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\(=16-\left(x^2-2xy+y^2\right)\)
\(=4^2-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x-y\right)\)
Ta có:
\(x^2+2xy+y^2-x-y-12=(x^2+2xy+y^2)-(x+y)-12\)
\(=(x+y)^2-(x+y)-12 \) \((*)\)
Đặt \(x+y=a\)
từ \((*)\Rightarrow a^2-a-12=(a^2+3a)-(4a+12)\)
\(=(a+3)(a-4)\)
Thay \(a=x+y\)
\(\Rightarrow (x+y+3)(x+y-4)\)
\(x^2-2xy+y^2-z^2\)
Áp dụng hằng đẳng thức:\(\left(a-b\right)^2=a^2-2ab+b^2\)
\(=\left(x-y\right)^2-z^2\)
Áp dụng hằng đẳng thức:\(a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
Bài giải:
a) x3 – 2x2 + x = x(x2 – 2x + 1) = x(x – 1)2
b) 2x2 + 4x + 2 – 2y2 = 2[(x2 + 2x + 1) – y2]
= 2[(x + 1)2 – y2]
= 2(x + 1 – y)(x + 1 + y)
c) 2xy – x2 – y2 + 16 = 16 – (x2 – 2xy + y2) = 42 – (x – y)2
= (4 – x + y)(4 + x – y)
a) \(x^3 - 2x^2 + x\) \(= x(x^2 - 2x + 1)\)
\(= x (x - 1 )^2\)
b) \(2x^2 + 4x + 2 - 2y^2\) \(= 2(x^2 + 2x + 1 - y^2)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1^2\right)-y^2\right]\)
\(= 2 (x+1-y) (x+1+y)\)
c) \(2xy - x^2 - y^2 + 16\) \(= - (x^2 - 2xy + y^2 - 4^2)\)
\(= - [(x^2 - 2xy + y^2) - 4^2]\)
\(= - [(x-y)^2 - 4^2 ]\)
\(= - (x - y - 4) (x- y + 4)\)
Cách 1: \(x^2-2xy+y^2+4x-4y-5=\left(y^2-xy+y\right)+\left(-xy+x^2-x\right)+\left(-5y+5x-5\right)\)
\(=y\left(y-x+1\right)-x\left(y-x+1\right)-5\left(y-x+1\right)=\left(y-x+1\right)\left(y-x-5\right)\)
Cách 2: \(x^2-2xy+y^2+4x-4y-5=\left(x^2+y^2+2^2-2xy+4x-4y\right)-9\)
\(=\left(y-x-2\right)^2-3^2=\left(y-x-2-3\right)\left(y-x-2+3\right)=\left(y-x-5\right)\left(y-x+1\right)\)
a, Ta có x2 - x - y2 - y
= ( x2 - y2 ) - ( x + y )
= ( x - y ).( x + y ) - ( x + y )
= ( x+ y ).( x - y -1 )
b, Ta có x2 - 2xy + y2 - z2
= ( x2 - 2xy + y2 ) - z2
= ( x - y )2 - z2
= ( x - y - z ).( x - y + z )
a) x2 - x - y2 - y = x2 - y2 - x - y
=(x - y) (x + y) - (x + y)
=(x + y) (x - y - 1)
b) x2 - 2xy + y2 - z2 = (x - y)2 - z2
=(x - y- z) (x - y + z)
a) x\(^2\)-x-y\(^2\)-y
=(x\(^2\)-y\(^2\)) - (x-y)
=xy(x-y) - (x-y)
=xy(x-y)
\(=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y+2xy\right)\left(x+y-2xy\right)\)
x2 -4x2y2 +y2 +2xy =( x2 +2xy +y2) -(2xy)2 =(x+y)2 -(2xy)2 =(x+y+2xy)(x+y-2xy)
16-2xy-x2-y2
= - (x2+2xy+y2-16)
=- ( ( x+y)2- 42 )
làm tiếp nha bạn !