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

\(=\left(x-y\right)\left(x^2+x\cdot y+y^2\right)\)

\(=x^3-y^3\)

Bài 2:

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

\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)

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

\(=x^3+8-2x^2+2=x^3-2x^2+10\)

\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

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

\(=\left(-2y\right)^2+4\left(y+2\right)\)

\(=4y^2+4y+8\)

2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)

3: \(B=4y^2+4y+8\)

\(=4y^2+4y+1+7\)

\(=\left(2y+1\right)^2+7>=7>0\forall y\)

=>B luôn dương với mọi y

Bài 1:

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

\(=x^3-x^2y+x^2+x^3+x^2y-x-y\)

\(=2x^3-x+x^2-y\)

6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)

\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)

\(=6x^2+23x-55-6x^2-84x-294\)

=-61x-349

`x/(x+y) + (2xy)/(x^2-y^2) - y(x+y)`

`= (x(x-y))/(x^2-y^2) + (2xy)/(x^2-y^2) - (y(x-y))/(x^2-y^2)`

`= (x^2 - xy + 2xy - xy + y^2)/(x^2-y^2)`

`= (x^2+y^2)/(x^2-y^2)`

\(\dfrac{x}{x+y}+\dfrac{2xy}{x^2-y^2}-\dfrac{y}{x+y}\)

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

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

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

\(=\dfrac{x^2+y^2}{x^2-y^2}\)

\(MTC:x^2-y^2=\left(x+y\right)\left(x-y\right)\\ =\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}+\dfrac{2xy}{x^2-y^2}-\dfrac{y\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}\\ =\dfrac{x\left(x-y\right)+2xy-y\left(x-y\right)}{x^2-y^2}\\ =\dfrac{x^2-xy+2xy-xy+y^2}{x^2-y^2}=\dfrac{x^2+y^2}{x^2-y^2}\)

Ta có: \(\left(x+y+1\right)\left(x+y-1\right)-\left(x-y\right)^2-4xy\)

\(=\left(x+y\right)^2-\left(x-y\right)^2-1-4xy\)

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

=-1

`(x+y+1)(x+y-1)-(x-y)^{2}-4xy`

`=(x+y)^{2}-1-(x-y)^{2}-4xy`

`=(x+y+x-y)(x+y-x+y)-1-4xy`

`=2x.2y-4xy-1`

`=4xy-4xy-1`

`=-1`

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

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

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

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

\(A=\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-z\right)\left(y-x\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)

\(=\frac{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

Phân tích tử thức ta có:

\(TS=x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\)

\(=x^2\left(z-y\right)-y^2\left[\left(z-y\right)+\left(y-x\right)\right]+z^2\left(y-x\right)\)

\(=x^2\left(z-y\right)-y^2\left(z-y\right)-y^2\left(y-x\right)+z^2\left(y-x\right)\)

\(=\left(z-y\right)\left(x^2-y^2\right)+\left(y-x\right)\left(z^2-y^2\right)\)

\(=\left(z-y\right)\left(x-y\right)\left(x+y\right)+\left(y-x\right)\left(z-y\right)\left(z+y\right)\)

\(=\left(x-y\right)\left(y-z\right)\left(-x-y+z+y\right)\)

\(=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

Vậy  \(A=1\)

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

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

Bài 4:

Ta có:

 VT=(x2+y2)2−(2xy)2VT=(x2+y2)2−(2xy)2

=(x2+y2−2xy)(x2+y2+2xy)=(x2+y2−2xy)(x2+y2+2xy)

=(x−y)2(x+y)2=VP=(x−y)2(x+y)2=VP

⇒đpcm⇒đpcm

Bexiu bị j ấy