow\left[{}\begin{matrix}tan2x=2\\tan2x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=arctan\left(2\right)+k180^0\\2x=-arctan\left(2\right)+k180^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}arctan\left(2\right)+k90^0\\x=-\frac{1}{2}arctan\left(2\right)+k90^0\end{matrix}\right.\)

Nghiệm trên nhận các giá trị \(k=\left\{0;1;2;3\right\}\) ; nghiệm dưới nhận các giá trị \(k=\left\{1;2;3;4\right\}\)

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

<=>\(\frac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\frac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\frac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\frac{1}{\left(x^2+2\right)\left(x^2+1\right)}=-1\)

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

<=>\(\frac{1}{x^2+1}-\frac{1}{x^2+5}=-1\)

<=>(x²+5)-(x²+1)=-(x²+1)(x²+5)

<=>4=-x⁴-6x²-5

<=>x⁴+6x²+9=0

<=>(x²+3)²=0

<=>x²+3=0

Do x²>0

=>x²+3>0 nên PT vô nghiệm