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e) Ta có: \(2\left|x-\dfrac{1}{2}\right|\ge0\forall x\)
\(\Leftrightarrow2\left|x-\dfrac{1}{2}\right|+2021\ge2021\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
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 !!!
[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
\(2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1< 3^{32}\)
Gợi ý: Sử dụng liên tục tính chất \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
2(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)
= (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)
= (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)
= (34 - 1)(34 + 1)(38 + 1)(316 + 1)
= (38 - 1)(38 + 1)(316 + 1)
= (316 - 1)(316 + 1)
= 332 - 1 < 332
a) \(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
\(=\left(a^2+\left(-b\right)^2+c^2-2ab+2ac-2bc\right)-\left(b^2-2bc+c^2\right)+2ab-2ac\)
\(=a^2+b^2+c^2-2ab+2ac-2bc-b^2+2bc-c^2+2ab-2ac\)
\(=a^2+b^2-b^2+c^2-c^2-2ab+2ab+2ac-2ac-2bc+2bc\)
\(=a^2\)
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:
\(A=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(2A=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^8-1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^{16}-1)(3^{16}+1)(3^{32}+1)\)
\(=(3^{32}-1)(3^{32}+1)=3^{64}-1\)
\(\Rightarrow A=\frac{3^{64}-1}{2}\)
Let A = (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
=> 2A = (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
= (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
= (34 - 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
= (38 - 1)(38 + 1)(316 + 1)(332 + 1)
= (316 - 1)(316 + 1)(332 + 1)
= (332 - 1)(332 + 1)
= 364 - 1
⇒A=364−12