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\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}=2\)
\(\Rightarrow1+\frac{1}{2.3}.2+\frac{1}{3.4}.2+...+\frac{1}{x\left(x+1\right)}.2=2\)
=> \(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=2\)
=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x\left(x+1\right)}=1\)
=> \(1-\frac{1}{x+1}=1\)
=> \(\frac{1}{x+1}=0\Rightarrow0\left(x+1\right)=1\Rightarrow x\in\varnothing\)
\(\frac{1}{1.2:2}+\frac{1}{2.3:2}+\frac{1}{3.4:2}+...+\frac{1}{x.\left(x+1\right):2}=2\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=2\)
\(2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=2\)
\(1-\frac{1}{x+1}=1\)
\(\frac{1}{x+1}=0\)
Vậy x vô nghiệm.
Bài 1: <Cho là câu a đi>:
a. \(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\)
\(\rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\)
\(\rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{49}{50}\)
\(\rightarrow1-\frac{1}{x+1}=\frac{49}{50}\)
\(\rightarrow\frac{1}{x+1}=1-\frac{49}{50}=\frac{1}{50}\)
\(\rightarrow x+1=50\rightarrow x=49\)
Vậy x = 49.
1.Ta có: \(\frac{x}{3}=-\frac{12}{9}\)
=> \(\frac{3x}{9}=-\frac{12}{9}\)
=> 3x = -12
=> x = -12 : 3
=> x = -4
\(\frac{4}{5}x-\frac{8}{5}=-\frac{1}{2}\)
=> \(\frac{4}{5}x=-\frac{1}{2}+\frac{8}{5}\)
=> \(\frac{4}{5}x=\frac{11}{10}\)
=> \(x=\frac{11}{10}:\frac{4}{5}\)
=> \(x=\frac{11}{8}\)
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}=2\)
=> \(1+\frac{1}{\frac{2\left(1+2\right)}{2}}+\frac{1}{\frac{3\left(1+3\right)}{2}}+....+\frac{1}{\frac{x\left(x+1\right)}{2}}=2\)
=> \(1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=2\)
=> \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=1\)
=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2}\)
=> \(\frac{1}{x+1}=0\Rightarrow x\in\varnothing\)
Bài làm :
Ta có :
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}=2\)
\(\Leftrightarrow1+\frac{1}{\frac{2\left(1+2\right)}{2}}+\frac{1}{\frac{3\left(1+3\right)}{2}}+....+\frac{1}{\frac{x\left(x+1\right)}{2}}=2\)
\(\Leftrightarrow1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=2\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=1\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{x+1}=0\)
=> Không tồn tại x