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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: \(=2x+x^3-5x^4\)
b: \(=\dfrac{8x^2+4x-7x-3}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{8x^2-3x-3}{\left(2x-1\right)\left(2x+1\right)}\)
\(\dfrac{x^2-50}{3x^2-9x}\div\dfrac{2x^2+10x}{x^2-9}\)
\(\Leftrightarrow\dfrac{x^2-50}{3x\left(x-3\right)}\div\dfrac{2x^2+10x}{\left(x-3\right)\left(x+3\right)}\)
MTC: 3x(x-3)(x+3)
\(\dfrac{(x^2-50)\left(x+3\right)}{3x\left(x-3\right)\left(x+3\right)}\div\dfrac{3x(2x^2+10x)}{3x\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\)(x2-50)(x+3):3x(2x2+10x)
\(\Rightarrow\)(x3+3x2-50x-150):6x3+30x2
a) Ta có: \(\dfrac{x^2-50}{3x^2-9x}:\dfrac{2x^2+10x}{x^2-9}\)
\(=\dfrac{x^2-50}{3x\left(x-3\right)}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{2x\left(x+5\right)}\)
\(=\dfrac{\left(x^2-50\right)\left(x+3\right)}{6x^2\left(x+5\right)}\)
b) Ta có: \(\dfrac{-3x^2}{2x+1}:\dfrac{-9}{4x^2-1}\)
\(=\dfrac{3x^2}{2x+1}\cdot\dfrac{\left(2x+1\right)\left(2x-1\right)}{9}\)
\(=\dfrac{x^2\left(2x-1\right)}{3}\)
a) (2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x(2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x
=4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x
=8x.5(2x+1)(2x−1)(2
b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)=\left(\dfrac{1}{x\left(x+1\right)}-\dfrac{x\left(2-x\right)}{x\left(x+1\right)}\right):\left(\dfrac{1}{x}+\dfrac{x^2}{x}-\dfrac{2x}{x}\right)=\left(\dfrac{1-2x+x^2}{x\left(x+1\right)}\right):\left(\dfrac{1+x^2-2x}{x}\right)=\left(\dfrac{\left(x-1\right)^2}{x\left(x+1\right)}\right)\cdot\left(\dfrac{x}{\left(x-1\right)^2}\right)=\dfrac{\left(x-1\right)^2\cdot x}{\left(x-1\right)^2\cdot x\cdot\left(x+1\right)}=\dfrac{1}{x+1}\)
`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`
\(\left(\dfrac{1}{2}x-1\right)\left(2x-3\right)=x^2-\dfrac{3}{2}x-2x+3=x^2-\dfrac{1}{2}x+3\)\(b,\left(x-7\right)\left(x-5\right)=x^2-5x-7x+35=x^2-12x+35\)\(c,\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)\left(4x-1\right)=\left(x^2-\dfrac{1}{4}\right)\left(4x-1\right)=4x^3-x^2-x+\dfrac{1}{4}\)
\(=\dfrac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}\cdot\dfrac{2-x}{1-2x}=\dfrac{4x}{1-2x}\)