Câu hỏi của tofu

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Mong mọi người giải giúp em với ạ em cảm ơn nhiều lắm ạ

Đoàn Đức Hà · Accepted Answer

d) $\left(2x+1\right)^2-2\left(x-3\right)\left(2x+1\right)+\left(x-3\right)^2=\left[\left(2x+1\right)-\left(x-3\right)\right]^2=\left(x+4\right)^2$

e) $3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{128}+1\right)$

$=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{128}+1\right)$

$=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{128}+1\right)$

$=\left(2^8-1\right)\left(2^8+1\right)...\left(2^{128}+1\right)$

$=...=\left(2^{128}-1\right)\left(2^{128}+1\right)=2^{256}-1$

Câu trả lời:

d) $\left(2x+1\right)^2-2\left(x-3\right)\left(2x+1\right)+\left(x-3\right)^2=\left[\left(2x+1\right)-\left(x-3\right)\right]^2=\left(x+4\right)^2$

e) $3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{128}+1\right)$

$=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{128}+1\right)$

$=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{128}+1\right)$

$=\left(2^8-1\right)\left(2^8+1\right)...\left(2^{128}+1\right)$

$=...=\left(2^{128}-1\right)\left(2^{128}+1\right)=2^{256}-1$