`@` `\text {Ans}`

`\downarrow`

\((x+y)(x-y)+(xy^4-x^3y^2) \div (xy^2) \)

`= x(x-y) + y(x-y) + xy^4 \div xy^2 - x^3y^2 \div xy^2`

`= x^2 - xy + xy - y^2 + y^2 - x^2`

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

`= 0`

cau 2 , n(2n-3)-2n(n+1)=2n^2-3n-2n^2-2n=-5n

-5chia het cho 5 nen nhan voi moi so nguyen deu chia het cho 5 suy ra n(2n-3)-2n(n+1)chia het cho 5

1,a) (x-1)(x^2+x+1)=x^3-1

VT=x³+x²+x-x²-x-1

=(x³-1)+(x²-x²)+(x-x)

=x³-1+0+0

=x³-1=VP (dpcm)

tương tự a

1,a) (x-1)(x^2+x+1)=x^3-1

VT=x³+x²+x-x²-x-1

=(x³-1)+(x²-x²)+(x-x)

=x³-1+0+0

=x³-1=VP (dpcm)

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

vậy (x-1)(\(x^2\)+x+1)=\(x^3\)-1

b) n(2n-3)-2n(n+1)

=n.2n -n.3 -2n.n-2n.1

=2\(n^2\)-3n-2\(n^2\)-2n

=-5n \(⋮\)5 với mọi số nguyên n

Vậy n(2n-3)-2n(n-1) chia hết cho 5 với mọi số nguyên n

Mày nhìn cái chóa j

h) \(=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]=3x\left(2y-3z\right)\left(x^2-10y+15z\right)\)

k) \(=\left(x+2\right)\left(3x-5\right)\)

l) \(=\left(18^2+3\right)\left(x+3\right)=327\left(x+3\right)\)

m) \(=7xy\left(2x-3y+4xy\right)\)

n) \(=2\left(x-y\right)\left(5x-4y\right)\)

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

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

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

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

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

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