Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=6x^4-9x^3+3x^2-4x^3+6x^2-2x+10x^2-15x+5\)
\(=6x^4-13x^3+19x^2-17x+5\)
b: \(=6x^4-\dfrac{9}{4}x^3-\dfrac{9}{2}x^2-\dfrac{8}{3}x^3+x^2+2x-\dfrac{20}{3}x^2+\dfrac{5}{2}x+5\)
\(=6x^4-\dfrac{59}{12}x^3-\dfrac{67}{6}x^2+\dfrac{9}{2}x+5\)
c: \(=3x^4-\dfrac{9}{8}x^3-\dfrac{3}{4}x^2+8x^3-3x^2-6x-\dfrac{4}{3}x^2+\dfrac{1}{2}x+1\)
\(=3x^4-\dfrac{55}{8}x^3-\dfrac{25}{12}x^2-\dfrac{11}{2}x+1\)
a, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)
⇔ \(6x^2-5x+3=2x-9x+6x^2\)
⇔ \(6x^2-5x+3-6x^2+9x-2x=0\)
⇔ \(2x+3=0\)
⇔ \(2x=-3\)
⇔ \(x=-\dfrac{3}{2}\)
b, \(\dfrac{2\left(x-4\right)}{4}-\dfrac{3+2x}{10}=x+\dfrac{1-x}{5}\)
⇔ \(\dfrac{20\left(x-4\right)}{4.10}-\dfrac{4\left(3+2x\right)}{4.10}=\dfrac{5x}{5}+\dfrac{1-x}{5}\)
⇔ \(\dfrac{20x-80}{40}-\dfrac{12+8x}{40}=\dfrac{5x+1-x}{5}\)
⇔ \(\dfrac{20x-80-12-8x}{40}=\dfrac{4x+1}{5}\)
⇔ \(\dfrac{12x-92}{40}-\dfrac{4x+1}{5}=0\)
⇔ \(\dfrac{12x-92}{40}-\dfrac{8\left(4x+1\right)}{40}=0\)
⇔ \(12x-92-8\left(4x+1\right)=0\)
⇔ 12x - 92 - 32x - 8 = 0
⇔ -100 - 20x = 0
⇔ 20x = -100
⇔ x = -100 : 20
⇔ x = -5
1/ \(8x^{n-1}\left(\dfrac{1}{2}x^{n+1}-\dfrac{3}{4}x\right)=4x^{2n}-6x^n\)
2/
a/ \(3x\left(2x-4\right)-2x\left(2x+5\right)=44\)
\(\Leftrightarrow6x^2-12x-4x^2-10x=44\)
\(\Leftrightarrow2x^2-22x=44\)
\(\Leftrightarrow2x^2-22x-44=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{11+\sqrt{209}}{2}\\x_2=\dfrac{11-\sqrt{209}}{2}\end{matrix}\right.\)
b/ \(x\left(5-3x\right)+3x\left(x+1\right)=40\)
\(\Leftrightarrow5x-3x^2+3x^2+3x=40\)
\(\Leftrightarrow8x=40\)
\(\Leftrightarrow x=5\)
Vậy....................
a: \(=xy^2+xy+x-y^3-y^2-y+\dfrac{2}{3}x^3y+\dfrac{1}{3}x^2y^3-2xy-y^3\)
\(=xy^2-xy+x-2y^3-y^2-y+\dfrac{2}{3}x^3y+\dfrac{1}{3}x^2y^3\)
b: \(=2x^3-4x^2+3x^3-3x^2-6x-15+5x^2\)
\(=5x^3-2x^2-6x-15\)
c: \(=x^2-4x+3+\left(x-4\right)\left(2x-1\right)-3x^3+2x-5\)
\(=-3x^3+x^2-2x-2+2x^2-x-8x+4\)
\(=-3x^3+3x^2-11x+2\)
a: \(=2x^2-x+5\)
b: \(=-\dfrac{3}{2}x^3+x^2-\dfrac{1}{2}x\)
c: \(=-x^3+\dfrac{3}{2}-2x\)
d: \(=-2x^2+4xy-6y^2\)
e: \(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)
a)\(\left(2x^2-3x\right)\left(5x^2-2x+1\right)\)
\(=2x^2\left(5x^2-2x+1\right)-3x\left(5x^2-2x+1\right)\)
\(=10x^4-4x^3+2x^2-15x^3+6x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
a. \(\left(2x^2-3x\right)\left(5x^2-2x+1\right)\)
\(=10x^4-4x^3+2x^2-15x^3+6x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
b. \(\left(2x^4-x^3+3x^2\right):\left(\frac{1}{3}x^2\right)\)
\(=\left(2x^4-x^3+3x^2\right).\frac{3}{x^2}\)
\(=0,6x^2-3x+0,9\)
`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`
hếp