= \(2\left(x^2+2x+1-y^2\right)=2\left[\left(x+1\right)^2-y^2\right]=2\left(x-y+1\right)\left(x+y+1\right)\)

ko pt đc nha bạn

thanks

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

1)

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)

2) Bạn xem lại đề!

x¹¹ + x¹⁰ + 1 

= ( x¹¹ - x⁹ + x⁸ - x⁶ + x⁵ - x³ + x² ) + ( x¹⁰ - x⁸ + x⁷ - x⁵ + x⁴ - x² + x ) + ( x⁹ - x⁷ + x⁶ - x⁴ + x³ - x + 1 )

= x² ( x⁹ - x⁷ + x⁶ - x⁴ + x³ - x + 1 ) + x(  ( x⁹ - x⁷ + x⁶ - x⁴ + x³ - x + 1 ) + 1( x⁹ - x⁷ + x⁶ - x⁴ + x³ - x + 1 )

= ( x² + x + 1 ) ( x⁹ - x⁷ + x⁶ - x⁴ + x³ - x + 1 )

Chỗ nào không hiểu thì ib nhé :)

x¹¹+x¹⁰+1

=x¹¹+x¹⁰+x⁹-x⁹-x⁸-x⁷+x⁸+x⁷+x⁶-x⁶-x⁵-x⁴+x⁵+x⁴+x³-x³-x²-x+x²+x+1

=(x¹¹+x¹⁰+x⁹)-(x⁹+x⁸+x⁷)+(x⁸+x⁷+x⁶)-(x⁶+x⁵+x⁴)+(x⁵+x⁴+x³)-(x³+x²+x)+(x²+x+1)

=x⁹(x²+x+1)-x⁷(x²+x+1)+x⁶(x²+x+1)-x⁴(x²+x+1)+x³(x²+x+1)-x(x²+x+1)+(x²+x+1)

=(x²+x+1)(x⁹-x⁷+x⁶-x⁴+x³-x+1)

Lời giải:

$y-x^2y-2xy^2-y^3=y(1-x^2-2xy-y^2)$

$=y[1-(x^2+2xy+y^2)]=y[1-(x+y)^2]=y(1-x-y)(1+x+y)$

b: \(\left(x^2+4\right)^2-16x^2\)

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

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

c: \(x^5-x^4+x^3-x^2\)

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

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

Lời giải:

a. Bạn xem lại đề

b. \((x^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)

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

c.

\(x^5-x^4+x^3-x^2=x^4(x-1)+x^2(x-1)=(x^4+x^2)(x-1)\)

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

a) 7x.(-y)+2(y-x)2

=>-7xy+4y-4x

b)(x^2+4)-16x^2

=>x^2+4-16x^2

=>-15x^2+4

c)x^5-x^4+x^3-x^2

=>x^4(x-1)+x^2(x-1)

=>(x^4+x^2)(x-1)

25n(n-1)-50(n-1) luôn chia hết cho 150 với mọi n là số nguyên

giúp mình chứng minh nha . Cám ơn mấy bạn