a)

\(\frac{7x-3}{x-1}=\frac{2}{3}\\ \Leftrightarrow\frac{21x-9}{3\cdot\left(x-1\right)}-\frac{2x-2}{3\cdot\left(x-1\right)}=0\\ \Leftrightarrow21x-9-2x+2=0\\ \Leftrightarrow19x-7=0\\ \Rightarrow x=\frac{7}{19}\)

Vậy \(x=\left\{\frac{7}{19}\right\}\) là nghiệm của phương trình.

b)

\(\frac{1}{x-2}+3=\frac{3-x}{x-2}\\ \Leftrightarrow\frac{1}{x-2}+\frac{3x-6}{x-2}-\frac{3-x}{x-2}=0\\ \Leftrightarrow1+3x-6-3+x=0\\ \Leftrightarrow4x-8=0\\ \Rightarrow x=\frac{8}{4}=2\)

Mà \(ĐKXĐ:x\ne2\\ \Rightarrow x\in\varnothing\)

Hay phương trình vô nghiệm.

c)

\(\frac{8-x}{x-7}-8=\frac{1}{x-7}\\ \Leftrightarrow\frac{8-x}{x-7}-\frac{8x-56}{x-7}-\frac{1}{x-7}=0\\ \Leftrightarrow8-x-8x+56-1=0\\ \Leftrightarrow63-9x=0\\ \Rightarrow x=\frac{63}{9}=7\)

Mà \(ĐKXĐ:x\ne7\\ \Rightarrow x\in\varnothing\)

Hay phương trình vô nghiệm.

d)

\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\\ \Leftrightarrow\frac{\left(x+5\right)^2}{x^2-25}-\frac{\left(x-5\right)^2}{x^2-25}-\frac{20}{x^2-25}=0\\ \Leftrightarrow\left(x+5\right)^2-\left(x-5\right)^2-20=0\\ \Leftrightarrow x^2+10x+25-x^2+10x-25-20=0\\ \Leftrightarrow20x-20=0\\ \Rightarrow x=1\)

Vậy \(x=1\) là nghiệm của phương trình.

e)

\(\frac{x}{2\cdot\left(x-3\right)}+\frac{x}{2\cdot\left(x+1\right)}=\frac{2x}{\left(x+1\right)\cdot\left(x-3\right)}\\ \Leftrightarrow\frac{x^2+x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}+\frac{x^2-3x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}-\frac{4x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}=0\\ \Leftrightarrow x^2+x+x^2-3x-4x=0\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\cdot\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Mà \(ĐKXĐ:x\ne\left\{-1;3\right\}\\ \Rightarrow x=0\)

Vật \(x=0\) là nghiệm của phương trình.

f)

\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\\ \Leftrightarrow\frac{12x+120}{12x\cdot\left(x+10\right)}+\frac{12x}{12x\cdot\left(x+10\right)}-\frac{x^2+10x}{12x\cdot\left(x+10\right)}=0\\ \Leftrightarrow12x+120+12x-x^2-10x=0\\ \Leftrightarrow14x+120-x^2=0\\ \Leftrightarrow x^2-14x-120=0\\ \Leftrightarrow x^2+6x-20x-120=0\\ \Leftrightarrow x\cdot\left(x+6\right)-20\cdot\left(x+6\right)=0\\ \Leftrightarrow\left(x-20\right)\cdot\left(x+6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-20=0\\x+6=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=20\\x=-6\end{matrix}\right.\)

Vậy \(x=\left\{20;-6\right\}\) là ngiệm của phương trình.

a/ ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)

Vậy..

b/ Ta có :

\(C=\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right).\frac{x^2-1}{5}\)

\(=\left(\frac{2x+1}{x-1}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x+1}\right).\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\left(\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\frac{2x^2+2x+x+1+8-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\frac{x^2+5x+8}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\frac{\left(x+\frac{5}{2}\right)^2+\frac{7}{4}}{5}\)

Vậy...

c/ Với mọi x ta có :

\(\left\{{}\begin{matrix}\left(x+\frac{5}{2}\right)^2+\frac{7}{4}>0\\5>0\end{matrix}\right.\)

\(\Leftrightarrow\frac{\left(x+\frac{5}{2}\right)^2+\frac{7}{4}}{5}>0\)

\(\Leftrightarrow C>0\left(đpcm\right)\)

I don't now

...............

.................

c) \(\left|2x-3\right|=4\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=4\\2x-3=-4\end{cases}}\)

\(TH:2x-3=4\)

\(\Leftrightarrow2x=4+3\)

\(\Leftrightarrow2x=7\)

\(\Leftrightarrow x=\frac{7}{2}\)

\(TH:2x-3=-4\)

\(\Leftrightarrow2x=-4+3\)

\(\Leftrightarrow2x=-1\)

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

Vậy \(x\in\left\{\frac{7}{2};\frac{-1}{2}\right\}\)

e) \(\frac{x-1}{x-3}>1\)

\(ĐKXĐ:x\ne3\)

\(\Leftrightarrow\frac{x-3+2}{x-3}>1\)

\(\Leftrightarrow\frac{x-3}{x-3}+\frac{2}{x-3}>1\)

\(\Leftrightarrow1+\frac{2}{x-3}>1\)

\(\Leftrightarrow\frac{2}{x-3}>0\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\)

\(x^3-6x^2+5x+12>0\\ < =>\left(x^3-5x-x+5x\right)+12>0\\ < =>\left[\left(x^3-x\right)-\left(5x-5x\right)\right]+12>0\\ < =>x^2+12>0\\ < =>x^2>-12\\ =>x\in R\\ BPTcóvôsốnghiem\)

a, Đẻ \(P< 1\)thì : 

\(P=\left(\frac{x}{x+2}+\frac{x}{x-2}-\frac{2}{x^2-4}\right).\frac{x-2}{2x+2}< 1\)

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

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

\(=\left(\frac{2x^4-10x^2+8}{x^4-8x^2+16}\right).\frac{x-2}{2x+2}=\left(2x^4-10x^2+8\right)\left(2x+2\right)=\left(x-2\right)\left(x^4-8x^2+16\right)\)

PT tương đương vs : \(\left(2x^4-10x^2+8\right)\left(2x+2\right)-\left(x-2\right)\left(x^4-8x^2+16\right)< 1\)

Khi đó pt trở thành : \(3x^5+6x^4-12x^3-36x^2+48< 1\)

Chắc vại đó ==