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Bài 1:
a: \(=2x^2-3xy+5y^2\)
b: \(=\dfrac{2x^3-10x^2-11x^2+55x+12x-60}{x-5}=2x^2-11x+12\)
c: \(=\dfrac{6x^3+3x^2-10x^2-5x+4x+2}{2x+1}=3x^2-5x+2\)
c: \(=\dfrac{\left(x+3\right)^2-y^2}{x+y+3}=x+3-y\)
Bài 3:
\(\Leftrightarrow x^3+64-x^3+25x=264\)
hay x=8
\(1,C=6x^2+23x-55-6x^2-23x-21=-76\\ 2,=\left(2x^4-x^2+2x^3-x-6x^2+6-3\right):\left(2x^2-1\right)\\ =\left[\left(2x^2-1\right)\left(x^2+x-6\right)-3\right]:\left(2x^2-1\right)\\ =x^2+x-6\left(dư.-3\right)\\ 3,\Leftrightarrow x^3+64-x^3+25x=264\\ \Leftrightarrow25x=200\Leftrightarrow x=8\)
\(a,=4x^2+3xy-y^2+4xy-4x^2=7xy-y^2\\ b,=x^2-9-x^3+3x+x^2-3=-x^3+2x^2+3x-12\\ c,=-2x^2+12x-18+5x^2+4x-1=3x^2+16x-19\\ d,=8x^3+1-3x^3+6x^2=5x^3+6x^2+1\\ e,=\left(3x^2+4x+15x+20\right):\left(3x+4\right)\\ =\left(3x+4\right)\left(x+5\right):\left(3x+4\right)\\ =x+5\\ f,=\left(x^3+4x^2-3x+3x^2+12x-9+3x+3\right):\left(x^2+4x-3\right)\\ =\left[\left(x^2+4x-3\right)\left(x+3\right)+3x+3\right]:\left(x^2+4x-3\right)\\ =x+3\left(dư.3x+3\right)\)
a)-(x-y)(x2+xy-1)=-(x3+x2y-x-x2y-xy2+y)
=-(x3-xy2-x+y)
=-x3+xy2+x-y
b)x2(x-1)-(x3+1)(x-y)=x3-x2-x3+x2y-x+y
=-x2+x2y-x+y
c)(3x-2)(2x-1)+(-5x-1)(3x+2)=6x2-3x-4x+2-15x2-10x-3x-2
=-9x2-20x
d) hình như bạn ghi lỗi
Bài 2: C=x(x2-y)-x2(x+y)+y(x2-x)
=x3-xy-x3-x2y+x2y-xy
=-2xy
Thay x=1/2,y=-1 vào C, ta có:
C=-2.1/2.(-1)=1
Vậy C=1 khi x=1/2 và y=-1.
a) \(\left(2x^3-x^2+5x\right):x\)
\(=\dfrac{2x^3-x^2+5x}{x}\)
\(=\dfrac{x\left(2x^2-x+5\right)}{x}\)
\(=2x^2-x+5\)
b) \(\left(3x^4-2x^3+x^2\right):\left(-2x\right)\)
\(=\dfrac{3x^4-2x^3+x^2}{-2x}\)
\(=\dfrac{2x\left(\dfrac{3}{2}x^3-x^2+\dfrac{1}{2}x\right)}{-2x}\)
\(=-\left(\dfrac{3}{2}x^3-x^2+\dfrac{1}{2}x\right)\)
\(=-\dfrac{3}{2}x^3+x^2-\dfrac{1}{2}x\)
c) \(\left(-2x^5+3x^2-4x^3\right):2x^2\)
\(=\dfrac{-2x^5+3x^2-4x^3}{2x^2}\)
\(=\dfrac{2x^2\left(-x^3+\dfrac{3}{2}-2x\right)}{2x^2}\)
\(=-x^3-2x+\dfrac{3}{2}\)
d) \(\left(x^3-2x^2y+3xy^2\right):\left(-\dfrac{1}{2}x\right)\)
\(=\dfrac{x^3-2x^2y+3xy^2}{-\dfrac{1}{2}x}\)
\(=\dfrac{\dfrac{1}{2}x\left(2x^2-4xy+6y^2\right)}{-\dfrac{1}{2}x}\)
\(=-\left(2x^2-4xy+6y^2\right)\)
\(=-2x^2+4xy-6y^2\)
e) \(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:5\left(x-y\right)^2\)
\(=\dfrac{3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2}{5\left(x-y\right)^2}\)
\(=\dfrac{5\left(x-y\right)^2\left[\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\right]}{5\left(x-y\right)^2}\)
\(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)
f) \(\left(3x^5y^2+4x^3y^3-5x^2y^4\right):2x^2y^2\)
\(=\dfrac{3x^5y^2+4x^3y^3-5x^2y^4}{2x^2y^2}\)
\(=\dfrac{2x^2y^2\left(\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\right)}{2x^2y^2}\)
\(=\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\)
a: \(=x^3-2x^5\)
b: \(=5x^2-x^3+5-x\)
e: \(=\left(x-y\right)^3=x^3-3x^2y+3xy^2-y^3\)
a: \(=\dfrac{6x^3+13x^2-5x}{2x+5}=\dfrac{6x^3+15x^2-2x^2-5x}{2x+5}=3x^2-x\)
b: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)
\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)
\(=x^2-2x+3\)
d: \(=\dfrac{\left(x+y\right)^2}{x+y}=x+y\)
b>x.(y-x)+y.(y-x) xong nhoms hai cais vaof nhau laf duocj
de ma