ĐKXĐ

x≠3 ; x≠-3

ĐKXĐ x≠3 ; x≠-3

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

=> (2x-1)(x-3)=(2x+1)(x+3)

⇔2x²-6x-x+3=2x²+6x+x+3

⇔2x²-2x²-7x-6x=3-3

⇔ -13x=0

⇔x=0 (tm)

vậy phương trình trên có tập n^o S={0}

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

=>x(x-2)=0

=>x=2

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-1\\x\ne0\end{matrix}\right.\)

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

ĐK: \(x,y\ne0\)

\(\left\{{}\begin{matrix}x-\dfrac{1}{x^3}=y-\dfrac{1}{y^3}\\\left(x-4y\right)\left(2x-y+4\right)=-36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y-\left(\dfrac{1}{x^3}-\dfrac{1}{y^3}\right)=0\\\left(x-4y\right)\left(2x-y+4\right)=-36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y+\dfrac{\left(x-y\right)\left(x^2+y^2+xy\right)}{x^3y^3}=0\\\left(x-4y\right)\left(2x-y+4\right)=-36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\left(x-y\right)\left(x^3y^3+x^2+y^2+xy\right)}{x^3y^3}=0\\\left(x-4y\right)\left(2x-y+4\right)=-36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\\left(x-3x\right)\left(2x-x+4\right)=-36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\-2x^2-8x=-36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x^2+4x-18=0\end{matrix}\right.\)

\(\Leftrightarrow x=y=-2\pm\sqrt{22}\left(tm\right)\)

ĐKXĐ: x\(\ne2\), \(x\ne1\)

\(\dfrac{2x-5}{x-2}-\dfrac{3x-5}{x-1}=-1\)

<=> \(\dfrac{\left(2x-5\right)\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}-\dfrac{\left(3x-5\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=\dfrac{-1.\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}\)

=> 2x²-2x-5x+5-3x²+6x+5x-10= -x²+2x-2+x

<=> 2x²-2x-5x+5-3x²+6x+5x-10+x²-2x+2-x=0

<=> x-3=0

<=> x=3 (thỏa mãn ĐKXĐ)

Vậy S=\(\left\{3\right\}\)

a. Bạn tự giải

b. \(\Leftrightarrow\left\{{}\begin{matrix}x-2y=4m-5\\4x+2y=6m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=4m-5\\5x=10m-5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2m-1\\y=-m+2\end{matrix}\right.\)

\(\dfrac{2}{x}-\dfrac{1}{y}=-1\Rightarrow\dfrac{2}{2m-1}-\dfrac{1}{-m+2}=-1\) (\(m\ne\left\{\dfrac{1}{2};2\right\}\))

\(\Leftrightarrow2\left(-m+2\right)-\left(2m-1\right)=\left(m-2\right)\left(2m-1\right)\)

\(\Leftrightarrow2m^2-m-3=0\Rightarrow\left[{}\begin{matrix}m=-1\\m=\dfrac{3}{2}\end{matrix}\right.\)

a: \(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)

Suy ra: \(x^2+2x-2=x-2\)

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

=>x(x+1)=0

=>x=0(loại) hoặc x=-1(nhận)

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

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

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

\(\Leftrightarrow4x^2-3x-1=0\)

\(\Leftrightarrow4x^2-4x+x-1=0\)

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

=>x=1(loại) hoặc x=-1/4(nhận)

b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)

=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)

=>x*(x+20)=400*6=2400

=>x^2+20x-2400=0

=>(x+60)(x-40)=0

=>x=-60 hoặc x=40

c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)

=>(2x+1)^2-(2x-1)^2=8

=>4x^2+4x+1-4x^2+4x-1=8

=>8x=8

=>x=1(nhận)

1: Sửa đề: 2/x+2

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

=>\(\dfrac{2x+1+2x-4}{x^2-4}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

=>4x-3=-3x-6

=>7x=-3

=>x=-3/7(nhận)

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

=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)

=>-6x^2+6=2(3x^2-10x+3)

=>-6x^2+6=6x^2-20x+6

=>-12x^2+20x=0

=>-4x(3x-5)=0

=>x=5/3(nhận) hoặc x=0(nhận)

3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)

=>x*19/6=35/12

=>x=35/38