Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c) \(5x^2+3y+15x+xy=5x\left(x+3\right)+y\left(x+3\right)=\left(x+3\right)\left(5x+y\right)\)
d) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3-y\right)\left(x+3+y\right)\)
e) \(x^2-y^2+2x+1=\left(x^2+2x+1\right)-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
f) \(x^2-2xy-9+y^2=\left(x^2-2xy+y^2\right)-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
c: \(5x^2+15x+3y+xy\)
\(=5x\left(x+3\right)+y\left(x+3\right)\)
\(=\left(x+3\right)\left(5x+y\right)\)
d: \(x^2+6x+9-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)
e: \(x^2+2x+1-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1-y\right)\left(x+1+y\right)\)
f: \(x^2-2xy+y^2-9\)
\(=\left(x-y\right)^2-9\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
\(1,\\ a,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ b,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ c,=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\\ d,=x\left(x-2y\right)+t\left(x-2y\right)=\left(x+t\right)\left(x-2y\right)\\ 2,\\ \Rightarrow x^2-4x+4-x^2+9=6\\ \Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\\ 3,\\ a,x^2+2x+2=\left(x+1\right)^2+1\ge1>0\\ b,-x^2+4x-5=-\left(x-2\right)^2-1\le-1< 0\)
x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử
= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung
= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung
a) \(\dfrac{1}{x^3-8}=\dfrac{1}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
\(\dfrac{3}{4-2x}=\dfrac{-3}{2\left(x-2\right)}=\dfrac{-3\left(x^2+2x+4\right)}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
b) \(\dfrac{x}{x^2-1}=\dfrac{x}{\left(x+1\right)\left(x-1\right)}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2\left(x-1\right)}\)
\(\dfrac{1}{x^2+2x+1}=\dfrac{1}{\left(x+1\right)^2}=\dfrac{x-1}{\left(x+1\right)^2\left(x-1\right)}\)
c) \(\dfrac{1}{x+2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{5}{2-x}=\dfrac{-5}{x-2}=\dfrac{-5\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)^2}\)
d) \(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{\left(x-y\right)^2}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{2x}{x^2-y^2}=\dfrac{2x}{\left(x+y\right)\left(x-y\right)}=\dfrac{6x\left(x-y\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{x^2-xy+y^2}{x^2-2xy+y^2}=\dfrac{x^2-xy+y^2}{\left(x-y\right)^2}=\dfrac{3\left(x^2-xy+y^2\right)\left(x+y\right)}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{3\left(x^3+y^3\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
a) \(=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
\(=\left(x-1\right)^2\left(x^2+x+1\right)\)
b) \(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
c) Đổi đề: \(a^2x+a^2y-7x-7y\)
\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)
d) \(=x^2\left(a-b\right)+y\left(a-b\right)=\left(a-b\right)\left(x^2+y\right)\)
e) \(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\left(x^2-x+1\right)\)
g) \(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
h) \(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)
i) \(=\left(x+1\right)^2-4=\left(x+1-2\right)\left(x+1+2\right)=\left(x-1\right)\left(x+3\right)\)
a\(x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
b)\(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
d)\(=a\left(x^2+y\right)-b\left(x^2+y\right)=\left(x^2+y\right)\left(x-b\right)\)
e)\(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
g)\(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
h)\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)
i)\(=\left(x-1\right)^2-4=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\)
a) \(x^2-y^2-3x+3y\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-3\right)\)
b) \(2x+2y-x^2+y^2\)
\(=2\left(x+y\right)-\left(x^2-y^2\right)\)
\(=2\left(x+y\right)-\left(x-y\right)\left(x+y\right)\)
\(=\left(x+y\right)\left(2-x+y\right)\)
c) \(x^2-16+y^2+2xy\)
\(=x^2+y^2+2xy-16\)
\(=\left(x+y\right)^2-16\)
\(=\left(x+y+4\right)\left(x+y-4\right)\)
a) \(x^2-y^2-3x+3y\)
\(=\left(ax+y\right)\left(ax-y\right)-3.\left(x-y\right)\)
b) \(2x+2y-x^2+y^2\)
\(=2\left(x+y\right)-\left(x+y\right)\left(x-y\right)\)
c) \(x^2-16+y^2+2xy\)
\(=\left(x+y\right)\left(x-y\right)+2xy-16\)
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
a) Cách 1.
Ta có 2xy + 3z + 6y + xz = (2xy + xz) + (3z + 6y)
= x(2 y + z)+3(z + 2 y) = (z + 2y)(x + 3).
Cách 2.
Ta có 2xy + 3z + 6y + xz = (2x1/ + 6y) + (3z + xz)
= 2y(x + 3) + z(3 + x) = (z + 2y)(x + 3).
b) Biến đổi được a 4 - 9 rt 3 + a 2 -9a = (a- 9)a( a 2 +1).
c) Biến đổi được 3 x 2 + 5y - 3xy + (-5x) = (x - y)(3x - 5).
d) Biến đổi được x 2 - (a + b)x + ab = (x- a)(x - b).
e) Ta có 4 x 2 - 4xy + y 2 – 9 t 2 = ( 2 x - y ) 2 - ( 3 t ) 2
= (2x - y - 3t )(2x - y + 31).
g) Ta có x 3 - 3 x 2 y + 3 xy 2 - y 3 - z 3
= ( x - y ) 3 - z 3 = (x - y - z)( x 2 + y 2 + z 2 - 2xy + xz - yz).
h) Ta có x 2 - y 2 + 8x + 6y+ 7 = ( x 2 +8x + 16) - ( y 2 - 6y+ 9)
= ( x + 4 ) 2 - ( y - 3 ) 2 =(x-y + 7)(x + y + l).
a: \(\dfrac{\left(x+1\right)}{x^2+2x-3}=\dfrac{\left(x+1\right)}{\left(x+3\right)\cdot\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+5\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)\left(x+5\right)}\)
\(\dfrac{-2x}{x^2+7x+10}=\dfrac{-2x}{\left(x+2\right)\left(x+5\right)}=\dfrac{-2x\left(x+3\right)\left(x-1\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x-1\right)}\)
b: \(\dfrac{x-y}{x^2+xy}=\dfrac{x-y}{x\left(x+y\right)}=\dfrac{y^2\left(x-y\right)}{xy^2\left(x+y\right)}\)
\(\dfrac{2x-3y}{xy^2}=\dfrac{\left(2x-3y\right)\left(x+y\right)}{xy^2\left(x+y\right)}\)
c: \(\dfrac{x-2y}{2}=\dfrac{\left(x-2y\right)\left(x-xy\right)}{2\left(x-xy\right)}\)
\(\dfrac{x^2+y^2}{2x-2xy}=\dfrac{x^2+y^2}{2\left(x-xy\right)}\)
1: \(x^2-xy+x-y\)
\(=x\left(x-y\right)+\left(x-y\right)\)
=(x-y)(x+1)
2: \(xy-2x-y^2+2y\)
=x(y-2)-y(y-2)
=(y-2)(x-y)
3: \(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2=\left(x+2+y\right)\left(x+2-y\right)\)
4: \(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2=\left(x+2+y\right)\left(x+2-y\right)\)
5: \(x^2-2xy+y^2-4\)
\(=\left(x^2-2xy+y^2\right)-4\)
\(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
6: \(xz+yz-5x-5y\)
=z(x+y)-5(x+y)
=(x+y)(z-5)
7: \(x^2-2xy+y^2-x+y\)
\(=\left(x-y\right)^2-\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-1\right)\)
8: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
=(x+y)(x-y-2)
9: \(x^2-1-2xy+2y\)
\(=\left(x-1\right)\left(x+1\right)-2y\left(x-1\right)\)
\(=\left(x-1\right)\left(x+1-2y\right)\)
10: \(x^2-3x+xy-3y\)
\(=x\left(x-3\right)+y\left(x-3\right)\)
\(=\left(x-3\right)\left(x+y\right)\)
8) x² - y² - 2x + 2y
= (x² - y²) - (2x - 2y)
= (x - y)(x + y) - 2(x - y)
= (x - y)(x + y - 2)