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\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{120}{121}=\frac{3.8.15...120}{4.9.16...121}\)
\(=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{\left(2.2\right).\left(3.3\right).\left(4.4\right)...\left(11.11\right)}\)
\(=\frac{\left(1.2.3...10\right).\left(3.4.5...12\right)}{\left(2.3.4...11\right).\left(2.3.4...11\right)}=\frac{1.12}{11.2}=\frac{6}{11}\)
ta có :
A=\(\left(-\frac{3}{4}\right)\left(-\frac{8}{9}\right)\left(-\frac{15}{16}\right)...\left(-\frac{120}{121}\right)\)(có 10 số hạng)
= \(\frac{3\cdot8\cdot15\cdot...\cdot120}{4\cdot9\cdot16\cdot...\cdot121}=\frac{\left(1.3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(10\cdot12\right)}{2^2\cdot3^2\cdot4^2\cdot...\cdot11^2}=\frac{\left(1\cdot2\cdot3\cdot...\cdot10\right)\left(3\cdot4\cdot5\cdot...\cdot12\right)}{\left(2\cdot3\cdot4\cdot..\cdot11\right)\left(2\cdot3\cdot4\cdot..\cdot11\right)}\)
=\(\frac{12}{11\cdot2}=\frac{12}{22}\)
b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)
\(M=1-\frac{1}{7}=\frac{6}{7}\)
Mình làm câu 1 thoi nha!
1.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
=>4x+(1/2+1/4+1/8+1/16)=1
<=>4x+15/16=1
=>4x=1/16
=>x=1/16:4=1/64
vậy x=1/64
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Rightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(\Rightarrow4x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right)=1\)
\(\Rightarrow4x+\left(1-\frac{1}{16}\right)=1\)
\(\Rightarrow4x+\frac{15}{16}=1\)
\(\Rightarrow4x=1-\frac{15}{16}\)
\(\Rightarrow x=\frac{1}{16}:4\)
\(\Rightarrow x=\frac{1}{64}\)
vậy \(x=\frac{1}{64}\)
