1, Ta có: \(x+y=9\Rightarrow\left(x+y\right)^2=81\)

\(\Rightarrow x^2+2xy+y^2=81\)

\(\Rightarrow x^2+y^2=45\)

\(\Rightarrow x^2+y^2-2xy=9\)

\(\Rightarrow\left(x-y\right)^2=9\Rightarrow\left[{}\begin{matrix}x-y=3\\x-y=-3\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}A=3.63=189\\A=-3.63=-189\end{matrix}\right.\)

Vậy...

a. \(\dfrac{\left(x-2\right)^2}{2x\left(x-2\right)}=\dfrac{x-2}{2x}\)

b. \(\dfrac{\dfrac{1}{2}-2}{2.\dfrac{1}{2}}=-1,5\)

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

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

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

b,

thay x=\(\dfrac{1}{2}\) vào bt M ta được:

\(1-\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^2=\dfrac{5}{8}\)

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

b) \(A=B+C\Rightarrow C=A-B\)

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

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

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

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

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

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

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

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

Vậy \(C=\dfrac{2}{x^2+x+1}\)

12344

a, 4/3

b=  2(x+1)(2x+1)/x^3-1

Bài này đã có tại đây:

Cho biểu thức:  \(A=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\)Với ... - Hoc24

a, ĐKXĐ: x≠±3

A=\(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)

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

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

A=\(\left(\dfrac{9-x^2}{x^2-9}+\dfrac{x^2-3x}{x^2-9}\right):\dfrac{3x^2}{x+3}\)

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

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

b, Thay x=\(-\dfrac{1}{2}\) (TMĐKXĐ) vào A ta có:

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

c, A<0 ⇔ \(\dfrac{-1}{x^2}< 0\) ⇔ x²>0 (Đúng với mọi x)

Vậy để A<0 thì x đúng với mọi giá trị (trừ ±3)

c: \(=\dfrac{\left(x+2\right)^2}{\left(x-5y\right)^2}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{\left(x+2\right)\left(x+5y\right)}{\left(x-5y\right)\left(x-2\right)}\)

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

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

b: Khi x=-5/3 thì \(A=\left(1+\dfrac{5}{3}\right)\left(\dfrac{25}{9}+1\right)=\dfrac{8}{3}\cdot\dfrac{34}{9}=\dfrac{272}{27}\)

c: Để A<0 thì 1-x<0

hay x>1