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

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

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

a: =>x>=0 và x^2+x=x^2

=>x=0

b: =>x>=2 và x^2-4x-3=x^2-4x+4

=>-3=4(loại)

\(a)ĐK:x\ge0\)

\(pt\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+x=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x=0\left(tm\right)\end{matrix}\right.\)

Vậy, pt có nghiệm duy nhất là x=0

\(b)ĐK:x\ge2+\sqrt{7}\)

\(pt\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x^2-4x-3=(x-2)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-4x-3=x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\-3=4\end{matrix}\right.\)(vô lý)

Vậy pt vô nghiệm

\(A=\left(x-y\right)^2-3\left(x-y\right)=10^2-3\cdot10=100-30=70\)

a. \(\left(2x+1\right)^2-4x\left(x-1\right)=4x^2+4x+1-4x^2+4x=8x+1\)

b. \(\left(x-2\right)\left(x+2\right)-\left(x-1\right)^2=x^2-4-x^2+2x-1=2x-5\)

a)\((2x+1)^2-4x(x-1)=4x^2+4x+1-4x^2+4x\)

=\(8x+ 1\)

b)\((X-2)(x+2)-(x-1)^2=X^2-4-(x^2-2x+1)\)

=\(X^2-4-x^2+2x-1=2x-5\)

a: \(\left(2x+1\right)^2-4x\left(x-1\right)\)

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

=8x+1

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

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

=2x-5

\(a,=\left(x^3+3x^2-x^2-3x+x+3\right):\left(x+3\right)\\ =\left(x+3\right)\left(x^2-x+1\right):\left(x+3\right)\\ =x^2-x+1\\ b,=\left(x^3+2x^2-x^2-2x+3x+6\right):\left(x+2\right)\\ =\left(x+2\right)\left(x^2-x+3\right):\left(x+2\right)\\ =x^2-x+3\)

a) \(2xy\left(3x^2-5xy+4y^2\right)=6x^3y-10x^2y^2+8xy^3\)

b) \(\left(x-3\right)^2+\left(x+5\right)\left(5-x\right)=x^2-6x+9+25-x^2=34-6x\)

a: \(2xy\left(3x^2-5xy+4y^2\right)=6x^3y-10x^2y^2+8xy^3\)

b: \(\left(x-3\right)^2+\left(x+5\right)\left(5-x\right)\)

\(=x^2-6x+9+25-x^2\)

=-6x+34

\(a,\left(x-2\right)\left(x+3\right)-x\left(x-5\right)=x^2-2x+3x-6-x^2+5x=6x-6\)

\(b,\dfrac{1}{x-2}+\dfrac{-2}{x+2}+\dfrac{2x-8}{x^2-4}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2-2x+4+2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{1}{x+2}\)

a)\(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4=x^4-y^4\)

b) \(x\left(3x-18\right)-3\left(x-4\right)\left(x-2\right)+8=3x^2-18x-3x^2+18x-24+8=-16\)

a) = x^2 - 9 - (x^2 + 3x - 10)

= -3x + 1

b) = 3x + 1 - 3x + 19

= 20

a: \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\)

\(=x^2-9-x^2-3x+10\)

\(=-3x+1\)

b: \(\dfrac{27x^3+1}{9x^2-3x+1}-\left(3x-19\right)\)

\(=3x+1-3x+19\)

=20

a: \(=x^2-4-x^2+x+8=x+4\)

a) (x-2)(x+2)-x(x-1)+8
= x²-4-x²+x+8
= (x²-x²)+(-4+8)+x
= 4+x
b) bn viết lại đề đi:v
đọc khó quá.