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a: \(\dfrac{9x^3y^2-4xy^2+5x}{2x}=\dfrac{9}{2}x^2y^2-2y^2+\dfrac{5}{2}\)

b: \(\left(\dfrac{3}{4}x^3y^6+\dfrac{6}{5}x^4y^3-\dfrac{9}{10}x^5y\right):\dfrac{-3}{5}x^3y\)

\(=y^5\cdot\left(\dfrac{3}{4}:\dfrac{-3}{5}\right)-xy^2\cdot\left(\dfrac{6}{5}:\dfrac{3}{5}\right)+\dfrac{9}{10}:\dfrac{3}{5}\cdot x^2\)

\(=\dfrac{-5}{4}y^5-2xy^2+\dfrac{3}{2}x^2\)

NV
30 tháng 10 2019

\(A=x^2-xy+\frac{y^2}{4}+\frac{3}{4}\left(y^2-4y+4\right)+2013\)

\(=\left(x-\frac{y}{2}\right)^2+\frac{3}{4}\left(y-2\right)^2+2013\ge2013\)

\(B\) đề thiếu

\(C\) đề sai, dấu của \(y^2\) là âm thì không tồn tại GTNN

\(P=-\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)+7\)

\(=-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\)

\(2Q=-4x^2-20y^2+12xy+8x-6y+4\)

\(=-\left(4x^2+9y^2+4-12xy-8x+12y\right)-11\left(y^2-\frac{6}{11}y+\frac{36}{121}\right)+\frac{97}{11}\)

\(=-\left(2x-3y-2\right)^2-11\left(y-\frac{3}{11}\right)^2+\frac{97}{11}\le\frac{97}{11}\)

\(\Rightarrow Q\le\frac{97}{22}\)

a) Ta có: \(4x^2-6x\)

\(=2x\left(2x-3\right)\)

b) Ta có: \(9x^4y^3+3x^2y^4\)

\(=3x^2y^3\left(3x^2+y\right)\)

c) Ta có: 3(x-y)-5x(y-x)

=3(x-y)+5x(x-y)

=(x-y)(3+5x)

d) Ta có: \(x^3-2x^2+5x\)

\(=x\left(x^2-2x+5\right)\)

e) Ta có: \(5\left(x+3y\right)-15x\left(x+3y\right)\)

\(=\left(x+3y\right)\left(5-15x\right)\)

\(=5\left(x+3y\right)\cdot\left(1-3x\right)\)

f) Ta có: \(2x^2\left(x+1\right)+4\left(x+1\right)\)

\(=\left(x+1\right)\left(2x^2+4\right)\)

\(=2\left(x+1\right)\left(x^2+2\right)\)

6 tháng 4 2020

\(a.\left(8x^4-4x^3+x^2\right):2x^2=4x^2-2x+\frac{1}{2}\)

\(b.\left(2x^4-x^3+3x^2\right):\left(-\frac{1}{3x^2}\right)=-6x^6+3x^5-9x^4\)

\(c.\left(-18x^3y^5+12x^2y^2-6xy^3\right):6xy=-3x^2y^4+2xy-y^2\)

\(d.\left(\frac{3}{4x^3y^6}+\frac{6}{5x^4y^5}-\frac{9}{10x^5y}\right):-\frac{3}{5x^3y}=-\frac{5}{4y^5}-\frac{2}{xy^4}-\frac{3}{2x^2}\)

6 tháng 4 2020

Thank you

8 tháng 8 2019

1/x^3 - 2x^2 - 9x + 18

= x\(^2\)( x - 2 ) - 9 ( x - 2 ) = ( x\(^2\) - 9 ) ( x - 2 )= ( x - 3 ) ( x +3 ) ( x - 2 )

2/3x^2 -5x - 3y^2 + 5y

= 3( x\(^2\) - y\(^2\) ) - 5 ( x - y ) = 3 ( x - y ) ( x + y ) - 5 ( x - y ) = ( x - y ) [ 3( x+ y ) - 5 ]

= ( x - y ) ( 3x + 3y - 5 )

3/49 - x^2 + 2xy - y^2

= 49 - ( x\(^2\) - 2xy + y\(^2\) ) = 49 - ( x - y )\(^2\) = ( 7 - x + y ) ( 7 + x - y )

5/ x^2 - 4x^2y^2 + 2xy

= x ( x - 4xy\(^2\) + 2y )

6/ 3x - 3y - x^2 + 2xy - y^2

= ( 3x - 3y ) - ( x\(^2\) - 2xy + y\(^2\) ) = 3 ( x - y ) - ( x - y )\(^2\) = ( x - y ) ( 3 - x + y )

a: \(\dfrac{-6x^3y^4+4x^4y^3}{2x^3y^3}\)

\(=\dfrac{-6x^3y^4}{2x^3y^3}+\dfrac{4x^4y^3}{2x^3y^3}\)

\(=-3y+2x\)

b: \(\dfrac{5x^4y^2-x^3y^2}{x^3y^2}=\dfrac{5x^4y^2}{x^3y^2}-\dfrac{x^3y^2}{x^3y^2}\)

\(=5x-1\)

c: \(\dfrac{27x^3y^5+9x^2y^4-6x^3y^3}{-3x^2y^3}\)

\(=-\dfrac{27x^3y^5}{3x^2y^3}-\dfrac{9x^2y^4}{3x^2y^3}+\dfrac{6x^3y^3}{3x^2y^3}\)

\(=-9xy^2-3y+2x\)

30 tháng 8 2023

a) \(\dfrac{-6x^3y^4+4x^4y^3}{2x^3y^3}\)

\(=\dfrac{2x^3y^3\cdot\left(-3y+2x\right)}{2x^3y^3}\)

\(=-3y+2x\)

\(=2x-3y\)

b) \(\dfrac{5x^4y^2-x^3y^2}{x^3y^2}\)

\(=\dfrac{5x\cdot x^3y^2-x^3y^2\cdot1}{x^3y^2}\)

\(=\dfrac{x^3y^2\cdot\left(5x-1\right)}{x^3y^2}\)

\(=5x-1\)

c) \(\dfrac{27x^3y^5+9x^2y^4-6x^3y^3}{-3x^2y^3}\)

\(=\dfrac{-3x^2y^3\cdot-9xy^2+-3x^2y^3\cdot-3y+-3x^2y^3\cdot2x}{-3x^2y^3}\)

\(=\dfrac{-3x^2y^3\cdot\left(-9xy^2-3y+2x\right)}{-3x^2y^3}\)

\(=-9xy^2-3x+2x\)

Đề bài là gì sao không ghi rõ?? 

8 tháng 9 2021

\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)