Question

Поднести до квадрата двочлен

Answer 1

#412
a) (m+2) 2= (m+2) (m+2) =m2+2m+2m+4=m2+4m+4
б) (2a+5x) 2= (2a+5x) (2a+5x)= 4a2+10ax+10ax+25x2=4a2+20ax+25x2
в) (3+a2) 2= (3+a2) (3+a2) =9+3a2+3a2+a4=9+6a2+a4
г) (x-1) 2=(x-1) (x-1) =x2-x-x+1=x2-2x+1
ґ) (2c-a) 2=(2c-a) (2c-a) =4c2-2ac-2ac+a2=4c2-4ac+a2
д) (5-x2) =(5-x2) (5-x2) =25-5x2-5x2+x4=25-10x2+x4
е) (1-ab) 2= (1-ab) (1-ab) =1-ab-ab+ab2=1-2ab+ab2
є) (cq-2p) 2= (cq-2p)(cq-2p)= cq2-2cqp-2cqp+4p2= cq2-4cqp+4p2
#413
а) (ax+5) 2=(ax+5)(ax+5)= ax2+5ax+5ax+25=ax2+10ax+25
б) (a+c2) 2= (a+c2)(a+c2)= a2+ac2+ac2+c4= a2+ac4+c4
в) (n+2a) 2= (n+2a)(n+2a)= n2+2an+2an+4a2= n2+ 4an+ 4a2
г) (3x+2y) 2= (3x+2y)(3x+2y)= 9x2+6xy+6xy+4y2=9x2+12xy+4y2
ґ) (5a+3b) 2= (5a+3b)(5a+3b)= 25a2+ 15ab+15ab+9b2= 25a2+30ab+9b2
д) (1+2abc) 2= (1+2abc)(1+2abc)= 1+2abc+2abc+4abc2= 1+4abc+4abc2
е) (4n+3c) 2= (4n+3c)(4n+3c)= 16n2+12nc+12nc+9c2= 16n2+ 24nc+9c2
є) (-2+5ac) 2= (-2+5ac)(-2+5ac)= 4-10ac-10ac+25ac2= 4-20ac+25ac2