a. (x+2y)2)(x+2y)2)­­ =x2+4xy+4y2=x2+4xy+4y2

b. (x−3y)(x+3y)(x−3y)(x+3y) =x2−(3y)2=x2−9y2=x2−(3y)2=x2−9y2

c. (5−x)2(5−x)2 =52−10x+x2=25−10x+x2

chú tự kỉ à ?

a, x/5-y/2

=> 3x/15=2y/4=3x-2y/15-4=44/11=4

+, x/5=4 => x=20

+, y/2=4 => y=8

c, 4x=3y

=> x/3=y/4=x-y=3-4=11/-1=-11

+, x/3=-11 => x=-33

+, y/4=-11 => y=-44

4x = 3y nên \(\frac{x}{3}\)=\(\frac{y}{4}\)

Theo tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x}{3}\)=\(\frac{y}{4}\)=\(\frac{x-y}{3-4}\)=\(\frac{11}{-1}\)= -11

+) \(\frac{x}{3}\)= -11

      x = -11x3=-33

+) \(\frac{y}{4}\)= -11

      y = -11x4 =-44

 Vậy x = -33 ; y = -44

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

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

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

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

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

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

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

Bài 12:

a) \(\left(\dfrac{1}{2}x+4\right)^2\)

\(=\left(\dfrac{1}{2}x\right)^2+2\cdot\dfrac{1}{2}x\cdot4+4^2\)

\(=\dfrac{1}{4}x^2+4x+16\)

b) \(\left(7x-5y\right)^2\)

\(=\left(7x\right)^2-2\cdot7x\cdot5y+\left(5y\right)^2\)

\(=49x^2-70xy+25y^2\)

c) \(\left(6x^2+y^2\right)\left(y^2-6x^2\right)\)

\(=\left(y^2+6x^2\right)\left(y^2-6x^2\right)\)

\(=y^4-36x^4\)

d) \(\left(x+2y\right)^2\)

\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)

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

e) \(\left(x-3y\right)\left(x+3y\right)\)

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

\(=x^2-9y^2\)

f) \(\left(5-x\right)^2\)

\(=5^2-2\cdot5\cdot x+x^2\)

\(=25-10x+x^2\)

\(11,\)

\(a,\left(7x+4\right)^2-\left(7x+4\right)\left(7x-4\right)\)

\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)

\(=\left(7x+4\right).8=56x+32\)

\(b,\left(x+2y\right)^2-6xy\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x+2y-6xy\right)\)

Bài `12`

`(1/2x+4)^2`

`=(1/2x)^2 + 2 . 1/2x.4 + 4^2`

`= 1/4 x^2 +4x + 16`

__

`(7x-5y)^2`

`=(7x)^2-2.7x.5y+(5y)^2`

`= 49x^2 -  70xy   + 25y^2`

__

`(6x^2+y^2)(y^2-6x^2)`

`=(y^2+6x^2)(y^2-6x^2)`

`=(y^2)^2 - (6x^2)^2`

`=y^4-36x^4`

__

`(x+2y)^2`

`=x^2+ 2.x.2y+(2y)^2`

`= x^2 + 4xy +4y^2`

__

`(x-3y)(x+3y)`

`=x^2 - (3y)^2`

`=x^2 - 9y^2`

__

`(5-x)^2`

`=5^2 -2.5.x+x^2`

`=25 - 10x+x^2`

Bài `11`

`(7x+4)^2 -(7x+4)(7x-4)`

`= (7x+4)(7x+4) -(7x+4)(7x-4)`

`=(7x+4)(7x+4-7x+4)`

`=8(7x+4)`

`= 56x+32`

__

`(x+2y)^2-6xy (x+2y)`

`= (x+2y) (x+2y-6xy)`

\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

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

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

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

c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2

=>A-B=-5x^2y^3-x^3y^2

d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2

\(12x^2y-18xy^2-30y^2=6y\left(2x^2-3xy-5y\right)\)

\(d,5\left(x-y\right)-y\left(x-y\right)=\left(5-y\right)\left(x-y\right)\)