x + 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 = 1
x + 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 = 1
x + 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 = 1
x + 1/1 - 1/7 = 1
x + 6/7 = 1
x = 1 - 6/7
x = 1/7
x + 1/2 + 1/6 + 1/20 + 1/30 + 1/42 = 1
x + 65/84 = 1
x = 1 - 65/84
x = 19/84
x + 1/2 + 1/6 + 1/12 + 1/20 + 1/30 × 1/42 = 1
<=> x + 6/7 = 1
<=> x = 1 - 6/7
<=> x = 1/7
Vậy x = 1/7
Dấu \(.\)là dấu nhân nhé
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=1\)
\(\Rightarrow x+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{5.6}+\frac{1}{6.7}=1\)
\(\Rightarrow x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}=1\)
\(\Rightarrow x+1-\frac{1}{7}=1\)
\(\Rightarrow x+\frac{6}{7}=1\)
\(\Rightarrow x=1-\frac{6}{7}\)
\(\Rightarrow x=\frac{1}{7}\)
Vậy \(x=\frac{1}{7}\)
~ Ủng hộ nhé
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=1\)
\(x+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}=1\)
\(x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}.....+\frac{1}{6}-\frac{1}{7}=1\)
\(x+1-\frac{1}{7}=1\)
\(x+\frac{6}{7}=1\)
\(x=1-\frac{6}{7}\)
\(x=\frac{1}{7}\)
Chắc vậy đó....
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+\frac{1}{42}=1\)
\(x+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=1\)
\(x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)=1\)
\(x+\left(1-\frac{1}{7}\right)=1\)
\(x+\frac{6}{7}=1\)
\(x=1-\frac{6}{7}\)
\(\Rightarrow x=\frac{1}{7}\)
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=1\)
\(x+\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}=1\)
\(x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}=1\)
\(x+1-\frac{1}{7}=1\)
\(x+\frac{6}{7}=1\)
\(x=1-\frac{6}{7}\)
\(x=\frac{1}{7}\)
\(x+\frac{1}{2}+...+\frac{1}{42}=\)\(1\)
\(\Leftrightarrow x+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{6.7}=1\)
\(\Leftrightarrow x+\left(\frac{1}{1}-\frac{1}{2}+...+\frac{1}{6}-\frac{1}{7}\right)=1\)
\(\Leftrightarrow x+\left(\frac{1}{1}-\frac{1}{7}\right)\)\(=1\)
\(\Leftrightarrow x+\frac{6}{7}=1\)
\(\Leftrightarrow x=1-\frac{6}{7}\)
\(\Leftrightarrow x=\frac{1}{7}\)
Vậy \(x=\frac{1}{7}\)
