Ta có:\(VP=\left(10a+5\right)^2=\left(10a\right)^2+2.10a.5+5^2\)

\(=100a^2+100a+25\)

\(=100a\left(a+1\right)+25\)

\(=VT\)

\(\Rightarrowđpcm\)

ta có: (10a+5)²=(10a)²+2.10a.5+5²

                      =100a²+100a+25

                      =100a.(a+1)+25

Ta có:\(1999.2001\)

\(=\left(2000-1\right)\left(2000+1\right)\)

\(=2000^2-1^2\)\(< 2000^2\)

\(\Rightarrow1999.2001< 2000^2\)

Ta có:

1999.2001=1999.(2000+1)=1999.2000+1999

2000^2=2000.2000=(1999+1).2000=1999.2000+2000

Vì 1999.2000+1999<1999.2000+2000 nên 1999.2001<2000^2

k giúp mk nhé

\(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y+x+y\right)\left(x+y-x+y\right)\)(Áp dụng hằng đẳng thức số 3)

\(=\left(2x+2y\right)2y\)

\(=4xy+4y^2\)

to k co biet

a) ab+bd-ac-cd

= (ab + bd) - (ac + cd)

= b(a+d) - c(a+d)

= (b-c)(a+d)

b) ax+by-ay-bx

= (ax - ay) - (bx - by)

= a(x-y) - b(x-y)

= (a-b)(x-y)

c) x²-xy-xy+y²

= x^2 - 2xy + y^2

= (x-y)^2

a) \(x^2-y^2=x^2-xy+xy-y^2=x.\left(x-y\right)+y.\left(x-y\right)=\left(x+y\right)\left(x-y\right)\)

b) \(\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=x\left(x^2-xy+y^2\right)+y\left(x^2-xy+y^2\right)\)

\(=x^3-x^2y+xy^2+x^2y-xy^2+y^3\)

\(=x^3+y^3\)