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\(a,=\left(x+y\right)^2-5^2=\left(x+y+5\right)\left(x+y-5\right)\)
\(b,=\left(a-1\right)^2-1-\left(2b+1\right)^2-1=\)
\(c,=\left(a-b\right)\left(a+b\right)-5\left(a+b\right)=\left(a+b\right)\left(a-b-5\right)\)
a,( x2 + 2xy + y2 ) -25
=( x + y )2 - 52
=( x + y + 5) ( x + y - 5)
b,
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)
c)5x-5y+ax-ay=5(x-y)+a(x-y)=(x-y)(5+a)
d)a3-a2x-ay+xy=a2(a-x)-y(a-x)=(a-x)(a2-y)
e)4x2-y2+4x+1=[(2x)2+2.2x.1+12]-y2=(2x+1)2-y2=(2x+1-y)(2x+1+y)
Chúc bạn học tốt!
\(a,\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)
\(=2x\left(4x+2\right)\)
\(=4x\left(2x+1\right)\)
\(b,6x-6y-x^2+xy\)
\(=\left(6x-6y\right)-\left(x^2-xy\right)\)
\(=6\left(x-y\right)-x\left(x-y\right)\)
\(=\left(x-y\right)\left(6-x\right)\)
Ta có
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)
5x^2 + 5xy - x - y
=5x.(x+y)-(x+y)
=(x+y)(5x-1)
7x - 6x^2 - 2
=-6x2+3x+4x-2
=-3x.(2x-1)+2.(2x-1)
=(2x-1)(2-3x)
Bài 1:
\(x^2-6x+9-y^2\)
\(=\left(x-3\right)^2-y^2\)
\(=\left(x-3+y\right)\left(x-3-y\right)\)
Bài 2:
\(x^2-x-12=0\)
\(\Leftrightarrow x^2-4x+3x-12=0\)
\(\Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+3=0\\x-4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=4\end{array}\right.\)
1. x2+6x-9-y2
=-(x2-6x+y2)-32
=-(x-y)2-32
=(-x+y-3)(-x+y+3)
a) 2a(x+y)+y+x = 2a(x+y)+(x+y)
= (2a+1)(x+y)
b) 2(x-y)^2-x+y = \(2\left(x-y\right)^2-\left(x-y\right)=\left(2x-2y-1\right)\left(x-y\right)\)