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Lời giải:
Áp dụng HĐT đáng nhớ \((a-b)(a+b)=a^2-b^2\). Ta có:
\(A=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(2A=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^8-1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^{16}-1)(3^{16}+1)(3^{32}+1)\)
\(=(3^{32}-1)(3^{32}+1)=3^{64}-1\)
\(\Rightarrow A=\frac{3^{64}-1}{2}\)
Lời giải:
Áp dụng HĐT đáng nhớ \((a-b)(a+b)=a^2-b^2\). Ta có:
\(A=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(2A=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^8-1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^{16}-1)(3^{16}+1)(3^{32}+1)\)
\(=(3^{32}-1)(3^{32}+1)=3^{64}-1\)
\(\Rightarrow A=\frac{3^{64}-1}{2}\)
[Toán 8] Rút gọn $ (3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)$ | HOCMAI Forum - Cộng đồng học sinh Việt Nam
a) \(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
\(=\left(a^2+\left(-b\right)^2+c^2-2ab+2ac-2bc\right)-\left(b^2-2bc+c^2\right)+2ab-2ac\)
\(=a^2+b^2+c^2-2ab+2ac-2bc-b^2+2bc-c^2+2ab-2ac\)
\(=a^2+b^2-b^2+c^2-c^2-2ab+2ab+2ac-2ac-2bc+2bc\)
\(=a^2\)
1: A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
=(3^4-1)(3^4+1)(3^8+1)(3^16+1)
=(3^8-1)(3^8+1)(3^16+1)
=(3^16-1)(3^16+1)
=3^32-1
2: B=(1-3^2)(1+3^2)*...*(1+3^16)
=(1-3^4)(1+3^4)(1+3^8)(1+3^16)
=1-3^32
1
\(A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)\\ =3^{32}-1\)
\(B=\left(1-3\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^2\right)\left(1+3^2\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^4\right)\left(1+3^4\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^8\right)\left(1+3^8\right)\left(3^{16}+1\right)\\ =\left(1-3^{16}\right)\left(1+3^{16}\right)=1-3^{32}\)
Áp dụng HĐT đáng nhớ :
\(\left(a-b\right)\left(a+b\right)=a^2-b^2\) . Ta có :
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)
\(\Rightarrow A=\frac{3^{64}-1}{2}\)
Chúc bạn học tốt !!!