Tính : 6\(\div2\left(1+1+1\right)\)
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\(6\div2\left(1+2\right)=6\div2.3\\ =6\div6\\ =1 \)
\(6\div2\times\left(1+2\right)=3\times\left(1+2\right)\\ =3\times3\\ =9\)
b, \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^{21-12}=\left(\frac{3}{7}\right)^9\)
a) \(25^3:5^2=5^6:5^2=5^4=625\)
b) \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c) \(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3-1+\frac{1}{8}=\frac{17}{8}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..+\frac{1}{x\left(x+1\right):2}=\frac{2018}{2019}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{x\left(x+1\right)}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{x\left(x+1\right)}\right)\)\(=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2018}{2019}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2018}{4038}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2018}{4038}=\frac{1}{4038}\)
\(\Rightarrow x+1=4038\)
\(\Rightarrow x=4037\)
\(\Leftrightarrow\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{x\left(x+1\right):2}=\dfrac{1991}{1993}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1991}{1993}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{1991}{1993}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1991}{3986}\)
=>1/x+1=1/1993
=>x+1=1993
hay x=1992
\(=\left(\frac{1}{2}+\frac{1}{3}\right):\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2}.1\right):2^3\)
\(=\frac{5}{6}:\frac{1}{6}+\frac{1}{2}:8\)
\(=5+\frac{1}{16}\)
\(=\frac{81}{16}\)
1
6:2(1+1+1)=9
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