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\(C=-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}-\frac{1}{100000}\)
\(10C=-1-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}\)
\(10C-C=\left(-1-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}\right)-\left(\frac{-1}{10}-\frac{1}{100}-...-\frac{1}{100000}\right)\)
\(9C=-1+\frac{1}{100000}\)
\(C=\frac{\frac{1}{100000}-1}{9}\)
\(\frac{69}{157}-\left(2+\left(3+4+5^{-1^{-1^{-1^{-1}}}}\right)\right)\)
\(=\frac{69}{157}-\left(2+\left(3+4+\frac{1}{5}\right)\right)\)
\(=\frac{69}{157}-\left(2+\frac{36}{5}\right)\)
\(=\frac{69}{157}-\frac{46}{5}\)
\(=\frac{-6877}{785}\)
chúc bạn học tốt
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{2}{100}-1=-\frac{49}{50}\)
Ta có : A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)
\(\Rightarrow\)3A = 1.2.(3-0)+2.3.(4-1)+3.4.(5-2).....n.(n+1).[(n+2)-(n-1)]
\(\Rightarrow\)3A= 1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+....+n.(n+1)(n+2)-(n-1)n(n+1)
\(\Rightarrow\)3A= (1.2.3-1.2.3)+(2.3.4-2.3.4)+....+[(n-1).n.(n+1)-(n-1)n(n+1)]+n.(n+1)(n+2)
\(\Rightarrow\)3A=n.(n+1)(n+2)
\(\Rightarrow\)A=\(\frac{\text{n.(n+1)(n+2)}}{3}\)
\(\left(\frac{1}{2}\right)^x=\frac{1}{64}\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^6\)
=> x=6
b, \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\Rightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Rightarrow5x+1=\frac{6}{7}\)
\(\Rightarrow5x=\frac{-1}{7}\)
\(\Rightarrow x=\frac{-1}{35}\)
=2 :))
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