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AH
Akai Haruma
Giáo viên
17 tháng 9 2021

Đề thiếu. $C$ phải có quan hệ gì với $A,B$ thì mới tìm được chứ bạn. Bạn coi lại.

Đề thiếu rồi bạn

10 tháng 9 2021

a) \(C=A+B=x^2-2y+xy+1+x^2+y-x^2y^2-1=2x^2-y+xy-x^2y^2\)

b) \(C+A=B\)

\(\Rightarrow C=B-A=x^2+y-x^2y^2-1-x^2+2y-xy-1=3y-x^2y^2-xy-2\)

10 tháng 9 2021

a, C= A+B= x- 2y + xy + 1+x+ y - x2y- 1

                 = (x2 +x 2) +(-2y +y) + xy -x2y2+(1-1)

                 = 2x2 -y +xy - x2y2

b, C+A=B => C = B- A= x+ y - x2y- 1-(x- 2y + xy + 1)

                                    = x+ y - x2y- 1-x2  + 2y - xy - 1)

                                     = (x2-x2)+(y+2y)-x2y2-xy+(-1-1)

                                      = 3y-x2y2-xy-2

Hoctot

17 tháng 7 2021

a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)

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

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

 

a) Ta có: \(x^4+2x^3-4x-4\)

\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)

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

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

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

25 tháng 6 2017

9 tháng 9 2021

\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)

9 tháng 9 2021

\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)

 

12 tháng 10 2023

a: \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)

b: \(4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)

c: \(2x-1-x^2\)

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

d: \(x^2+x+\dfrac{1}{4}=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)

e: \(9-x^2=3^2-x^2=\left(3-x\right)\left(3+x\right)\)

g: \(\left(x+5\right)^2-4x^2=\left(x+5+2x\right)\left(x+5-2x\right)\)

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

h: \(\left(x+1\right)^2-\left(2x-1\right)^2\)

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

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

i: \(=x^2y^2-4xy+4-3\)

\(=\left(xy-2\right)^2-3=\left(xy-2-\sqrt{3}\right)\left(xy-2+\sqrt{3}\right)\)

k: \(=y^2-\left(x-1\right)^2\)

\(=\left(y-x+1\right)\left(y+x-1\right)\)

l: \(=x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=\left(x+2\right)^3\)

m: \(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2-y^3=\left(2x-y\right)^3\)

4 tháng 9 2016

x2 + y2 - x2y2 + xy - x - y = (x2-x) + (y2-y) + (-x2y2 + xy) = x(x+1) + y(y+1) + xy(xy+1) = ( x+ y+ xy)( x + 1 + y + 1 + xy + 1)

2 tháng 3 2020

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

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

\(=x\left(x+1\right)+y\left(y+1\right)+xy\left(xy+1\right)\)

\(=\left(x+y+xy\right)\left(x+1+y+1+xy+1\right)\)

18 tháng 7 2021

a) \(x^2-2x-4y^2-4y=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)

b) \(x^3+2x^2+2x+1=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=\left(x+1\right)\left(x^2-x+1+2x\right)=\left(x+1\right)\left(x^2+x+1\right)\)

c) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)

d) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)

a) Ta có: \(x^2-2x-4y^2-4y\)

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

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

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

b) Ta có: \(x^3+2x^2+2x+1\)

\(=\left(x^3+1\right)+2x\left(x+1\right)\)

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

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

29 tháng 10 2023

a) \(x^2+2x+1=x^2+2\cdot x\cdot1+1^2=\left(x+1\right)^2\)

b) \(x^2-4x+4=x^2-2\cdot x\cdot2+2^2=\left(x-2\right)^2\)

c) \(x^2+6xy+9y^2=x^2+2\cdot x\cdot3y+\left(3y\right)^2=\left(x+3y\right)^2\)

d) \(z^2-z+\dfrac{1}{4}=z^2-2\cdot z\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(z-\dfrac{1}{2}\right)^2\)

e) \(25x^2-10x+1=\left(5x\right)^2-2\cdot5x\cdot1+1^2=\left(5x-1\right)^2\)

25 tháng 12 2018

a) (x - y)(x + y + 3).                    b) (x + y - 2xy)(2 + y + 2xy).

c) x 2 (x + l)( x 3  -  x 2  + 2).              d) (x – 1 - y)[ ( x   -   1 ) 2   +   ( x   -   1 ) y   +   y 2 ].