K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

Bài 1:

a: ĐKXĐ: \(x+4\ne0\)

=>\(x\ne-4\)

b: ĐKXĐ: \(2x-1\ne0\)

=>\(2x\ne1\)

=>\(x\ne\dfrac{1}{2}\)

c: ĐKXĐ: \(x\left(y-3\right)\ne0\)

=>\(\left\{{}\begin{matrix}x\ne0\\y-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\y\ne3\end{matrix}\right.\)

d: ĐKXĐ: \(x^2-4y^2\ne0\)

=>\(\left(x-2y\right)\left(x+2y\right)\ne0\)

=>\(x\ne\pm2y\)

e: ĐKXĐ: \(\left(5-x\right)\left(y+2\right)\ne0\)

=>\(\left\{{}\begin{matrix}x\ne5\\y\ne-2\end{matrix}\right.\)

 Bài 2:

a: \(\dfrac{-12x^3y^2}{-20x^2y^2}=\dfrac{12x^3y^2}{20x^2y^2}=\dfrac{12x^3y^2:4x^2y^2}{20x^2y^2:4x^2y^2}=\dfrac{3x}{5}\)

b: \(\dfrac{x^2+xy-x-y}{x^2-xy-x+y}\)

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

\(=\dfrac{x\left(x+y\right)-\left(x+y\right)}{x\left(x-y\right)-\left(x-y\right)}=\dfrac{\left(x+y\right)\left(x-1\right)}{\left(x-y\right)\left(x-1\right)}\)

\(=\dfrac{x+y}{x-y}\)

c: \(\dfrac{7x^2-7xy}{y^2-x^2}\)

\(=\dfrac{7x\left(x-y\right)}{\left(y-x\right)\left(y+x\right)}\)

\(=\dfrac{-7x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{-7x}{x+y}\)
d: \(\dfrac{7x^2+14x+7}{3x^2+3x}\)

\(=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)

\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)

e: \(\dfrac{3y-2-3xy+2x}{1-3x-x^3+3x^2}\)

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

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

g: \(\dfrac{x^2+7x+12}{x^2+5x+6}\)

\(=\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x+3\right)\left(x+2\right)}\)

\(=\dfrac{x+4}{x+2}\)

 

10 tháng 6 2018

Bài 1:

a) \(x^2+10x+26+y^2+2y=(x^2+10x+25)+(y^2+2y+1)\)

..................................................= \(\left(x+5\right)^2+\left(y+1\right)^2\)

b) \(z^2-6z+5-t^2-4t=(z^2-6t+9)-(t^2+4t+4)\)

............................................= \(\left(z-3\right)^2-\left(t+2\right)^2\)

c) \(x^2-2xy+2y^2+2y+1=(x^2-2xy+y^2)+(y^2+2y+1)\)

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

d) \(4x^2-12x-y^2+2y+8=\left(4x^2-12x+9\right)-\left(y^2-2y+1\right)\)

.................................................= \(\left(2x-3\right)^2-\left(y-1\right)^2\)

10 tháng 6 2018

Bài 2:

a) \(\left(x+y+4\right)\left(x+y-4\right)=\left(x+y\right)^2-16\)

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

c) \(\left(y+2z-3\right)\left(y-2z+3\right)=y^2-\left(2z-3\right)^2\)

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

16 tháng 7 2017

giải luôn nhé 

A= -2x+4y-6z+3x+6y-6-3z

  =x+10y-9z-6

B=4x-6y+8z-4x+12y-4z-5z+5x

  =5x+6y-z

chúc bạn hk giỏi!!!

16 tháng 7 2017

A = \(-2\left(x-2y+3z\right)-3\left(-x-2y+2\right)-3z\)

A = \(-2x+4y-6z+3x+6y-6-3z\)

A = \(\left(-2x+3x\right)+\left(4y+6y\right)-\left(6z-3z\right)-6\)

A = \(-x+10y-2z-6\)

B = \(2\left(2x-3y+4z\right)-4\left(x-3y+z\right)-5\left(z-x\right)\)

B = \(4x-6y+8z-4x+12y-4z-5z+5x\)

B = \(\left(4x-4x+5x\right)-\left(6y+12y\right)+\left(8z-4z-5z\right)\)

B = \(5x-18y-1z\)

3 tháng 9 2016

1a/ z2 - 6z + 5 - t2 - 4t = z2 - 2 . 3z + 32 - 4 - t2 - 4t = (z2 - 2 . 3z + 32) - (22 + 2 . 2t + t2) = (z - 3)2 - (2 + t)2

b/ x2 - 2xy + 2y2 + 2y2 + 1 = x2 - 2xy + y2 + y2 + 2y + 1 = (x2 - 2xy + y2) + (y2 + 2y + 1) = (x - y)2 + (y + 1)2

c/ 4x2 - 12x - y2 + 2y + 8 = (2x)2 - 12x - y2 + 2y + 32 - 1 = [ (2x)2 - 2 . 3 . 2x + 32 ] - (y2 - 2y + 1) = (2x - 3)2 - (y - 1)2

3 tháng 9 2016

2a/ (x + y + 4)(x + y - 4) = x2 + xy - 4x + xy + y2 - 4y + 4x + 4y + 16 = x2 + (xy + xy) + (-4x + 4x) + (-4y + 4y) + y2 + 16

= x2 + 2xy + y2 + 42 = (x + y)2 + 42

b/ (x - y + 6)(x + y - 6) = x2 + xy - 6x - xy - y2 + 6y + 6x + 6y - 36 = x2 + (xy - xy) + (-6x + 6x) + (6y + 6y) - y2 - 36

= x2 - y2 + 12y - 62 = x2 - (y2 - 12y + 62) = x2 - (y2 - 2 . 6y + 62) = x2 - (y - 6)2

c/ (y + 2z - 3)(y - 2z - 3) = y2 -2yz - 3y + 2yz - 4z2 - 6z - 3y + 6z + 9 = y2 + (-2yz + 2yz) + (-3y - 3y) + (-6z + 6z) - 4z2 + 9

= y2 - 6y - 4z2 + 9 = (y2 - 6y + 9) - 4z2 = (y - 3)2 - (2z)2

d/ (x + 2y + 3z)(2y + 3z - x) = 2xy + 3xz - x2 + 4y2 + 6yz - 2xy + 6yz + 9z2 - 3xz = (2xy - 2xy) + (3xz - 3xz) - x2 + (6yz + 6yz) + 9z2 + 4y2

= -x2 + 4y2 + 12yz + 9z2 = (4y2 + 12yz + 9z2) - x2 = [ (2y)2 + 2 . 2 . 3yz + (3z)2 ] - x2 = (2y + 3z)2 - x2

a: \(=\left(a+b\right)^2-\left(c+d\right)^2\)

b: \(=\left(a-d\right)^2-\left(b-c\right)^2\)

c: \(=\left(x+3z\right)^2-4y^2\)

d: \(=\left(a^2-9\right)\left(a^2+9\right)=a^4-81\)

e: \(=\left(a-5\right)^2\cdot\left(a+5\right)^2=\left(a^2-25\right)^2\)

Bài 3:

a: =>(2x-7)(x-2)=0

=>x=7/2 hoặc x=2

b: =>(x-1)(x+2)=0

=>x=1 hoặc x=-2

d: =>2x+3=0

hay x=-3/2

4 tháng 8 2016

A = (3x-2)^2-(x+3)^2

= 9x^2 - 12x + 4 - x^2 - 6x - 9

= 8x^2 - 18x - 5

B = (5x+3)^2+(x-2)^2

= 25x^2 + 30x + 9 + x^2 - 4x + 4

= 26x^2 +26x +13

C = (2x+y-3)^2-(x+2y+3)^2

= (2x + y)^2 - 6(2x + y) + 9 - (x + 2y)^2 - 6(x + 2y) - 9

= 4x^2 + 4xy + y^2 - 12x - 6y - x^2 - 4xy - 4y^2 - 6x - 12y 

= 3x^2 - 3y^2 -18x - 18y

D = (x+2y+3z)^2 -(x-2y-3z)^2

= (x + 2y)^2 + 6z(x + 2y) + 9z^2 - (x - 2y)^2 + 6z(x - 2y) - 9z^2

= x^2 + 4xy + y^2 + 6xz + 12yz - x^2 + 4xy - y^2 + 6xz - 12yz 

= 8xy + 12xz 

4 tháng 8 2016

A= (3x-2)^2-(x+3)^2

= (9x^2 - 12x + 4) - (x^2 + 6x +9)

= 8x^2 - 6x + 13

31 tháng 8 2016

Bạn phải bấm rõ mình mới giúp dc, nhìn vào ko hỉu lắm

31 tháng 8 2016

OK

ok