Bài 1: Rút gọn
a) -3x5 y 4 + 3x2 y 3 – 7x2 y 3 + 5x5 y 4
b) 5x – 7xy2 + 3x - 1 2 𝑥𝑦 2
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\(a,=4x^2+3xy-y^2+4xy-4x^2=7xy-y^2\\ b,=x^2-9-x^3+3x+x^2-3=-x^3+2x^2+3x-12\\ c,=-2x^2+12x-18+5x^2+4x-1=3x^2+16x-19\\ d,=8x^3+1-3x^3+6x^2=5x^3+6x^2+1\\ e,=\left(3x^2+4x+15x+20\right):\left(3x+4\right)\\ =\left(3x+4\right)\left(x+5\right):\left(3x+4\right)\\ =x+5\\ f,=\left(x^3+4x^2-3x+3x^2+12x-9+3x+3\right):\left(x^2+4x-3\right)\\ =\left[\left(x^2+4x-3\right)\left(x+3\right)+3x+3\right]:\left(x^2+4x-3\right)\\ =x+3\left(dư.3x+3\right)\)
a) \(=\dfrac{-x^2+2xy-y^2}{x^2-xy}=\dfrac{-\left(x-y\right)^2}{x\left(x-y\right)}=\dfrac{-x+y}{x}\)
b) \(=\dfrac{x^4-1}{2x-2}=\dfrac{\left(x^2-1\right)\left(x^2+1\right)}{2\left(x-1\right)}=\dfrac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{2\left(x-1\right)}=\dfrac{x^3+x^2+x+1}{2}\)
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) \(\left(3x^5y^2+4x^3y^3-5x^2y^4\right):2x^2y^2\)
\(=\dfrac{3x^5y^2+4x^3y^3-5x^2y^4}{2x^2y^2}\)
\(=\dfrac{2x^2y^2\left(\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\right)}{2x^2y^2}\)
\(=\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\)
mình làm bài 2 trước nha:
a) y.(a-b)+a.(y-b)=a.y-b.y+a.y-b.y
=(a.y+a.y)-(b.y+b.y)
=2.a.y-2.b.y
=2.y.(a-b)
b)x2.(x+y)-y.(x2-y2)=x3+x2.y-x2y+y3=x3+y3
a) \(\left(5x+3y\right)\left(5x-3y\right)+\left(4x-3y\right)^2\)
\(=25x^2-9y^2+16x^2-24xy+9y^2\)
\(=41x^2-24xy\)
b) \(\left(2x-3y\right)^3-\left(3x+2y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27xy^2-27x^3-54x^2y-36xy^2-8y^3\)
\(=-19x^3-90x^2y+18xy^2-35y^3\)
\(A=\left(-\dfrac{2}{3}x^3y^4\right)^2.\left(-3x^5y^2\right)^3\)
\(A=\left(\dfrac{4}{9}x^6y^8\right).\left(-27x^{15}y^6\right)\)
\(A=\left(\dfrac{4}{9}.-27\right)\left(x^6.x^{15}\right)\left(y^8.y^{16}\right)\)
\(A=-12x^{21}y^{24}\)
\(\text{Hệ số:-12}\)
\(\text{Bậc:45}\)
\(B=\left(3x^2y\right).\left(-\dfrac{1}{3}x^3y\right).\left(-\dfrac{1}{4}x^3y^4\right)\)
\(B=\left(3.-\dfrac{1}{3}.-\dfrac{1}{4}\right).\left(x^2.x^3.x^3\right).\left(y.y.y^4\right)\)
\(B=\dfrac{1}{4}x^8y^6\)
\(\text{Hệ số:}\dfrac{1}{4}\)
\(\text{Bậc:14}\)
bài 1 : a. x^3 +27 -54-x^3 =-27
b. 8x^3 +y^3 -8x^3 +y^3 =2y^3
c. (2x-1+2x+2)(2x-1-2x-2)=(4x+1).(-3)=-12x-3
d. a^3 +b^3 +3ab(a+b) -3ab(a+b)=a^3+b^3
Bài 1:
\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)
Bài 2:
\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)