1)\(n^2\left(n-1\right)\left(n+1\right)-\left(n^2+2\right)\left(n^2-2\right)=n^2\left(n^2-1\right)-\left(n^4-4\right)=n^4-n^2-n^4+4\)

\(=-n^2+4\)

2)\(\left(y+3\right)\left(y-3\right)\left(y^2+9\right)-\left(y^2-4\right)\left(y^2+4\right)=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-16\right)\)

\(=y^4-81-y^4+16=-65\)

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

\(=x^2+6x+9-4y^2-x^2+4y^2=6x+9\)

4)\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

5)\(\left(a+b-c\right)^2=a^2+b^2+c^2+2ab-2bc-2ac\)

6)\(\left(a-b-c\right)^2=a^2+b^2+c^2-2ab+2bc-2ac\)

Học tốt nha bạn !

Bài 2: Tính giá trị của biểu thức:
a) P= 1/3 x^2 y + xy^2 - xy + 1/2 xy^2 - 5xy - 1/3 x^2 y (1)

Tại x = 0,5; y = 1

Thay \(x=0,5 ; y=1\) vào biểu thức (1) , ta có :

P= \(\dfrac{1}{3} . 0,5^2.1+0,5.1^2-0,5.1+\dfrac{1}{2}. 0,5.1^2-5.0,5.1-\dfrac{1}{3}.0,5^2.1\)

P= \(=\dfrac{1}{12}+\dfrac{1}{2} -0,5+\dfrac{1}{4} -\dfrac{5}{2} - \dfrac{1}{12}\)

P= \(= \dfrac{-9}{4}\)

Vậy \(P =\dfrac{-9}{4}\)

Bài 1:

Ta có:

\(a+b+c=0\\ \Leftrightarrow a^3+b^3+c^3+3\left(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2abc\right)=0\\ \Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\\ \Leftrightarrow a^3+b^3+c^3-3abc=0\\ \Leftrightarrow a^3+b^3+c^3=3abc\left(dpcm\right)\)

a: \(=y^2-9\)

b: \(=m^3+n^3\)

c: \(=8-a^3\)

d: \(=\left(a-b-c-a+b-c\right)\left(a-b-c+a-b+c\right)\)

\(=-2c\cdot\left(2a-2b\right)\)

\(=-4ac+4bc\)

f: \(=\left(1-x^3\right)\left(1+x^3\right)=1-x^6\)

a) \(\left(x+y\right)^2=x^2+y^2+2xy\Rightarrow4=10+2xy\Leftrightarrow xy=-3\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=2^3+3.3.2=26\)

b) \(\left(x-y\right)^2=x^2+y^2-2xy\Rightarrow m^2=n-2xy\Leftrightarrow xy=\frac{n-m^2}{2}\)

\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=m^3+3.m.\frac{n-m^2}{2}=\frac{3mn}{2}-\frac{m^3}{2}\)