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

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

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

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

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

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

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

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

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

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

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

<=> \(4.4=x^2+8x+12\)

<=> \(x^2+8x-4=0\)

<=> ...

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

\(2x^3-2x^2+18x\Leftrightarrow2x\left(x^2-x+9\right)\)

\(2x^3-2x^2+18x=2x\left(x^2-x+1\right)\)

1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)

\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)

\(=x^2+4x+3-\dfrac{12}{2x+3}\)

c: =>\(\dfrac{2x-1}{\left(x+5\right)\left(x-1\right)}+\dfrac{x-2}{\left(x-1\right)\left(x-9\right)}=\dfrac{3x-12}{\left(x-9\right)\left(x+5\right)}\)

=>(2x-1)(x-9)+(x-2)(x+5)=(3x-12)(x-1)

=>2x^2-19x+9+x^2+3x-10=3x^2-15x+12

=>-16x-1=-15x+12

=>-x=13

=>x=-13

Điều kiện: \(2x-3\ge0\Leftrightarrow x\ge\dfrac{3}{2}\)

\(3\sqrt{2x-3}+2\sqrt{8x-12}=\sqrt{18x-27}+9\)

\(3\sqrt{2x-3}+2\sqrt{4\left(2x-3\right)}-\sqrt{9\left(2x-3\right)}=9\)

\(3\sqrt{2x-3}+4\sqrt{2x-3}-3\sqrt{2x-3}=9\)

\(4\sqrt{2x-3}=9\)

\(x\ge\dfrac{3}{2}\)\(\Rightarrow16\left(2x-3\right)=81\)

\(2x-3=\dfrac{81}{16}\Leftrightarrow x=\dfrac{\dfrac{81}{16}+3}{2}=\dfrac{129}{32}\)