Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(2x^2-3x+5\right)\left(x^2-8x-2\right)\)
\(=2x^4-16x^3-4x^2+24x^2-3x^3+6x+5x^2-40x-10\)
\(=2x^4-\left(16x^3+3x^3\right)-\left(4x^2-24x^2-5x^2\right)+\left(6x-40x\right)-10\)
\(=2x^4-19x^3+25x^2-34x-10\)
a) \(\left(x^2-1\right)\left(x^2+2x\right)=x^4+2x^3-x^2-2x\)
b) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)=6x^2-3x+4x-2\left(3-x\right)\)
\(=6x^2-3x+4x-6+2x\)
\(=6x^2+3x-6\)
c) \(\left(x+3\right)\left(x^2+3x-5\right)=x^3+3x^2+3x^2+9x-5x-15\)
\(=x^3+6x^2+4x-15\)
d) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+x^2-x^2-x+x+1\)
\(=x^3+1\)
e) \(\left(2x^3-3x-1\right)\left(5x+2\right)=10x^4-15x^2-5x+4x^3-6x-2\)
\(=10x^4+4x^3-15x^2-11x-2\)
f) \(\left(x^2-2x+3\right)\left(x-4\right)=x^3-2x^2+3x-4x^2+8x-12\)
\(=x^3-6x^2+11x-12\)
a) ( 3x + 2y - 1 )( x - 5 ) - ( x - 2 )2y
= 3x(x - 5) + 2y(x - 5) - 1(x - 5) - ( 2xy - 4y )
= 3x2 - 15x + 2xy - 10y - x + 5 - 2xy + 4y
= 3x2 - 16x - 6y + 5
b) ( 3x - 2 )( 3x + 2 ) - ( 2x + 1 )( 4x + 3 )
= [ ( 3x )2 - 22 ] - ( 8x2 + 10x + 3 )
= 9x2 - 4 - 8x2 - 10x - 3
= x2 - 10 - 7
3x(x+1)(x-1)-(2x-3)2 =3x(x2-1)-4x2+12x-9=3x3-3x-4x2+12x-9=3x3-4x2+9x-9
3(x+1)+(x+1)2=(x+1)(1+x+1)=(x+1)(x+2)=x2+2x+x+2=x2+3x+2
the end
a. \(=\frac{x+1}{2.\left(x+3\right)}+\frac{2x+3}{x.\left(x+3\right)}=\frac{x^2+x+4x+6}{2x.\left(x+3\right)}=\frac{x^2+5x+6}{2x.\left(x+3\right)}=\frac{\left(x+2\right).\left(x+3\right)}{2x.\left(x+3\right)}=\frac{x+2}{2x}\)
b. =\(\frac{2.\left(x+3\right)}{x.\left(3x-1\right)}.\frac{-\left(3x-1\right)}{x.\left(x+3\right)}=\frac{-2}{x^2}\)
Chắc chắn đúng, mik nhaaaaaa
( 2x - 1 )( 3x + 2 )( 3 - x )
= ( 6x2 - 3x + 4x - 2 )( 3 - x )
= ( 6x2 + x - 2 )( 3 - x )
= 18x2 + 3x - 6 - 6x3 - x2 + 2x
= - ( 6x3 + 18x2 - 5x + 6 )
hình như cái cuôi bạn sai thì phải nó biến đổi thé này chú
=18x^2+3x-6-6x^3-x^2+2x
=-6x^3+(18x^2-x^2)+(2x+3x)-6
=-6x^3+17x^2+5x-6
a/\(\left(x-1\right)\left(x^5+x^4+x^3+x^2+x+1\right).\)
\(=\left(x-1\right)\left[\left(x^5+x^4+x^3\right)+\left(x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left[x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(=\left(x^2-1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
Câu b/ quên làm ạ :> Bù nè
b/ \(2\left(3x-1\right)\left(2x+5\right)-\left(4x-1\right)\left(3x-2\right)\)
\(=2\left(6x^2+15x-2x-5\right)-\left(12x^2-8x-3x+2\right)\)
\(=2\left(6x^2+13x-5\right)-\left(12x^2-11x+2\right)\)
\(=12x^2+26x-10-\left(12x^2-11x+2\right)\)
\(=12x^2+26x-10-12x^2+11x-2\)
\(=37x-12\)
\(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\\ =\left(6x^2+x-2\right)\left(3-x\right)\\ =18x^2+3x-6-6x^3-x^2+2x\\ =-6x^3+17x^2+5x-6\)
(2x - 1) . (3x + 2) . (3 - x)
= (6x2 + 4x - 3x - 2) . (3 - x)
= 18x2 + 12x - 9x - 6 - 6x3 - 4x2 + 3x2 + 2x
= -6x3 + 17x2 + 5x - 6