=>\(\dfrac{8}{x}+\dfrac{8}{y}+\dfrac{20}{3}\cdot\dfrac{1}{y}=1\)

=>\(\dfrac{8}{x}+\dfrac{44}{3y}=1\)

=>\(\dfrac{24y+44x}{3xy}=1\)

=>44x+24y=3xy

=>44x+24y-3xy=0

=>44x-3y(x-8)=0

=>44x-352-3y(x-8)=352

=>(x-8)(44-3y)=352

=>\(\left(x-8;44-3y\right)\in\left\{\left(32;11\right)\left(44;8\right);\left(176;2\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(40;11\right);\left(52;12\right);\left(184;14\right)\right\}\)

a: \(\Leftrightarrow x^2+x+1-3x^2=2x\left(x-1\right)\)

=>-2x^2+x+1-2x^2+2x=0

=>-4x^2+3x+1=0

=>4x^2-3x-1=0

=>4x^2-4x+x-1=0

=>(x-1)(4x+1)=0

=>x=1(loại) hoặc x=-1/4(nhận)

b: \(\Leftrightarrow\dfrac{440}{x-2}-\dfrac{440}{x}=1\)

=>x(x-2)=440x-440x+880

=>x^2-2x-880=0

=>\(x=1\pm\sqrt{881}\)

c: \(\Leftrightarrow\dfrac{x+5+x}{x\left(x+5\right)}=\dfrac{1}{6}\)

=>x^2+5x=6(2x+5)

=>x^2+5x-12x-30=0

=>x^2-7x-30=0

=>(x-10)(x+3)=0

=>x=10 hoặc x=-3

d: =>(x-1)(x+1)-x=2x-1

=>x^2-1-x=2x-1

=>x^2-x-2x=0

=>x(x-3)=0

=>x=0(loại) hoặc x=3(nhận)

ĐKXĐ: \(\left\{{}\begin{matrix}2x\ne0\\2\left(25-x\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne25\end{matrix}\right.\)

\(\dfrac{1}{2x}+\dfrac{1}{2\left(25-x\right)}=\dfrac{1}{12}\\ \Leftrightarrow\dfrac{25-x+x}{2x\left(25-x\right)}=\dfrac{1}{12}\\ \Leftrightarrow\dfrac{25}{-2x^2+50x}=\dfrac{1}{12}\\ \Leftrightarrow-2x^2+50x=300\\ \Leftrightarrow-2x^2+50x-300=0\\ \Leftrightarrow\left[{}\begin{matrix}x=15\left(tm\right)\\x=10\left(tm\right)\end{matrix}\right.\)

Vậy...

x=90/7

\(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-x\)

\(\Leftrightarrow\dfrac{2x}{6}-\dfrac{2x+1}{6}=\dfrac{x}{6}-\dfrac{6x}{6}\)

\(\Leftrightarrow2x-2x+1=x-6x\)

\(\Leftrightarrow1=-5x\)

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

ĐKXĐ: \(x\ne\pm1\)

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

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

\(\Leftrightarrow4x=x^3+3\)

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

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

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

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

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

\(\Rightarrow x=\dfrac{-1\pm\sqrt{13}}{2}\)

=>x/30-x/40=3/4

=>x/120=3/4

=>x=90

\(\dfrac{x}{30}=\dfrac{x}{40}+\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{4x}{120}=\dfrac{3x}{120}+\dfrac{90}{120}\)

\(\Leftrightarrow4x=3x+90\)

\(\Leftrightarrow4x-3x-90=0\)

\(\Leftrightarrow x-90=0\)

\(\Leftrightarrow x=90\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{90\right\}\)

refer

Δ=(2m-2)^2-4(m+1)

=4m^2-8m+4-4m-4

=4m^2-12m

Để phương trình co hai nghiệm thì 4m^2-12m>0

=>m>3 hoặc m<0

x1/x2+x2/x1=4

=>x1^2+x2^2=4x1x2

=>(x1+x2)^2-2x1x2=4x1x2

=>(2m-2)^2-6(m+1)=0

=>4m^2-8m+4-6m-6=0

=>4m^2-14m-2=0

=>\(m=\dfrac{7\pm\sqrt{57}}{2}\)