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a) Ta có: ( x2 -1 )( x2 + 2x )
= x2( x2 + 2x ) - ( x2 + 2x )
= x4 + 2x3 - x2 - 2x
b) Ta có ( x + 3 )( x2 + 3x -5 )
= x( x2 + 3x -5 ) + 3( x2 + 3x -5 )
= x3 + 3x2 - 5x + 3x2 + 9x - 15
= x3 + 6x2 + 4x - 15
c) Ta có ( x -2y )( x2y2 - xy + 2y )
= x( x2y2 - xy + 2y ) - 2y( x2y2 - xy + 2y )
= x3y2 - x2y + 2xy - 2x2y3 + 2xy2 - 4y2
d) Ta có ( 1/2xy -1 )( x3 -2x -6 )
= 1/2xy( x3 -2x -6 ) - ( x3 -2x -6 )
= 1/2x4y - x2y - 3xy - x3 + 2x + 6
a, \(=5x^3-2x^2y-5x^2y+2xy^2+5x-2y\)
\(=5x^3-7x^2y+2xy^2+5x-2y\)
b, \(=\left(x^2-1\right)\left(x+2\right)\)
\(=x^3-x+2x^2-2\)
c, đề không rõ
d, đề không rõ
P/s có gì bạn tham khảo các thanh công cụ ở trên để đánh cô hỏi cho rõ nha
c: \(=\dfrac{x^3+2x^2+x^2+2x-10x-20}{x+2}\)
\(=x^2+x-10\)
a: \(=\dfrac{5}{3}x^2-x+\dfrac{1}{3}\)
b: \(=-5y-9+xy\)
\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)
\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)
a: \(=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}\)
b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)\left(2x-y\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)}{\left(2x-y\right)^2}\)
c: \(=\dfrac{x+3}{x+2}\cdot\dfrac{2x-1}{3\left(x+3\right)}\cdot\dfrac{2\left(x+2\right)}{2\left(2x-1\right)}\)
=1/3
d: \(=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{2x\left(2x-3\right)}{3\left(x+1\right)}=\dfrac{2x\left(2x-3\right)}{3\left(x+2\right)}\)
a: \(=25x^4-10x^3+5x^2\)
c: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)
a. (5\(x\) – 2\(y\))(\(x^2\) –\(x\) \(y\) + 1)
= 5\(x\).(\(x^2\) –\(x\) \(y\) + 1) – 2\(y\)(\(x^2\) – \(x\) \(y\)+ 1)
= (5 \(x^3\) – 5\(x^2\)\(y\) + 5\(x\)) – (2\(x\)2\(y\) – 2\(x\)\(y\)2 + 2\(y\))
= 5\(x\)3 – 5\(x\)2\(y\) + 5\(x\) – 2\(x\)2\(y\) + 2\(x\)\(y\)2 – 2\(y\)
= 5\(x\)3 – 7\(x\)2\(y\) + 5\(x\) + 2\(x\)\(y\)2 – 2\(y\)
b. (\(x\) – 1)(\(x\) + 1)(\(x\) + 2)
= (\(x^2\) + \(x\) – \(x\) – 1)(\(x\) + 2)
= (\(x^2\) – 1)(\(x\) + 2)
= \(x^2\)( \(x\) + 2) – 1.(\(x\) +2)
= \(x^3\) + 2\(x\) – \(x\) – 2
c. \(\frac{1}{2}\).\(x\) 2\(y\)2(2\(x\) + \(y\))(2\(x\) – \(y\))
= \(\frac{1}{2}\).\(x\)2\(y\)2 (4\(x\)2 – 2\(x\) \(y\)+ 2\(x\) \(y\)– \(y\)2)
= \(\frac{1}{2}\).\(x\)2\(y\)2 (4\(x\)2 – \(y\)2)
= \(\frac{1}{2}\).\(x\)2.\(y\)2.4\(x\)2 + \(\frac{1}{2}\).\(x\)2\(y\)2. (-\(y\)2)
= 2\(x\)4\(y\)2 - \(\frac{1}{2}\).\(x\)2\(y\)4