Cho a là 1 số nguyên sao cho a.căn 3 là 1 số hữu tỷ. Chứng minh rằng a = 0
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a/
\(2xA=\frac{3-1}{1x2x3}+\frac{4-2}{2x3x4}+\frac{5-3}{2x4x5}+...+\frac{7-5}{5x6x7}\)
\(2xA=\frac{1}{1x2}-\frac{1}{2x3}+\frac{1}{2x3}-\frac{1}{3x4}+\frac{1}{3x4}-\frac{1}{4x5}+...+\frac{1}{5x6}-\frac{1}{6x7}\)
\(2xA=\frac{1}{2}-\frac{1}{6x7}=\frac{2}{6x7}\Rightarrow A=\frac{1}{6x7}=\frac{1}{42}\)
b/
\(2xB=\frac{5-1}{1x3x5}+\frac{7-3}{3x5x7}+\frac{9-5}{5x7x9}+...+\frac{13-9}{9x11x13}\) Làm tiếp tương tự câu a
c/
\(C=\frac{1x3x5+2x1x2x3x2x5+4x1x4x3x4x5+7x1x7x3x7x5}{1x5x7+2x1x2x5x2x7+4x1x4x5x4x7+7x1x7x5x7x7}=\)
\(=\frac{1x3x5x\left(1+2x2x2+4x4x4+7x7x7\right)}{1x5x7x\left(1+2x2x2+4x4x4+7x7x7\right)}=\frac{1x3x5}{1x5x7}=\frac{3}{7}\)
d/
\(E=\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{30}\right)+...+\left(1-\frac{1}{110}\right)=\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+...+\frac{1}{10x11}\right)=\)
\(=8-\left(\frac{4-3}{3x4}+\frac{5-4}{4x5}+\frac{6-5}{5x6}+...+\frac{11-10}{10x11}\right)=\)
\(=8-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{11}\right)=\)
\(=8-\left(\frac{1}{3}-\frac{1}{11}\right)=8-\frac{8}{33}=\frac{256}{33}\)
g/
\(3xB=\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+...+\frac{1}{486}+\frac{1}{1458}\)
\(2xB=3xB-B=\frac{3}{2}-\frac{1}{1458}\Rightarrow B=\frac{3}{4}-\frac{1}{2916}\)
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B1 :
\(\frac{x}{3}=\frac{y}{6}=\frac{xy}{3\times6}=\frac{162}{18}=9\)
---> x = 3.9 = 27
---> y = 6.9 = 54
B2 :
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=\frac{xyz}{2\times3\times5}=\frac{-240}{30}=-8\)
---> x = -8.2 = -16
---> y = -8.3 = -24
---> z = -8.5 = -40
xin tiick