a: \(\Leftrightarrow\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+...+\dfrac{1}{49}-\dfrac{1}{57}\right)+2x-2=\dfrac{2}{3}x+\dfrac{7}{3}+\dfrac{5}{4}x-2\)

\(\Leftrightarrow\dfrac{56}{57}+2x-2=\dfrac{23}{12}x+\dfrac{1}{3}\)

=>1/12x=77/57

=>x=308/19

b: =>(x^2-4)(x^2-10)=72

=>x^4-14x^2+40-72=0

=>x^4-14x^2-32=0

=>(x^2-16)(x^2+2)=0

=>x^2-16=0

=>x^2=16

=>x=4 hoặc x=-4

\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}+2\right)+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}+2\right)=\left(x-5\right)^2-5\)

\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}\right)+20+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)^2-10\left(x^2+\dfrac{1}{x^2}\right)=\left(x-5\right)^2-5\)

\(\Leftrightarrow\left(x-5\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)

\(\dfrac{1}{x+3}+\dfrac{8}{\left(x+1\right)\left(x-3\right)}=\dfrac{2x}{x^2-2x-3}\)

* x² - 2x - 3 = x²- 3x + x - 3 = x(x-3 ) + ( x - 3) = ( x - 3 ) (  x + 1 )

\(\Leftrightarrow\dfrac{1}{x+3}+\dfrac{8}{\left(x+1\right)\left(x-3\right)}=\dfrac{2x}{\left(x-3\right)\left(x+1\right)}\left(ĐKXĐ:x\ne\pm3;x\ne-1\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)+8\left(x+3\right)=2x\left(x+3\right)\)

\(\Leftrightarrow x^2-2x+1+8x+24=2x^2+6x\)

\(\Leftrightarrow-x^2+25=0\)

\(\Leftrightarrow x^2-25=0\Leftrightarrow\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

Vậy \(S=\left\{-5;5\right\}\)

\(ĐK:x\ne2;x\ne-3\\ PT\Leftrightarrow\left(x-2\right)\left(x+3\right)+2\left(x+3\right)=10\left(x-2\right)+50\\ \Leftrightarrow x^2+x-6+2x+6=10x-20+50\\ \Leftrightarrow x^2-13x-30=0\\ \Leftrightarrow x^2-15x+2x-30=0\\ \Leftrightarrow\left(x-15\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=15\\x=-2\end{matrix}\right.\left(tm\right)\)

a: \(\Leftrightarrow x^2+x-6+2x-6=10x-20+50\)

\(\Leftrightarrow x^2+3x-12-10x-30=0\)

\(\Leftrightarrow x^2-7x-42=0\)

\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot\left(-42\right)=217>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{217}}{2}\\x_2=\dfrac{7+\sqrt{217}}{2}\end{matrix}\right.\)

b: \(\Leftrightarrow x^2-3x+5=-x^2+4\)

\(\Leftrightarrow2x^2-3x+1=0\)

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

hay \(x\in\left\{\dfrac{1}{2};1\right\}\)

\(\Leftrightarrow\dfrac{\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2}{\left(x^2-1\right)^2}-\dfrac{11\left(x^4-5x^2+4\right)}{\left(x^2-1\right)^2}=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2-11\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)^2-6x\left(x^2+2\right)+9x^2+\left(x^2+2\right)^2+6x\left(x^2+2\right)+9x^2-11\left(x^4-5x^2+4\right)=0\)

\(\Leftrightarrow2\left(x^2+2\right)^2+18x^2-11x^4+55x^2-44=0\)

\(\Leftrightarrow2\left(x^4+4x^2+4\right)-11x^4+73x^2-44=0\)

=>\(-9x^4+81x^2-36=0\)

=>9x^4-81x^2+36=0

=>x^4-9x^2+4=0

=>\(x^2=\dfrac{9\pm\sqrt{65}}{2}\)

=>\(x=\pm\sqrt{\dfrac{9\pm\sqrt{65}}{2}}\)

\(\dfrac{140}{x}+5=\dfrac{\left(140+10\right)}{x-1}\left(x\ne0,x\ne1\right)\)

\(\Leftrightarrow\dfrac{140+5x}{x}=\dfrac{150}{x-1}\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(140+5x\right)=150x\)

\(\Leftrightarrow140x+5x^2-140-5x-150x=0\)

\(\Leftrightarrow5x^2-15x-140=0\)

\(\Leftrightarrow x^2-3x-28=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(N\right)\\x=-4\left(N\right)\end{matrix}\right.\)

\(S=\left\{7,-4\right\}\)

ĐK: `x \ne 0 ; x \ne -1`

`140/x+5=150/(x-1)`

`<=>(140+5x)/x=150/(x-1)`

`<=>(140x+5x)(x-1)=150x`

`<=>5x^2+135x-140=150x`

`<=>5x^2-15x-140=0`

`<=>` \(\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)

Vậy...

Với \(x\ne0;x\ne1\), ta có:

(Mình không viết lại đề nữa nhé!)

\(\Leftrightarrow\dfrac{140+5x}{x}=\dfrac{150}{x-1}\)

\(\Rightarrow\left(140+5x\right)\left(x-1\right)=150x\)

\(\Leftrightarrow140x-140+5x^2-5x=150x\)

\(\Leftrightarrow5x^2-15x-140=0\)

\(\Leftrightarrow x^2-3x-28=0\)

\(\Leftrightarrow x^2+4x-7x-28=0\)

\(\Leftrightarrow x\left(x+4\right)-7\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=7\end{matrix}\right.\) (TM)

Vậy...

\(\Leftrightarrow\dfrac{x^3+8}{2}=\dfrac{\left(x+2\right)^3}{8}\)

\(\Leftrightarrow4x^3+32=\left(x+2\right)^3\)

\(\Leftrightarrow4\left(x+2\right)\left(x^2-2x+4\right)=\left(x+2\right)^3\)

\(\Leftrightarrow\left(x+2\right)\left(4x^2-8x+16\right)-\left(x+2\right)^3=0\)

\(\Leftrightarrow\left(x+2\right)\left(4x^2-8x+16-x^2-4x-4\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(3x^2-12x+12\right)=0\)

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

hay \(x\in\left\{2;-2\right\}\)