a) \(B=\left(x^2+2x+1\right)+\left(y^2-2.2.y+2^2\right)=\left(x+1\right)^2+\left(y-2\right)^2\)
thay x=99 và y=102 vào B ta có:
\(B=\left(99+1\right)^2+\left(102-2\right)^2=100^2-100^2=0\)
b)
b) \(2x^2+16x+32-2y^2=2\left(x^2+8x+16-y^2\right)=2\left(\left(x+4\right)^2-y^2\right)=2\left(x+4-y\right)\left(x+4+y\right)\)
\(x^2-3x+2x-6=x\left(x-3\right)+2\left(x-3\right)=\left(x-3\right)\left(x+2\right)=0\)
nếu x-3=0=>x=3
nếu x+2=0=>x=-2
nhớ k cả 3 nhé thanks bn nhìu
a)
B= x^2+2x+1+y^2-4y+4
= (x+1)^2+(y-2)^2
=(x+1+y-2)*(x+1-y+2)
thay x=99 va y=102 vao bieu thuc
(99+1+102-2)*(99+1-102+2)=-4*0=0
b)2x^2-2y^2+16x+32
=2(x^2-y^2)+16(x+2)
=2(x-y)(x+y)+16(x+2)
c) x^2-3x+2x-6=0
x(x-3)+2(x+3)=0
(x+3)(x+2)=0
=> x+3=0 hoac x+2=0
x+3=0
x = -3
x+2=0
x =-2