a)Ta có : \(\dfrac{x+1}{1-x}\)( giữ nguyên )

\(\dfrac{x^2-2}{1-x}\)( giữ nguyên )

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

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

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

\(\dfrac{2x-3x^2}{x^3-1}\)(giữ nguyên )

c) MTC = ( x+ 2)²(x - 2)²

Do đó , ta có : \(\dfrac{1}{x^2+4x+4}=\dfrac{1}{\left(x+2\right)^2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)^2\left(x-2\right)^2}\)

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

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

d) MTC = xyz( x - y)( y - z)( x - z)

Do đó , ta có : \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}=\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(\dfrac{1}{y\left(y-x\right)\left(y-z\right)}=\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(\dfrac{1}{z\left(z-x\right)\left(z-y\right)}=\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

Cộng các phân thức lại ta có :

\(\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

= \(\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(A=\dfrac{1}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+9\right)}+\dfrac{1}{\left(x+9\right)\left(x+11\right)}\)\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}+\dfrac{1}{x+9}-\dfrac{1}{x+11}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+11}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{x+11}{\left(x+1\right)\left(x+11\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+11\right)}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{x+11-x-1}{\left(x+1\right)\left(x+11\right)}\right)=\dfrac{1}{2}.\dfrac{10}{\left(x+1\right)\left(x+11\right)}=\dfrac{10}{2\left(x+1\right)\left(x+11\right)}\)

a) 3x+2(x-5)=-x+2

<=> 3x+2x+x=2+10

<=>6x=12

<=>x=2

b) 3x²-2x=0

<=>x(3x-2)=0

<=>\(\left[{}\begin{matrix}x=0\\3x-2=0\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

c) \(\dfrac{2x}{3}\)+\(\dfrac{x-4}{6}\)=2-\(\dfrac{x}{2}\)

<=>\(\dfrac{8x+2x-8}{12}\)=\(\dfrac{24-6x}{12}\)

<=> 8x+2x-8=24-6x

<=>8x+2x+6x=24+8

<=>16x=32

<=>x=2

d) \(\dfrac{x-2}{x+2}\)-\(\dfrac{3}{x-2}\)= -\(\dfrac{2\left(x-11\right)}{4-x^2}\) ( ĐKXĐ: x\(\ne\)\(\pm\)2)

<=> \(\dfrac{\left(x-2\right)^2-3\left(x+2\right)}{x^2-4}\)=\(\dfrac{2\left(x-11\right)}{x^2-4}\)

=> (x-2)²-3(x+2)=2(x-11)

<=> x²-4x+4-3x-6=2x-22

<=> x²-4x-3x-2x=-22-4+6

<=> x-9x+20=0

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

<=>\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\) ( thỏa mãn diều kiện )

d) (x²+1)(x²-4x+4)=0

=> x²-4x+4=0 (x²+1\(\ge\)1 với mọi x)

=>(x-2)² =0

=>x=2

Cảm ơn bạn nhăn Ngọc Vô Tâm

a: \(A=2x-3-5x+2-3x+1=-6x=-6\cdot\dfrac{-2}{3}=4\)

b: \(B=x^{2n-2n+3}=x^3=\left(-3\right)^3=-27\)

\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)

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

\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)

\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)

\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)

Ta có: \(\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}\)=\(\dfrac{a}{x-1}\)+\(\dfrac{bx+c}{x^2+1}\)

<=>\(\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}\)=\(\dfrac{a\left(x^2+1\right)+\left(bx+c\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+1\right)}\)

=>a(x²+1)+(bx+c)(x-1)=x²+2x-1

<=>ax²+a+bx²-bx+cx-c=x²+2x-1

<=>(a+b)x²+(c-b)x-(c-a)=x²+2x-1

=>\(\left\{{}\begin{matrix}a+b=1\\c-b=2\\c-a=1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}a+b=1\\b=c-2\\a=c-1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}c-1+c-2=1\\b=c-2\\a=c-1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}c=2\\b=0\\a=1\end{matrix}\right.\)

Vậy a=1,b=0,c=2