\(\frac{12x+1}{6x-2}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(9x^2-1\right)}\)đkxđ \(x\ne\pm\frac{1}{3}\)

\(\Leftrightarrow72x^2+6x+24x+2-108x^2+60x+36x-20-108x+36x^2+9=0\)

\(\Leftrightarrow18x-9=0\)

\(\Leftrightarrow18x=9\)

\(\Leftrightarrow x=\frac{1}{2}\left(tm\right)\)

a)\(-ĐKXĐ:\hept{\begin{cases}x-14\ne0;x-13\ne0\\x-9\ne0\\x-11\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne14;x\ne13\\x\ne9\\x\ne11\end{cases}}\)

 - Ta có : \(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}\)

\(\Leftrightarrow\frac{2}{x-14}-\frac{5}{x-13}-\frac{2}{x-9}+\frac{5}{x-11}=0\)

\(\Leftrightarrow\left(\frac{2}{x-14}-\frac{2}{x-9}\right)-\left(\frac{5}{x-13}-\frac{5}{x-11}\right)=0\)

\(\Leftrightarrow2\left(\frac{1}{x-14}-\frac{1}{x-9}\right)-5\left(\frac{1}{x-13}-\frac{1}{x-11}\right)=0\)\(\Leftrightarrow2.\frac{\left(x-9\right)-\left(x-14\right)}{\left(x-9\right)\left(x-14\right)}-5.\frac{\left(x-11\right)-\left(x-13\right)}{\left(x-11\right)\left(x-13\right)}=0\)

\(\Leftrightarrow2.\frac{5}{\left(x-9\right)\left(x-14\right)}-5.\frac{2}{\left(x-11\right)\left(x-13\right)}=0\)

\(\Leftrightarrow\frac{10}{\left(x-9\right)\left(x-14\right)}-\frac{10}{\left(x-11\right)\left(x-13\right)}=0\)

\(\Leftrightarrow10\left[\frac{1}{\left(x-9\right)\left(x-14\right)}-\frac{1}{\left(x-11\right)\left(x-13\right)}\right]=0\)

\(\Leftrightarrow\frac{\left(x-11\right)\left(x-13\right)}{\left(x-9\right)\left(x-14\right)\left(x-11\right)\left(x-13\right)}-\frac{\left(x-9\right)\left(x-14\right)}{\left(x-9\right)\left(x-14\right)\left(x-11\right)\left(x-13\right)}=\) \(0\)

\(\Leftrightarrow\left(x-11\right)\left(x-13\right)-\left(x-9\right)\left(x-14\right)=0\)

\(\Leftrightarrow x^2-24x+143-x^2+23x-126=0\)

\(\Leftrightarrow-x+17=0\Leftrightarrow-x=-17\Leftrightarrow x=17\)

Vậy pt có tập nghiệm S = { 17 }

P/s: Mk làm hơi lòng vòng, bn thông cảm nhé !

a) ĐKXĐ: \(x\notin\pm\frac{1}{3}\)

Ta có: \(\frac{12x+1}{6x-2}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(9x^2-1\right)}\)

\(\Leftrightarrow\frac{12x+1}{2\left(3x-1\right)}-\frac{9x-5}{3x+1}=\frac{9\left(12x-4x^2-1\right)}{4\left(9x^2-1\right)}\)

\(\Leftrightarrow\frac{2\left(12x+1\right)\left(3x+1\right)}{4\left(3x-1\right)\left(3x+1\right)}-\frac{4\left(9x-5\right)\left(3x-1\right)}{4\left(3x+1\right)\left(3x-1\right)}=\frac{9\left(12x-4x^2-1\right)}{4\left(3x+1\right)\left(3x-1\right)}\)

\(\Leftrightarrow72x^2+30x+2-\left(108x^2-96x+20\right)=108x-36x^2-9\)

\(\Leftrightarrow72x^2+30x+2-108x^2+96x-20-108x+36x^2+9=0\)

\(\Leftrightarrow18x-9=0\)

\(\Leftrightarrow9\left(2x-1\right)=0\)

mà 9≠0

nên 2x-1=0

⇔2x=1

hay \(x=\frac{1}{2}\)(tm)

Vậy: \(x=\frac{1}{2}\)

b)ĐKXĐ: x≠0

Ta có: \(x+\frac{1}{x}=x^2+\frac{1}{x^2}\)

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

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

\(\Leftrightarrow x^3+x-x^4-1=0\)

\(\Leftrightarrow x^3\left(1-x\right)+\left(x-1\right)=0\)

\(\Leftrightarrow x^3\left(1-x\right)-\left(1-x\right)=0\)

\(\Leftrightarrow\left(1-x\right)\left(x^3-1\right)=0\)

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

\(\Leftrightarrow-\left(x-1\right)^2\cdot\left(x^2+x+1\right)=0\)(1)

Ta có: \(x^2+x+1=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)(2)

Từ (1) và (2) suy ra x-1=0

hay x=1(tm)

Vậy: x=1

c) ĐKXĐ: x≠0

Ta có: \(\frac{1}{x}+2=\left(\frac{1}{x}+2\right)\left(x^2+2\right)\)

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

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

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

Ta có: 1≠0

x≠0

Do đó: \(\frac{1}{x}\ne0\)

\(\Leftrightarrow\frac{1}{x}+2\ne0\)(4)

Từ (3) và (4) suy ra x=0(ktm)

Vậy: x∈∅

d) ĐKXĐ: x≠0

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

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

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

\(\Leftrightarrow2x\cdot\left(2+\frac{2}{x}\right)=0\)

\(\Leftrightarrow4x\left(1+\frac{1}{x}\right)=0\)

mà 4≠0

và x≠0

nên \(1+\frac{1}{x}=0\)

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

hay x=-1(tm)

Vậy: x=-1

c/ đk: x khác 1; x khác -3

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

\(\Rightarrow\left(3x+1\right)\left(x+3\right)+\left(2x+5\right)\left(x-1\right)+4=x^2+2x-3\)

\(\Leftrightarrow3x^2+10x+3+2x^2+3x-5+4=x^2+2x-3\)

\(\Leftrightarrow4x^2+11x+5=0\)

\(\Leftrightarrow\left(4x^2+2\cdot2x\cdot\dfrac{11}{4}+\dfrac{121}{16}\right)-\dfrac{41}{16}=0\)

\(\Leftrightarrow\left(2x+\dfrac{11}{4}\right)^2=\dfrac{41}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{11}{4}=\dfrac{\sqrt{41}}{4}\\2x+\dfrac{11}{4}=-\dfrac{\sqrt{41}}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{41}}{8}\\x=\dfrac{-11-\sqrt{41}}{8}\end{matrix}\right.\)

Vậy.........

d/ \(\dfrac{12x+1}{6x-2}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(9x^2-1\right)}\)

đk: \(x\ne\pm\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{12x+1}{2\left(3x-1\right)}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(3x-1\right)\left(3x+1\right)}\)

\(\Rightarrow2\left(12x+1\right)\left(3x+1\right)-4\left(9x-5\right)\left(3x-1\right)=108x-36x^2-9\)

\(\Leftrightarrow72x^2+24x+6x+2-108x^2+36x-60x-20-108x+36x^2+9=0\)

\(\Leftrightarrow-102x-9=0\)

\(\Leftrightarrow-102x=9\Leftrightarrow x=-\dfrac{3}{34}\)(TM)

Vậy.........

a/ \(\left(x+1\right)^2\left(x+2\right)+\left(x+1\right)^2\left(x-2\right)=-24\)

\(\Leftrightarrow\left(x+1\right)^2\left(x+2+x-2\right)=-24\)

\(\Leftrightarrow2x\left(x^2+2x+1\right)=-24\)

\(\Leftrightarrow2x^3+4x^2+2x+24=0\)

\(\Leftrightarrow2x^3-2x^2+8x+6x^2-6x+24=0\)

\(\Leftrightarrow x\left(2x^2-2x+8\right)+3\left(2x^2-2x+8\right)=0\)

\(\Leftrightarrow\left(2x^2-2x+8\right)\left(x+3\right)=0\)

\(\Leftrightarrow2\left(x^2-x+4\right)\left(x+3\right)=0\)

Ta thấy: \(x^2-x+4=\left(x^2-2x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{15}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0\)

=> x+ 3 = 0 <=> x= -3

Vậy......

b/ \(2x^3+3x^2+6x+5=0\)

\(\Leftrightarrow2x^3+x^2+5x+2x^2+x+5=0\)

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

\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)

Ta thấy: \(2x^2+x+5=\left(\sqrt{2}x+2\cdot\sqrt{2}x\cdot\dfrac{\sqrt{2}}{4}+\dfrac{1}{8}\right)+\dfrac{39}{8}=\left(\sqrt{2}x+\dfrac{\sqrt{2}}{4}\right)^2+\dfrac{39}{8}>0\)

=> x + 1 = 0 <=> x = -1

Vậy....

bt2.

A=[2(4x^2+4x+5)-2]/(4x^2+4x+5)

=2-2/[(4x+1)^2+4]

A>=2-2/4=3/2

khi x=-1/4