Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a: \(2x^3+x^2-13x+6\)
\(=2x^3-4x^2+5x^2-10x-3x+6\)
\(=\left(x-2\right)\left(2x^2+5x-3\right)\)
\(=\left(x-2\right)\left(2x^2+6x-x-3\right)\)
\(=\left(x-2\right)\left(x+3\right)\left(2x-1\right)\)
b: \(2x^2+y^2-6x+2xy-2y+5=0\)
\(\Leftrightarrow x^2+2xy+y^2+x^2-4x+4-2x-2y+1=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-2\right)^2-2\left(x+y\right)+1=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(x+y-1\right)^2=0\)
=>x-2=0 và x+y-1=0
=>x=2 và y=-1
a) \(=x^2+2xy+y^2-x^2+y^2=2xy+2y^2=2y\left(x+y\right)\)
b) \(=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
c) \(=3\left[\left(x^2+2xy+y^2\right)-z^2\right]=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)
d) \(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
e) \(=\left(x-3\right)\left(x^2+3x+9\right)-2x\left(x-3\right)=\left(x-3\right)\left(x^2+x+9\right)\)
f) \(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)=\left(x+5\right)\left(x^2-6x+25\right)\)
a) \(\left(x+y\right)^2-\left(x^2-y^2\right)\)
\(=x^2+2xy+y^2-x^2+y^2\)
\(=2y^2+2xy\)
\(=2y\left(x+y\right)\)
c) \(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-x^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y+z\right)\left(x+y-z\right)\)
d) \(\left(2xy+1\right)^2-\left(2x+y\right)^2\)
\(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
\(=\left[\left(2xy+2x\right)+\left(y+1\right)\right]\left[\left(2xy-2x\right)-\left(y-1\right)\right]\)
\(=\left[2x\left(y+1\right)+\left(y+1\right)\right]\left[2x\left(y-1\right)-\left(y-1\right)\right]\)
\(=\left(2x+1\right)\left(y+1\right)\left(2x-1\right)\left(y-1\right)\)
\(=\left(4x^2-1\right)\left(y^2-1\right)\)
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
a) \(=2xy^2\left(x^2+8x+15\right)\)
\(=2xy^2\left[\left(x^2+8x+16\right)-1\right]\)
\(=2xy^2\left[\left(x+4\right)^2-1\right]\)
\(=2xy^2\left(x+4+1\right)\left(x+4-1\right)\)
\(=2xy^2\left(x+5\right)\left(x-3\right)\)
mấy câu sau tự làm nha :*
b,=(x^2-10x+25)-4
=(x-5)^2-2^2
=(x-5-2)(x-5+2)
=(x-7)(x-3)
a) Đặt x - y = a. Ta có:
(x - y)2 + 4(x - y) - 12
= a2 + 4a - 12
= (a2 - 2a) + (6a - 12)
= a(a - 2) + 6(a - 2)
= (a + 6)(a - 2)
= (x + y + 6)(x + y - 2)
a)\(\left(x-y\right)^2+4\left(x-y\right)-12=\left(x-y\right)^2-2\left(x-y\right)+6\left(x-y\right)-12\)
....................................................\(=\left(x-y\right)\left(x-y-2\right)+6\left(x-y-2\right)\)
....................................................\(=\left(x-y+6\right)\left(x-y-2\right)\)
b)\(x^2+y^2+3x-3y-2xy-10\)
\(=\left(x-y\right)^2+3\left(x-y\right)-10\)
\(=\left(x-y\right)\left(x-y+3\right)-10\)
\(=z\left(z+3\right)-10\) với \(z=x-y\)
\(=z^2+3z-10\)
\(=z^2-2z+5z-10\)
\(=z\left(z-2\right)+5\left(z-2\right)\)
\(=\left(z+5\right)\left(z-2\right)\)
f)\(x^2-6x-16=x^2+2x-8x-16=x\left(x+2\right)-8\left(x+2\right)=\left(x-8\right)\left(x+2\right)\)
g)\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=y\left(y+2\right)-24\) với \(y=x^2+7x+10\)
\(=y^2+2y-24\)
\(=y^2-4y+6y-24\)
\(=y\left(y-4\right)+6\left(y-4\right)\)
\(=\left(y+6\right)\left(y-4\right)\)
h)\(\left(x^2+6x+5\right)\left(x^2+10x+21\right)+15\)
\(=\left(x^2+x+5x+5\right)\left(x^2+3x+7x+21\right)+15\)
\(=\left[x\left(x+1\right)+5\left(x+1\right)\right]\left[x\left(x+3\right)+7\left(x+3\right)\right]+15\)
\(=\left(x+1\right)\left(x+5\right)\left(x+7\right)\left(x+3\right)+15\)
\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
\(=y\left(y+8\right)+15\) với \(y=x^2+8x+7\)
\(=y^2+8y+15\)
\(=y^2+3y+5y+15\)
\(=y\left(y+3\right)+5\left(y+3\right)\)
\(=\left(y+3\right)\left(y+5\right)\)