a: A+2xy^2-x^2y-B=3x^2y-4xy^2

=>A-B=3x^2y-4xy^2-2xy^2+x^2y=4x^2y-6xy^2

=>A=4x^2y; B=6xy^2

b: 5xy^2-A-6x^2y+B=-7xy^2+8x^2y

=>-A+B=-7xy^2+8x^2y-5xy^2+6x^2y=14x^2y-12xy^2

=>A=12xy^2; B=14x^2y

c: 5xy^3-A-5/8x^3y+B=2+1/4xy^3-7/6x^3y

=>-A+B=2+1/4xy^3-7/6x^3y-5xy^3+5/8x^3y

=>B-A=-19/4xy^3-13/24x^3y+2

=>B=-19/4xy^3; A=13/24x^3y-2

Có : \(2x^2+9x-11=0\)

\(2x^2-2x+11x-11=0\)

\(\Rightarrow2x\left(x-1\right)+11\left(x-1\right)=0\)

\(\Rightarrow\left(2x+11\right).\left(x-1\right)=0\)

=> 2x + 11 =0 hoặc x-1 = 0

=> x = \(\dfrac{-11}{2}\)hoặc x =1

Ta có  M = x³ + x²y - 2x² - xy - y² +3y + x + 2017

               = x²(x + y - 2) - y(x + y - 2) + x + y - 2 + 2019

thay x + y - 2 = 0 vào M ta có :  M = x².0 - y.0 + 0 + 2019

                                                      = 2019

\(M=x^3+x^2y-2x^2-xy-y^2+3y+x+2017\)

\(=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(y+x-2\right)+2019\)

\(=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-2\right)+2019\)

\(=\left(x+y-2\right)\left(x^2-y+1\right)+2019\)

Thay \(x+y-2=0\)vào đa thức ta được:

\(M=0.\left(x^2-y+1\right)+2019=2019\)

bai 1

=ax5-x5-9xy-4xy-7x

=ax5-(5x+7x)-(9xy+4xy)

=5ax-12x-13xy

2

M=4a+ab-2b+2a-2b+ab

=6a+2ab-4b

n=6a+2b-ab+2a

=8a+2b-ab

m-n=6a+2ab-4b-8a-2b+ab

=3ab-2a-6b

\(N=2x-2x^2-5=-2\left(x^2-x+\dfrac{5}{2}\right)\)

\(=-2\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{9}{4}\right)\)

\(=-2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\right]\)

\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le\dfrac{-9}{2}\)

Dấu " = " khi \(-2\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(MAX_N=\dfrac{-9}{2}\) khi \(x=\dfrac{1}{2}\)

Đặt A(x)=0

ta được:

2x-7+(x-14)=0

x+x-7+x-7-7=0

(x-7)+(x-7)+(x-7)=0

3(x-7)=0

x-7=0

x=7

Vậy x=7 là nghiệm của A(x). ticks mình nhe

A(x) = 2x - 7 + (x - 14)

Để đa thức A(x) có nghiệm thì A(x) = 0

hay 2x - 7 + (x - 14) = 0

\(\Leftrightarrow\) 2x - 7 + x - 14 = 0

\(\Leftrightarrow\) 3x - 21 = 0

\(\Leftrightarrow\) 3x = 21

\(\Leftrightarrow\) x = 7

Vậy x = 7 là nghiệm của đa thức A(x)

A(x) = 0 \(\Rightarrow\) 2x - 7 + ( x - 14 ) = 0

\(\Rightarrow\) 2x -7 + x -14 = 0

\(\Rightarrow\) ( 2x + x ) - ( 7 + 14 ) = 0

\(\Rightarrow\) 3x - 21 = 0 \(\Rightarrow\) 3x = 21 \(\Rightarrow\) x = 7

\(a,x^3y^2-xy^2=xy^2\left(x^2-1\right)=xy^2\left(x-1\right)\left(x+1\right)\\ b,2x^3y^2+4x^2y^2+2xy^2=2xy^2\left(x^2+2x+1\right)=2xy^2\left(x+1\right)^2\\ c,3x^3y-12x^2y+12xy=2xy\left(x^2-4x+4\right)=2xy\left(x-2\right)^2\\ d,6x^3y+12x^2y^2+6xy^3=6xy\left(x^2+2xy+y^2\right)=6xy\left(x+y\right)^2\\ e,x^2\left(x-y\right)+y^2\left(y-x\right)=\left(x^2-y^2\right)\left(x-y\right)=\left(x-y\right)^2\left(x+y\right)\\ f,9x^2\left(x-2\right)-4y^2\left(x-2\right)=\left(9x^2-4y^2\right)\left(x-2\right)=\left(3x-2y\right)\left(3x+2y\right)\left(x-2\right)\)

Tick plz

a: \(x^3y^2-xy^2=xy^2\left(x^2-1\right)=xy^2\left(x-1\right)\left(x+1\right)\)

b: \(2x^3y^2+4x^2y^2+2xy^2=2xy^2\left(x^2+2x+1\right)=2xy^2\cdot\left(x+1\right)^2\)

c: \(3x^3y-12x^2y+12xy=3xy\left(x^2-4x+4\right)=3xy\cdot\left(x-2\right)^2\)

d: \(6x^3y+12x^2y^2+6xy^3=6xy\left(x^2+2xy+y^2\right)=6xy\cdot\left(x+y\right)^2\)

e: \(x^2\left(x-y\right)+y^2\left(y-x\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)

f: \(9x^2\left(x-2\right)-4y^2\left(x-2\right)=\left(x-2\right)\left(3x-2y\right)\left(3x+2y\right)\)

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