\(\left(2x-5\right)\left(x-3\right)+\left(2x-5\right)^2=0\)

\(\Rightarrow\left(2x-5\right)\left(x-3+2x-5\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(3x-8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-5=0\\3x-8=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{8}{3}\end{cases}}\)

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

\(\Leftrightarrow\frac{18x-30+8x-12}{24}=\frac{x-3}{3}\)

\(\Leftrightarrow\frac{26x-42}{24}=\frac{x-3}{3}\)

\(\Leftrightarrow78x-126=24x-72\)

Chuyển vế các kiểu

a)\(ĐKXĐ:x\ne\pm1\)

\(\frac{x+1}{x-1}+\frac{x-1}{x+1}=\frac{16}{x^2-1}\)

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

\(\Rightarrow\left(x+1\right)^2+\left(x-1\right)^2=16\)

\(\Rightarrow\left(x^2+2x+1\right)+\left(x^2-2x+1\right)=16\)

\(\Rightarrow2x^2+2=16\Rightarrow x^2+1=8\Rightarrow x^2=7\)

\(\Rightarrow x=\pm\sqrt{7}\)

c)\(ĐKXĐ:x\ne-2\)

 \(\frac{12}{8+x^3}=1+\frac{1}{x+2}\)

\(\Rightarrow\frac{12}{8+x^3}=\frac{x+3}{x+2}\)

\(\Rightarrow\frac{12}{8+x^3}=\frac{\left(x+3\right)\left(x^2-2x+4\right)}{x^3+8}\)

\(\Rightarrow\left(x+3\right)\left(x^2-2x+4\right)=12\)

\(\Rightarrow x^3-5x^2+10x-12=12\)

\(\Rightarrow x^3-5x^2+10x=0\)

\(\Rightarrow x\left(x^2-5x+10\right)=0\)

Vì \(\left(x^2-5x+10\right)>0\)nên x = 0

Vậy x = 0

Từ đề bài ta có \(f\left(x\right)=A\left(x\right).\left(x-3\right)+2\Rightarrow f\left(3\right)=2\)

\(f\left(x\right)=B\left(x\right).\left(x+4\right)+9\Rightarrow f\left(-4\right)=9\)

\(f\left(x\right)=\left(x^2+3\right).\left(x^2+x-12\right)+\left(x^2+3\right).\left(ax+b\right)=\left(x^2+3\right).\left(x-3\right).\left(x+4\right)+\left(x^2+3\right).\left(ax+b\right)\left(1\right)\)Từ (1).Ta có \(f\left(3\right)=\left(3^2+3\right)\left(3a+b\right)=36a+12b\Rightarrow36a+12b=2\)

\(f\left(-4\right)=\left(\left(-4\right)^2+3\right)\left(-4a+b\right)=-76a+19b\Rightarrow-76a+19b=9\)

Giải hệ phương trình ẩn a,b ta tìm được a,b.Từ đó thế vào (1).Ta tìm được f(x)

Phương trình đầu bài tương đương với 
\(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)\(\Leftrightarrow\frac{x+43+57}{57}+\frac{x+46+54}{54}=\frac{x+49+51}{51}+\frac{x+52+48}{48}\)\(\Leftrightarrow\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)

\(\Leftrightarrow\orbr{\begin{cases}x+100=0\\\frac{1}{57}+\frac{1}{54}=\frac{1}{51}+\frac{1}{48}\left(sai\right)\end{cases}\Leftrightarrow x+100=0\Leftrightarrow x=-100}\)

Vậy phương trình có nghiệm duy nhất là x=-100

<=> \(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)

<=> \(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)

<=> \(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)

vi \(\frac{1}{57}< \frac{1}{51};\frac{1}{54}< \frac{1}{48}\Rightarrow\frac{1}{57}-\frac{1}{51}+\frac{1}{54}-\frac{1}{48}< 0\)

=> x+100=0 => x= -100

vay pt co nghiem \(x=-100\)

a) \(ĐKXĐ:\hept{\begin{cases}x\ne0;x\ne2\\x\ne-1\end{cases}}\)

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

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

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

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

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

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

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

\(\Leftrightarrow Q=\frac{x-1}{x+1}\)

b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(ktm\right)\\x=-\frac{1}{2}\left(tm\right)\end{cases}}\)

Thay \(x=-\frac{1}{2}\)vào Q, ta được :

\(Q=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}\)

\(\Leftrightarrow Q=\frac{-\frac{3}{2}}{\frac{1}{2}}\)

\(\Leftrightarrow Q=-3\)

c) Để \(Q\inℤ\)

\(\Leftrightarrow x-1⋮x+1\)

\(\Leftrightarrow x+1-2⋮x+1\)

\(\Leftrightarrow2⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Leftrightarrow x\in\left\{-2;0;-3;1\right\}\)

Vậy để \(Q\inℤ\Leftrightarrow x\in\left\{-2;0;-3;1\right\}\)