Bài 1:
a) \(\left(a+b\right)^2-\left(a-b\right)^2\)
\(=\left(a+b+\left(a-b\right)\right).\left(a+b-\left(a-b\right)\right)\)
\(=2a.2b\)
\(=4ab\)
Câu 1:
a) (a +b )2 - ( a -b )2
=a2+b2-a2+b2
=2b2
b) (a + b )3- ( a - b )3 - 2b3
=a3+b3-a+b3-2b3
=a3-a
c) ( x+y+z)2 - 2(x+y+z)(x+y) + (x + y )2
=x2+xy+xz+xy+y2+yz+xz+yz+z2-2.(x2+xy+xz+xy+y2+yz)+x2+xy+xy+y2
=x2+y2+z2+2xy+2xz+2yz-2x2-2y2-4xy-2xz-2yz+x2+2xy+y2
=0
Câu 2:
a) x2 + 4x + 4
Tại x = 98 ta có:
982+4.98+4=98.(4+98)+4
=98.102+4
=9996+4
=10000
b) x3 + 3x2 + 3x +1
Tại x=99 ta có:
993+3.(992)+3.99+1=992.(99+3)+3.99+1
=992.102+298
=999702+298
=1000000
bài 2:
x\(^2\)+4x+4
=x\(^2\)+2.x.2+2\(^2\)
=(x+2)\(^2\)
Thay x=98 vào biểu thức :
ta có :(98+2)\(^2\)=100\(^2\)=1000
Bai1
a)(a+b)^2- ( a-b )^2
=[( a+ b)-(a- b)] .[( a+ b)+(a- b)]
=( a+ b- a+ b)(a+ b+ a- b)
