Câu hỏi của Núi non tình yêu thuần k...

Question

Rút gọn các biểu thức sau:

$a,\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2$

$b,\left(3x^2-y\right)^2-\left(2x^2+y\right)^2$

Phạm Phương Anh · Accepted Answer

$a,\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2$

$=\left(3x+1-3x+5\right)^2$

$=6^2$

$=36$

$b,\left(3x^2-y\right)^2-\left(2x^2+y\right)^2$

$=\left(3x^2-y+2x^2+y\right)\left(3x^2-y-2x^2-y\right)$

$=\left(5x^2\right)\left(x^2-2y\right)$

$=5x^4-10x^2y$Câu trả lời: $a,\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2$

$=\left(3x+1-3x+5\right)^2$

$=6^2$

$=36$

$b,\left(3x^2-y\right)^2-\left(2x^2+y\right)^2$

$=\left(3x^2-y+2x^2+y\right)\left(3x^2-y-2x^2-y\right)$

$=\left(5x^2\right)\left(x^2-2y\right)$

$=5x^4-10x^2y$

Shinichi Kudo · Answer

a) $\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2$

$=\left[\left(3x+1\right)-\left(3x-5\right)\right]^2$

$=\left(3x+1-3x+5\right)^2$

$=6^2=36$

b) $\left(3x^2-y\right)^2-\left(2x^2+y\right)^2$

$=\left(3x^2-y-2x^2-y\right)\left(3x^2-y+2x^2+y\right)$

$=5x^2\left(x^2-2y\right)$

$=5x^4-10y$Câu trả lời: a) $\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2$

$=\left[\left(3x+1\right)-\left(3x-5\right)\right]^2$

$=\left(3x+1-3x+5\right)^2$

$=6^2=36$

b) $\left(3x^2-y\right)^2-\left(2x^2+y\right)^2$

$=\left(3x^2-y-2x^2-y\right)\left(3x^2-y+2x^2+y\right)$

$=5x^2\left(x^2-2y\right)$

$=5x^4-10y$

Phép nhân và phép chia các đa thức