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Question

rút gọn các biểu thức saua, $\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2$b, $\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)$

Nguyễn Lê Phước Thịnh · Accepted Answer

a: $\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2$$=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2$$=\left(2x+1+2x-1\right)^2=\left(4x\right)^2=16x^2$b: $\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)$$=x^3+2x^2-x-2-x^3+8$$=2x^2-x+6$Câu trả lời: a: $\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2$$=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2$$=\left(2x+1+2x-1\right)^2=\left(4x\right)^2=16x^2$b: $\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)$$=x^3+2x^2-x-2-x^3+8$$=2x^2-x+6$

HT.Phong (9A5) · Answer

a) $\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2$$=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2$$=\left[\left(2x+1\right)+\left(2x-1\right)\right]^2$$=\left(2x+1+2x-1\right)^2$$=\left(4x\right)^2$$=16x^2$b) $\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)$$=\left(x^3+2x^2-x-2\right)-\left(x^3-8\right)$$=x^3+2x^2-x-2-x^3+8$$=2x^2-x+6$Câu trả lời: a) $\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2$$=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2$$=\left[\left(2x+1\right)+\left(2x-1\right)\right]^2$$=\left(2x+1+2x-1\right)^2$$=\left(4x\right)^2$$=16x^2$b) $\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)$$=\left(x^3+2x^2-x-2\right)-\left(x^3-8\right)$$=x^3+2x^2-x-2-x^3+8$$=2x^2-x+6$

Toru · Answer

$a,\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2$$=\left(2x+1\right)^2+2\cdot\left[\left(2x\right)^2-1^2\right]+\left(2x-1\right)^2$$=\left(2x+1\right)^2+2\cdot\left(2x+1\right)\cdot\left(2x-1\right)+\left(2x-1\right)^2$$=\left[\left(2x+1\right)+\left(2x-1\right)\right]^2$$=\left(2x+1+2x-1\right)^2$$=\left(4x\right)^2$$=16x^2$$---$$b,\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)$$=x^2\left(x+2\right)-\left(x+2\right)-\left(x^3-2^3\right)$$=x^3+2x^2-x-2-x^3+8$$=2x^2-x+6$Câu trả lời: $a,\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2$$=\left(2x+1\right)^2+2\cdot\left[\left(2x\right)^2-1^2\right]+\left(2x-1\right)^2$$=\left(2x+1\right)^2+2\cdot\left(2x+1\right)\cdot\left(2x-1\right)+\left(2x-1\right)^2$$=\left[\left(2x+1\right)+\left(2x-1\right)\right]^2$$=\left(2x+1+2x-1\right)^2$$=\left(4x\right)^2$$=16x^2$$---$$b,\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)$$=x^2\left(x+2\right)-\left(x+2\right)-\left(x^3-2^3\right)$$=x^3+2x^2-x-2-x^3+8$$=2x^2-x+6$

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