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#)Giải :
Đặt \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{5}-\frac{1}{10}\)
\(A=\frac{1}{10}\)
a, \(\frac{6+x}{33}=\frac{7}{11}\)
\(\Leftrightarrow\left(6+x\right).11=7.33\)
\(\Leftrightarrow66+11x=231\)
\(\Leftrightarrow11x=231-66\)
\(\Leftrightarrow x=\frac{165}{11}=15\)
Vậy x = 15.
b,\(\frac{12+x}{43-x}=\frac{2}{3}\)
\(\Leftrightarrow3.\left(12+x\right)=2\left(43-x\right)\)
\(\Leftrightarrow36+3x=86-2x\)
\(\Leftrightarrow3x+2x=-36+86\)
\(\Leftrightarrow5x=50\)
\(\Leftrightarrow x=10\)
Vây x = 10.
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2012\)
Vậy x = 2012
\(\frac{43}{20}+\frac{17}{6}:\left(\frac{5}{8}+\frac{7}{30}\right)\)
\(=\frac{43}{20}+\frac{17}{6}\cdot\frac{120}{103}\)
\(=\frac{43}{20}+\frac{340}{103}\)
\(=\frac{11229}{2060}\)
= 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + ...... + 1/512 - 1/1024
=1 - 1/1024
=1023/1024
k nha
a)\(\frac{x}{17}=\frac{60}{204}=\frac{5}{17}\Rightarrow x=5\)
b)\(\frac{6+x}{33}=\frac{7}{11}\Rightarrow11\left(6+x\right)=7.33\Rightarrow11.6+11x=231\Rightarrow66+11x=231\)
\(\Rightarrow11x=231-66\Rightarrow11x=165\Rightarrow x=\frac{165}{11}=15\)
c)\(\frac{12+x}{43-x}=\frac{2}{3}\Rightarrow2\left(43-x\right)=3\left(12+x\right)\Rightarrow2.43-2x=3.12+3x\)
\(86-2x=36+3x\Rightarrow86-36=3x+2x\Rightarrow50=5x\Rightarrow x=\frac{50}{5}=10\)
\(6\frac{1}{7}x1\frac{6}{43}\)
= \(\frac{43}{7}x\frac{49}{43}\)
= \(\frac{49}{7}\)
= 7
43/7 * 49/43= 7