Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2
b: =x^3+3x^2-2x-3x^2-9x+6
=x^3-11x+6
c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)
\(=2x^2-3x-1+\dfrac{5}{2x+1}\)
a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)
\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)
\(=2x^5-16x^3-2x^5-x^3\)
\(=-17x^3\)
b) \(\left(x+3\right)\left(x^2+3x-2\right)\)
\(=x^3+3x^2-2x+3x^2+9x-6\)
\(=x^3+6x^2+7x-6\)
c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)
\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)
\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)
`a)3x(2x^2-3x+4)`
`=6x^3-9x^2+12x`
______________________________________________
`b)(x+3)^2+(3x-2)(x+4)`
`=x^2+6x+9+3x^2+12x-2x-8`
`=4x^2+16x+1`
______________________________________________
`c)[2x-4]/[x-1]+[2x+2]/[x^2-1]` `ĐK: x \ne +-1`
`=[(2x-4)(x+1)+2x+2]/[(x-1)(x+1)]`
`=[2x^2+2x-4x-4+2x+2]/[(x-1)(x+1)]`
`=[2x^2-2]/[x^2-1]`
`=2`
\(\frac{x^2-3x-x+3}{x-3}=\frac{x\left(x-3\right)-\left(x-3\right)}{x-3}=\frac{\left(x-3\right)\left(x-1\right)}{x-3}=x-1\)( ĐK: \(x\ne3\))
\(\frac{2x^3-5x^2-4x+3}{2x-1}=\frac{\left(2x^3-x^2\right)-\left(4x^2-2x\right)-\left(6x-3\right)}{2x-1}=\frac{x^2\left(2x-1\right)-2x\left(2x-1\right)-3\left(2x-1\right)}{2x-1}=\frac{\left(2x-1\right)\left(x^2-2x-3\right)}{2x-1}=x^2-2x-3\)( ĐK: \(x\ne\frac{1}{2}\))
Tham khảo nhé~
A/\(\left(2x^3+y^2-7xy\right)4xy^2.\)
\(=8x^4y^2+4xy^4-28x^2y^3\)
B/\(\left(2x^3-x-1\right)\left(5x-2\right)\)
\(=10x^4-5x^2-5x-4x^3+2x+2\)
\(=10x^4-5x^3-3x-4x^3+2\)
C/\(\left(2x^2-3\right)\left(4x^4+6x^2+9\right)\)
\(=\left(2x^2-3\right)\left(2x+3\right)^2\)
D/\(\left(3x^2-2y\right)^3-\left(2x^2-y\right)^3\)
( Bài này áp dụng hằng đẳng thức là làm được ạ )
\(\frac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)
\(=\frac{x^2\left(3x^2-2x+1\right)-2x\left(3x^2-2x+1\right)-5\left(3x^2-2x+1\right)}{3x^2-2x+1}\)
\(=\frac{\left(3x^2-2x+1\right)\cdot\left(x^2-2x-5\right)}{3x^2-2x+1}\)
\(=x^2-2x-5\)
\(\frac{2x^3-9x^2+19x-15}{x^2-3x+5}\)
\(=\frac{2x\left(x^2-3x+5\right)-3\left(x^2-3x+5\right)}{x^2-3x+5}\)
\(=\frac{\left(x^2-3x+5\right)\left(2x-3\right)}{x^2-3x+5}\)
\(=2x-3\)
a: \(=2x+x^3-5x^4\)
b: \(=\dfrac{8x^2+4x-7x-3}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{8x^2-3x-3}{\left(2x-1\right)\left(2x+1\right)}\)
a) \(\left(2x-3\right)\left(x^2-2x+1\right)+2\left(2-x\right)^3\)
\(=2x\left(x^2-2x+1\right)-3\left(x^2-2x+1\right)+2\left(2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\right)\)
\(=2x^3-4x^2+2x-3x^2+6x-3+2\left(8-12x+6x^2-x^3\right)\)
\(=2x^3-4x^2+2x-3x^2+6x-3+16-24x+12x^2-2x^3\)
\(=\left(2x^3-2x^3\right)+\left(-4x^2-3x^2+12x^2\right)+\left(2x+6x-24x\right)+\left(-3+16\right)\)
\(=5x^2-16x+13\)
b)
Vậy \(\left(2x^3-7x^2+2x+3\right):\left(x^2-4x+3\right)=2x+1\)
Câu b thêm dấu " - " ở chỗ 2x3 - 7x2 + 2x +3 và 2x3 - 8x2 + 6x nhé :)))