Tính giá trị của biểu thức:
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}-1\right)\)
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Những câu hỏi liên quan
6 tháng 10 2019
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 !!!
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 3 2022
Xét đa thức bậc 8: \(P\left(x\right)=x^8+\dfrac{x^3-x}{2}\)
Ta có, \(P\left(x\right)-P\left(-x\right)=x^8+\dfrac{x^3-x}{2}-\left(-x\right)^8-\dfrac{\left(-x\right)^3-\left(-x\right)}{2}=x^3-x\)
Thay \(x=1;2;3;4\) đều thỏa mãn
\(\Rightarrow P\left(5\right)-P\left(-5\right)=5^3-5=120\)
4 tháng 8 2016
[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
VV
22 tháng 7 2021
B = \(\left(-1\dfrac{1}{6}\right)\) : \(\left(\dfrac{-10}{3}+\dfrac{9}{4}\right)\) - \(\left(-\dfrac{3}{8}\right)\) : \(\left(8-\dfrac{51}{8}\right)\)
B = \(\dfrac{-7}{6}\) : \(\dfrac{-13}{12}\) - \(\left(-\dfrac{3}{8}\right)\) : \(\dfrac{13}{8}\)
B = \(\dfrac{14}{13}\) - \(\dfrac{-3}{13}\)
B = \(\dfrac{17}{13}\)
Y
21 tháng 6 2023
\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)
\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)
\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)
\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)
\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)
\(=7-\dfrac{26}{5}\)
\(=\dfrac{9}{5}\)
\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)
\(=\dfrac{2}{3}+\dfrac{21}{8}\)
\(=\dfrac{79}{24}\)
\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)
\(=\dfrac{31}{4}:\dfrac{49}{8}\)
\(=\dfrac{62}{49}\)
\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)
17 tháng 7 2019
\(2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1< 3^{32}\)
Gợi ý: Sử dụng liên tục tính chất \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
17 tháng 7 2019
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
18 tháng 6 2018
a) \(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
\(=\left(a^2+\left(-b\right)^2+c^2-2ab+2ac-2bc\right)-\left(b^2-2bc+c^2\right)+2ab-2ac\)
\(=a^2+b^2+c^2-2ab+2ac-2bc-b^2+2bc-c^2+2ab-2ac\)
\(=a^2+b^2-b^2+c^2-c^2-2ab+2ab+2ac-2ac-2bc+2bc\)
\(=a^2\)
AH
Akai Haruma
Giáo viên
6 tháng 10 2019
Lời giải:
Áp dụng HĐT đáng nhớ \((a-b)(a+b)=a^2-b^2\). Ta có:
\(A=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(2A=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^8-1)(3^8+1)(3^{16}+1)(3^{32}+1)\)
\(=(3^{16}-1)(3^{16}+1)(3^{32}+1)\)
\(=(3^{32}-1)(3^{32}+1)=3^{64}-1\)
\(\Rightarrow A=\frac{3^{64}-1}{2}\)
AH
Akai Haruma
Giáo viên
3 tháng 10 2019
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