a) \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)

<=> \(\frac{x}{4}+\frac{5}{4}-\frac{2x}{3}+1=\frac{6x}{8}-\frac{1}{8}+\frac{2x}{12}-\frac{1}{12}\)

<=> \(-\frac{4}{3}x=-\frac{59}{24}\)

<=> \(x=\frac{59}{32}\)

Vậy S = { 59/32}

b) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

<=> \(\frac{x^2+14x+40}{12}-\frac{-x^2-2x+8}{4}=\frac{x^2+8x-20}{3}\)

<=> \(\left(\frac{x^2}{12}+\frac{x^2}{4}-\frac{x^2}{3}\right)+\left(\frac{14}{12}x+\frac{2}{4}x-\frac{8}{3}x\right)=-\frac{20}{8}+\frac{8}{4}-\frac{40}{12}\)

<=> \(-x=-8\)

<=> x = 8 

Vậy S = { 8 }

lm trên cymath.com

ĐK: x khác 0

Đặt: \(\frac{x^4}{2x^2+1}=t>0\Rightarrow\frac{2x^2+1}{x^4}=\frac{1}{t}\)

Ta có phương trình: \(t+\frac{1}{t}=2\Leftrightarrow t^2-2t+1=0\Leftrightarrow\left(t-1\right)^2=0\Leftrightarrow t=1\)

Với t = 1 ta có: \(\frac{x^4}{2x^2+1}=1\)<=> \(x^4-2x^2-1=0\Leftrightarrow\orbr{\begin{cases}x^2=1+\sqrt{2}\\x^2=1-\sqrt{2}\left(loai\right)\end{cases}}\)

khi đó: \(x=\pm\sqrt{1+\sqrt{2}}\)tm 

Vậy....

Một cách làm khác :

\(ĐKXĐ:x\ne0\)

Với điều kiện trên thì : \(\hept{\begin{cases}x^4>0\\2x^2+1>0\end{cases}}\)

Do đó áp dụng BĐT AM - GM cho 2 số dương ta có :

\(\frac{x^4}{2x^2+1}+\frac{2x^2+1}{x^4}\ge2.\sqrt{\frac{x^4}{2x^2+1}\cdot\frac{2x^2+1}{x^4}}=2\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{x^4}{2x^2+1}=\frac{2x^2+1}{x^4}\)

\(\Leftrightarrow x^4=2x^2+1\)

\(\Leftrightarrow x^4-2x^2+1=2\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-1=\sqrt{2}\\x^2-1=-\sqrt{2}\end{cases}}\)

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

\(\Leftrightarrow x=\pm\sqrt{\sqrt{2}+1}\)( Thỏa mãn ĐKXĐ )

Vậy pt đã cho có tập nghiệm \(S=\left\{\pm\sqrt{\sqrt{2}+1}\right\}\)

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

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

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

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

Xét (1), đặt \(a=x^2-2x+3\) pt trở thành:

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

\(\Leftrightarrow\frac{a-3}{a^2-1}=\frac{1}{a}\Leftrightarrow a^2-3a=a^2-1\Leftrightarrow3a=1\Rightarrow a=\frac{1}{3}\)

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

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

Vậy \(x=1\)

Đkxđ: \(\left\{{}\begin{matrix}x\ne0\\x\ne2\\x\ne-2\end{matrix}\right.\)

\(\frac{12}{x^2-4}+\frac{1}{2x-x^2}=\frac{4+x}{x\left(x+2\right)}\)\(\Leftrightarrow \dfrac{12x}{x(x-2)(x+2)}+\dfrac{(x+2)}{x(2-x)(x+2}-\dfrac{4x(x-2)}{x(x+2)(x-2)}=0\)

\(\Leftrightarrow \dfrac{12x-(x+2)-4x(x-2)}{x(x-2)(x+2)}=0\)

\(\Leftrightarrow -4x^2+11x=0\)\(\Leftrightarrow x\left(-4x+11\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\-4x+11=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{11}{4}\end{matrix}\right.\)

KL:........................

\(x\ne\left\{-4;-3;-2;-1\right\}\)

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

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

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

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

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

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

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

\(\Leftrightarrow\left(x+2\right)\left(x+4\right)+\left(x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow2x^2+10x+11=0\Rightarrow x=\frac{-5\pm\sqrt{3}}{2}\)

1) Ta có: x-4=2x+4

\(\Leftrightarrow x-4-2x-4=0\)

\(\Leftrightarrow-x-8=0\)

\(\Leftrightarrow-x=8\)

hay x=-8

Vậy: S={8}

2) Ta có: \(\frac{2x-1}{2}-\frac{x}{3}=x-\frac{x}{6}\)

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

\(\Leftrightarrow3\left(2x-1\right)-2x-6x+x=0\)

\(\Leftrightarrow6x-3-2x-6x+x=0\)

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

\(\Leftrightarrow-x=3\)

hay x=-3

Vậy: S={-3}

3) ĐKXĐ: \(x\notin\left\{\frac{-1}{2};3\right\}\)

Ta có: \(\frac{x+3}{2x+1}-\frac{x}{x-3}=\frac{3x^2+x+9}{\left(2x+1\right)\left(x-3\right)}\)

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

Suy ra: \(x^2-9-\left(2x^2+x\right)-3x^2-x-9=0\)

\(\Leftrightarrow-2x^2-x-18-2x^2-x=0\)

\(\Leftrightarrow-4x^2-2x-18=0\)

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

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

\(\Leftrightarrow x^2+2\cdot x\cdot\frac{1}{4}+\frac{1}{16}+\frac{59}{80}=0\)

\(\Leftrightarrow\left(x+\frac{1}{4}\right)^2+\frac{59}{80}=0\)(vô lý)

Vậy: S=\(\varnothing\)

4) Ta có: \(\frac{2x}{3}+\frac{2x-1}{6}=4-\frac{x}{3}\)

\(\Leftrightarrow\frac{4x}{6}+\frac{2x-1}{6}=\frac{24}{6}-\frac{2x}{6}\)

\(\Leftrightarrow4x+2x-1=24-2x\)

\(\Leftrightarrow6x-1-24+2x=0\)

\(\Leftrightarrow8x-25=0\)

\(\Leftrightarrow8x=25\)

hay \(x=\frac{25}{8}\)

Vậy: \(S=\left\{\frac{25}{8}\right\}\)