Question

va rog mult ex 1 !!!! 100 pct

Answer 1

$a).(2a+3b)^2=(2a)^2+2 \cdot 2a \cdot 3b+(3b)^2=4a^2+12ab+9b^2 \\ b).(2x^2+1)^2=(2x^2)^2+2 \cdot 2x^2 \cdot 1+1^2=4x^4+4 x^2+1 \\ c).(6x^3y^2+5x^2y)^2=(6x^3y^2)^2+2 \cdot 6x^3y^2 \cdot 5x^2y+(5x^2y)^2= \\ =36x^6y^4+60x^5y^3+25x^4y^2$
$d).(3 \frac{2}{3} a+1 \frac{2}{5} b)^2=( \frac{11}{3} a+ \frac{7}{5} b)^2=( \frac{11}{3} a)^2+2 \cdot \frac{11}{3} a \cdot \frac{7}{5} b+( \frac{7}{5} b)^2= \\ \\ = \frac{121}{9} a^2+ \frac{154}{15} ab+ \frac{59}{25} b^2= \frac{3025a^2+2310ab+441b^2}{225}$

$e).(1,2x^2y+0,5x^3y^2)^2=(1,2x^2y)^2+2 \cdot 1,2x^2y \cdot 0,5x^3y^2+(0,5x^3y^2)^2 \\ =1,44x^4y^2+1,2x^5y^3+0,25x^6y^4$
$f).(3x-2y)^2=(3x)^2-2 \cdot 3x \cdot 2y+(2y)^2=9x^2-12xy+4y^2$
$g).(5a^2b-2ab)^2=(5a^2b)^2-2 \cdot 5a^2b \cdot 2ab+(2ab)^2= \\ =25a^4b^2-20a^3b^2+4a^2b^2$
$h).( \frac{2}{3} x^3y^4-2 \frac{1}{2} x^5y)^2=( \frac{2}{3} x^3y^4- \frac{5}{2} x^5y)^2= \\ \\ =( \frac{2}{3} x^3y^4)^2-2 \cdot \frac{2}{3} x^3y^4 \cdot \frac{5}{2} x^5y+( \frac{5}{2} x^5y)^2= \\ \\ = \frac{4}{9} x^6y^8- \frac{20}{6} x^8y^5+ \frac{25}{4} x^1^0y^2= \frac{16x^6y^8-120x^8y^5+225x^1^0y^2}{36}$