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Ta có: \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{3^{32}-1}{2}\)
Rút gọn: (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(3-1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(32-1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(34-1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(38-1)(38 + 1)(316 + 1)(332 + 1)
A=(316-1)(316 + 1)(332 + 1)
A=(332 - 1)(332 + 1)
A=364-1
=>A=(364-1) /2
đầu bài là như này đúng không hả bạn
\(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)\)\(=\frac{3}{4}\)
Ta có :\(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)\)\(=\frac{3}{4}\)
\(\frac{2}{3}:\left(x-1\right)\)\(=\frac{1}{4}\)
\(\left(x-1\right)\)\(=\frac{8}{3}\)
\(x=\frac{11}{3}\)
\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2020-2019}{2019.2020}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(=\frac{1}{2}-\frac{1}{2020}=\frac{1009}{2020}\)
\(\dfrac{3}{4}:\left(2\dfrac{4}{9}\right)-\left|-3x+2\dfrac{2}{3}\right|=\dfrac{3}{4}\)
\(\Leftrightarrow\left|-3x+\dfrac{8}{3}\right|=\dfrac{3}{4}-\dfrac{3}{4}\cdot\dfrac{9}{22}\)
\(\Leftrightarrow\left|3x-\dfrac{8}{3}\right|=\dfrac{3}{4}-\dfrac{27}{88}=\dfrac{39}{88}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{8}{3}=\dfrac{39}{88}\\3x-\dfrac{8}{3}=-\dfrac{39}{88}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{821}{792}\\x=\dfrac{587}{792}\end{matrix}\right.\)
a)\(\frac{1}{5}x-\frac{1}{3}=\frac{2}{4}\left(x+2\right)\)
<=>\(\frac{1}{5}x-\frac{1}{3}=\frac{2}{4}x+1\)
<=>\(-\frac{3}{10}x=\frac{4}{3}\)
<=>\(x=-\frac{40}{9}\)
b)\(\frac{5}{4}\left(x-3\right)=4+\frac{3}{2}x\)
<=>\(\frac{5}{4}x-\frac{15}{4}=4+\frac{3}{2}x\)
<=>\(-\frac{1}{4}x=\frac{31}{4}\)
<=>\(x=-31\)
c)\(\frac{5}{4}\left(x-3\right)=\frac{3}{2}\left(x+4\right)\)
<=>\(\frac{5}{4}x-\frac{15}{4}=\frac{3}{2}x+6\)
<=>\(-\frac{1}{4}x=\frac{9}{4}\)
<=>x=-9
\(\left(3-\frac{2}{3}+\frac{4}{3}\right):\left(2\frac{1}{3}-2,5\right)^2\)
\(=\left(\frac{7}{3}+\frac{4}{3}\right):\left(2\frac{1}{3}-2\frac{1}{2}\right)^2\)
\(=\frac{11}{3}:\left(-\frac{1}{6}\right)^2\)
\(=\frac{11}{3}:\frac{1}{36}\)
\(=\frac{11}{3}x\frac{36}{1}\)
\(=\frac{396}{3}\)
\(=132\)
(3-2/3+4/3):(2 1/3-2,5)2
= -17/6:121/36
=-36/121