phân tích đa thức thành nhân tử
6x^2+4xy-y^2
8x^2+3xy-4y^2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(-\left(x+2y\right)^2\)
\(-\left(x-3\right)^2\)
\(\left(3-5x\right)^2\)
\(-x^2-4xy-4y^2=-\left(x+2y\right)^2\)
\(-x^2+6x-9=-\left(x-3\right)^2\)
\(25x^2-30x+9=\left(5x-3\right)^2\)
a, x2+2xy+y2+2x+2y-15
<=> (x+y )2+2(x+y)+1-16
Đặt x+y =a
<=> a2+2a+1-42
<=> (a+1)2-42
<=> (a+5)(a-3) =>( x+y+5)(x+y-3)
b, x2-4xy+4y2-2x-4y-35
<=> (x-2y)2-2(x-2y)+1-36
Đặt (x-2y) =b
=> b2-2b+1-62
<=> (b-1)2-62
<=> (b-7)(b+5)=> (x-2y-7)(x-2y+5)
c,
a,A= x^2+2xy+y^2+2x+2y-15
= (x+y)^2+(x+y)-15
Đặt x+y=a, ta có:
A=a^2+2a-15
=a^2+2a+1-16
=(a+1)^2-4^2
=(a+1+4)(a+1-4)
=(a+5)(a-3)
Thay a=x+y, ta có: A=(x+y+5)(x+y-3).
\(x^2+4y^2+9-4xy-6x+12y\)
\(=\left(x^2-4xy+4y^2\right)+\left(-6x+12y\right)+9\)
\(=\left(x-2y\right)^2-6\left(x-2y\right)+9\)
\(=\left(x-2y-3\right)^2\)
3 - 6x + 3x^2
= 3 ( 1 - 2x + x^2 )
= 3( 1 - x )^2
b, x^2 - 4xy + 4y^2
= ( x)^2 + 2.x.2y + (2y)^2
= ( x+ 2y)^2
1: =(16x^2-8x+1)-y^2
=(4x-1)^2-y^2
=(4x-1-y)(4x-1+y)
2: =(x^2-2xy+y^2)-z^2
=(x-y)^2-z^2
=(x-y-z)(x-y+z)
3: =(x^2+4xy+4y^2)-16
=(x+2y)^2-4^2
=(x+2y-4)(x+2y+4)
4: =(x^2-4xy+4y^2)-16
=(x-2y)^2-4^2
=(x-2y-4)(x-2y+4)
a)\(6x-9-x^2\)
\(=-\left(x^2+6x+9\right)\)
\(=-\left(x+3\right)^2\)
b)\(x^2+4y^2+4xy\)
\(=\left(x+2y\right)^2\)
c)\(x^2+8x+16\)
\(=\left(x+4\right)^2\)
d)\(9x^2-12xy+4y^2\)
\(=\left(3x-2y\right)^2\)
e)\(-25x^2y^2+10xy-1\)
\(=-\left(25x^2y^2-10xy+1\right)\)
\(=-\left(5xy-1\right)^2\)
f)\(4x^2-4x+1\)
\(=\left(2x-1\right)^2\)
j)\(x^2+6x+9\)
\(=\left(x+3\right)^2\)
h)\(9x^2-6x+1\)
\(=\left(3x-1\right)^2\)
#H
a, 6x - 9 - x2 = - x2 + 6x - 9 = - (x2 - 6x + 9) = - (x - 3)2
b, x2 + 4y2 + 4xy = x2 + 2. x . 2y + (2y)2 = (x + 2y)2
c, x2 + 8x + 16 = x2 + 2 . x . 4 + 42 = (x + 4)2
d, 9x2 - 12xy + 4y2 = (3x)2 - 2 . 3x . 2y + (2y)2 = (3x - 2y)2
e, - 25x2y2 + 10xy - 1 = - (25x2y2 - 10xy + 1) = - [(5xy)2 - 2 . 5xy + 1] = - (5xy - 1)2
f, 4x2 - 4x + 1 = (2x)2 - 2 . 2x + 1 = (2x - 1)2
j, x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
h, 9x2 - 6x + 1 = (3x)2 - 2 . 3x + 1 = (3x - 1)2
\(6x^2+4xy-y^2\)
\(=6\left(x^2+\dfrac{2}{3}xy-\dfrac{1}{6}y^2\right)\)
\(=6\left(x^2+2\cdot x\cdot\dfrac{1}{3}y+\dfrac{1}{9}y^2-\dfrac{5}{18}y^2\right)\)
\(=6\left[\left(x+\dfrac{1}{3}y\right)^2-\left(\dfrac{y\sqrt{5}}{3\sqrt{2}}\right)^2\right]\)
\(=6\left(x+\dfrac{1}{3}y-\dfrac{y\sqrt{10}}{6}\right)\left(x+\dfrac{1}{3}y+\dfrac{y\sqrt{10}}{6}\right)\)
b: \(8x^2+3xy-4y^2\)
\(=8\left(x^2+\dfrac{3}{8}xy-\dfrac{1}{2}y^2\right)\)
\(=8\left(x^2+2\cdot x\cdot\dfrac{3}{16}y+\dfrac{9}{256}y^2-\dfrac{137}{256}y^2\right)\)
\(=8\left[\left(x+\dfrac{3}{16}y\right)^2-\left(\dfrac{y\sqrt{137}}{16}\right)^2\right]\)
\(=8\left(x+\dfrac{3}{16}y-\dfrac{y\sqrt{137}}{16}\right)\left(x+\dfrac{3}{16}y+\dfrac{y\sqrt{137}}{16}\right)\)