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Question

Rut gon cac bieu thuc sau

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

T=(3+1)$\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)$

Aki Tsuki · Accepted Answer

$P=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16$

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$T=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)$

$\Rightarrow2T=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)$

$2T=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)$

$2T=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)$

$2T=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)$

$2T=\left(3^8-1\right)\left(3^8+1\right)=3^{16}-1$

$\Rightarrow T=\dfrac{3^{16}-1}{2}=21523360$Câu trả lời: $P=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16$

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