Question

(1/4а+в)^2 = (3с+5к)^2= 73^2=. (3с+1/2д)^2= (м+н)^2=​

Answer 1

$(\frac{1}{4}a+b)^2=(\frac{1}{4}a)^2+2*\frac{1}{4} a*b+b^2=\frac{1}{16}a^2+\frac{1}{2}ab+b^2\\\\(3c+5k)^2=(3c)^2+2*3c*5k+(5k)^2=9c^2+30ck+25k^2\\\\73^2=(70+3)^2=70^2+2*70*3+3^2=4900+420+9=5329\\\\(3c+\frac{1}{2}d)^2=(3c)^2+2*3c*\frac{1}{2}d+(\frac{1}{2}d)^2=9c^2+3cd+\frac{1}{4}d^2\\\\(m+n)^2=m^2+2mn+n^2$