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

\(A=2\left[\left(x-y\right)^3+3xy\left(x-y\right)\right]-3\left[\left(x-y\right)^2+4xy\right]\)

\(A=2\left[2^3+3xy.2\right]-3\left[2^2+4xy\right]\)

\(A=2\left[28+6xy\right]-3\left[4+4xy\right]\)

\(A=56+12xy-12-12xy=56-12=44\)

\(A=\dfrac{\left(a+b\right)\left(-x-y\right)-\left(a-y\right)\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)

\(=\dfrac{a\left(-x-y\right)+b\left(-x-y\right)-a\left(b-x\right)+y\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)

\(=\dfrac{-ax-ay-bx-by-ab+ax+by-xy}{abxy\left(xy+ay+ab+by\right)}\)

\(=\dfrac{-ay-bx-ab-xy}{abxy\left(xy+ay+ab+by\right)}\)

\(=\dfrac{-xy+ay+ab+by}{abxy\left(xy+ay+ab+by\right)}=\dfrac{-1}{abxy}\)

Với \(a=\dfrac{1}{3};b=-2;x=\dfrac{3}{2};y=1\)

\(\Rightarrow A=\dfrac{-1}{\dfrac{1}{3}.\left(-2\right).\dfrac{3}{2}.1}=-1\)

\(x+y+1=0\\ \Leftrightarrow x+y=-1\)

Thay x+y=-1 vào C ta có:

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

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

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

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

\(\Rightarrow C=0+0+1\)

\(\Rightarrow C=1\)

\(x+y+1=0\) =>\(x+y=-1\)

- Thay \(x+y=-1\) vào C ta được:

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

\(=-x^2+y^2+x^2-y^2-2+3\)=1

Sao bạn doanh doanh nhắn chữ "hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh" quài vậy ?

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

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

\(=3x^2+3y^2-2.1\left(x^2-xy+y^2\right)\)

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

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

\(B=x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\)

\(=x^3+y^3+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2.1\)

\(=x^3+y^3+3xy\left(x+y\right)^2-6x^2y^2+6x^2y^2\)

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

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

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

a, A = (x-1)(x+6) (x+2)(x+3)

= (x^2 + 5x -6 ) (x^2 + 5x + 6)

Đặt t = x^2 +5x 

A= (t-6)(t+6)

= t^2 - 36

GTNN của A là -36 khi và ck t= 0

<=> x^2 +5x = 0

<=> x=0 hoặc x=-5

Vậy...

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

\(\Rightarrow C=\dfrac{1}{2}\left[\left(\dfrac{1}{2}\right)^2+1\right]-\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}-1\right)-1\left[\left(\dfrac{1}{2}\right)^2-\dfrac{1}{2}\right]\)

\(\Rightarrow C=\dfrac{1}{2}\left(\dfrac{1}{4}+1\right)-\dfrac{1}{4}\left(-\dfrac{1}{2}\right)-\left(\dfrac{1}{4}-\dfrac{1}{2}\right)\)

\(\Rightarrow C=\dfrac{1}{2}.\dfrac{5}{4}+\dfrac{1}{8}-\left(-\dfrac{1}{4}\right)\)

\(\Rightarrow C=\dfrac{5}{8}+\dfrac{1}{8}+\dfrac{1}{4}\)

\(\Rightarrow C=1\)

\(4,x=\dfrac{1}{2},y=-100\)

\(\Rightarrow D=\dfrac{1}{2}\left[\left(\dfrac{1}{2}\right)^2+100\right]-\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}-100\right)-100\left[\left(\dfrac{1}{2}\right)^2-\dfrac{1}{2}\right]\)

\(\Rightarrow D=\dfrac{1}{2}\left(\dfrac{1}{4}+100\right)-\dfrac{1}{4}\left(-\dfrac{199}{2}\right)-100\left(\dfrac{1}{4}-\dfrac{1}{2}\right)\)

\(\Rightarrow D=\dfrac{1}{2}.\dfrac{401}{4}+\dfrac{199}{8}-100.\left(-\dfrac{1}{4}\right)\)

\(\Rightarrow D=\dfrac{401}{8}+\dfrac{199}{8}+25\)

\(\Rightarrow D=100\)

3: C=x^3-xy-x^3-x^2y+x^2y-xy

=-2xy=-2*1/2*(-1)=1

4: D=x^3-xy-x^3-x^2y+x^2y-xy

=-2xy

=-2*1/2*(-100)=100

\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)

\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)

\(x=0;y=0\Leftrightarrow B=0\)

Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)

Vậy \(A\ne B\)