Lm giúp mình vs ^^ a) (2x^2+3)^2 b) (y^2x-3ab)^2 c) (x^2-6z) . (x^2 +6z) 13/08/2021 Bởi Gianna Lm giúp mình vs ^^ a) (2x^2+3)^2 b) (y^2x-3ab)^2 c) (x^2-6z) . (x^2 +6z)
$\ a)$ $\ (2x²+3)²$ $\ = (2x²)² +2x².2 .3 +3²$ $\ =$ $4x^{4}$ $\ +12x² + 9$ $\ b)$ $\ ($ $y^{2x}$ $\ -3ab)²$ $\ =$ $\ ($ $y^{2x}$ $\ )$ $\ ²$ $\ -2.3ab .$ $y^{2x}$ $\ + (3ab)²$ $\ =$ $y^{4x}$ $-6aby^{2x}$ $\ + 9a²b²$ $\ c)$ $\ ($ $x^{2-6z}$ $\ )$ $\ . ($ $x^{2+6z}$ $\ )$ $\ = ($ $x^{2}$ $\ )²$ $\ -($ $6z)^{2}$ $\ =$ $x^{4}$ $\ -36z²$ Bình luận
Đáp án: Giải thích các bước giải: a) (2x^2+3)^2 =(2x²)² +2x².2 .3 +3² =4x^4 +12x² +9 b) (y^2x-3ab)^2 =(y^2x)² -2.3ab .y^2x +(3ab)² =y^4x -6aby^2x +9a²b² c) (x^2-6z) . (x^2 +6z) =(x²)² -(6z)² =x^4 -36z² Bình luận
$\ a)$ $\ (2x²+3)²$
$\ = (2x²)² +2x².2 .3 +3²$
$\ =$ $4x^{4}$ $\ +12x² + 9$
$\ b)$ $\ ($ $y^{2x}$ $\ -3ab)²$
$\ =$ $\ ($ $y^{2x}$ $\ )$ $\ ²$ $\ -2.3ab .$ $y^{2x}$ $\ + (3ab)²$
$\ =$ $y^{4x}$ $-6aby^{2x}$ $\ + 9a²b²$
$\ c)$ $\ ($ $x^{2-6z}$ $\ )$ $\ . ($ $x^{2+6z}$ $\ )$
$\ = ($ $x^{2}$ $\ )²$ $\ -($ $6z)^{2}$
$\ =$ $x^{4}$ $\ -36z²$
Đáp án:
Giải thích các bước giải:
a) (2x^2+3)^2
=(2x²)² +2x².2 .3 +3²
=4x^4 +12x² +9
b) (y^2x-3ab)^2
=(y^2x)² -2.3ab .y^2x +(3ab)²
=y^4x -6aby^2x +9a²b²
c) (x^2-6z) . (x^2 +6z)
=(x²)² -(6z)²
=x^4 -36z²