\(\frac{1}{2x^2+10x+12}+\frac{1}{2x^2+14x+24}+\frac{1}{2x^2+18x+40}+\frac{1}{2x^2+22x+60}=\frac{1}{8}\)

<=> \(\frac{1}{2x^2+6x+4x+12}+\frac{1}{2x^2+6x+8x+24}+\frac{1}{2x^2+8x+10x+40}+\frac{1}{2x^2+12x+10x+60}=\frac{1}{8}\)

<=> \(\frac{1}{2x\left(x+3\right)+4\left(x+3\right)}+\frac{1}{2x\left(x+3\right)+8\left(x+3\right)}+\frac{1}{2x\left(x+4\right)+10\left(x+4\right)}+\frac{1}{2x\left(x+6\right)+10\left(x+6\right)}=\frac{1}{8}\)

<=> \(\frac{1}{\left(x+3\right)\left(2x+4\right)}+\frac{1}{\left(x+3\right)\left(2x+8\right)}+\frac{1}{\left(x+4\right)\left(2x+10\right)}+\frac{1}{\left(x+6\right)\left(2x+10\right)}=\frac{1}{8}\)

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

<=> \(\frac{1}{2}.\left[\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\right]=\frac{1}{8}\)

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

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

<=> \(\frac{4\left(x+6\right)-4\left(x+2\right)}{4\left(x+2\right)\left(x+6\right)}=\frac{\left(x+2\right)\left(x+6\right)}{4\left(x+2\right)\left(x+6\right)}\)

<=> \(4\left(x+6\right)-4\left(x+2\right)=\left(x+2\right)\left(x+6\right)\)

<=> \(4\left(x+6-x-2\right)=x^2+8x+12\)

<=> \(4.4=x^2+8x+12\)

<=> \(x^2+8x-4=0\)

<=> ...

Đến đây bạn tự giải tiếp. Mình bấm máy 570ES PLUS II thì ra nghiệm \(x\approx0,47\).

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

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

\(\Leftrightarrow\frac{4x^2+4x+1-4x^2+4x-1-8}{4x^2-1}=0\)

\(\Leftrightarrow\frac{8x-8}{4x^2-1}=0\)

\(\Rightarrow8x-8=0\)

\(\Rightarrow x=1\)

tick mình nha!

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

\(\Leftrightarrow\left(2x+1\right)^2-\left(2x-1\right)^2=9\)

\(\Leftrightarrow4x^2+4x+1-4x^2+4x+1=9\)

\(\Leftrightarrow8x=7\)

Vậy x=7/8

Bạn sửa lại đề dùm mình nha, sai đề hơi nhiều đó.

ĐKXĐ:\(x\ne0;2\)

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

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

=[x(x-2)/2(x²+4)-2x²/(4+x²)(2-x)][x(x-2)(x+1)/x³]

={[x(x-2)(2-x)-4x² ]/2(2-x)(4+x²)} .[x(x-2)(x+1)/x³ ]

=[-x(x²+4)/2(2-x)(4+x²)].[x(x-2)(x+1)/x³ ]

=-x.x(x-2)(x+1)/2(2-x)x³

=(x+1)/2x

\(a,\frac{1-2x}{8}=\frac{-4}{2\left(2x-1\right)}\)

\(\Rightarrow2\left(1-2x\right)\left(2x-1\right)=-32\)

\(\Rightarrow2\left(2x-1\right)\left(2x-1\right)=32\)

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

\(\Rightarrow\orbr{\begin{cases}2x-1=4\\2x-1=-4\end{cases}\Rightarrow\orbr{\begin{cases}2x=5\\2x=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}}\)

\(b,\frac{-2}{x-1}=\frac{1-x}{\frac{8}{25}}\)

\(\Leftrightarrow(x-1)(1-x)=-\frac{16}{25}\)

\(\Leftrightarrow-(x-1)^2=-\frac{16}{25}\)

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

\(\Leftrightarrow\orbr{\begin{cases}-x+1=-\frac{4}{5}\\-x+1=\frac{4}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{5}\\x=\frac{1}{5}\end{cases}}\)