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\(\left(\frac{11}{1.12}+\frac{11}{12.23}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+...+\frac{1}{89}-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{100}\right)+x=\frac{5}{3}\Leftrightarrow\frac{99}{100}+x=\frac{5}{3}\Leftrightarrow x=\frac{5}{3}-\frac{99}{100}\)
\(\Leftrightarrow x=\frac{297}{300}=\frac{ }{100}\)
a) \(\left(\dfrac{11}{12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow\left(\dfrac{11}{1.12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow\left(1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{34}+...+\dfrac{1}{89}-\dfrac{1}{100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow1-\dfrac{1}{100}+x=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{5}{3}-1+\dfrac{1}{100}\)
\(\Rightarrow x=\dfrac{500}{300}-\dfrac{300}{300}+\dfrac{3}{300}\)
\(\Rightarrow x=\dfrac{203}{300}\)
b) \(\left(\dfrac{5}{11.16}+\dfrac{5}{16.21}+...+\dfrac{5}{19.24}\right)-x+\dfrac{1}{3}=\dfrac{7}{3}\)
=>\(\left(\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{19}-\dfrac{1}{24}\right)-x=\dfrac{7}{3}-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{11}-\dfrac{1}{24}-x=2\)
\(\Rightarrow-x=2-\dfrac{1}{11}+\dfrac{1}{24}\)
\(\Rightarrow-x=\dfrac{528}{264}-\dfrac{24}{264}+\dfrac{11}{264}\)
\(\Rightarrow x=\dfrac{515}{264}\)
c) Câu hỏi của Đàm Chu Hữu An - Toán lớp 6 - Học toán với OnlineMath
. Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\)
\(\dfrac{1}{12}>\dfrac{1}{20}\)
.................
\(\dfrac{1}{19}>\dfrac{1}{20}\)
\(\dfrac{1}{20}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{11}+\dfrac{1}{12}+......+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+.....+\dfrac{1}{20}\)
\(\Leftrightarrow S>\dfrac{1}{20}.10\)
\(\Leftrightarrow S>\dfrac{1}{2}\)
2. \(\dfrac{x}{12}=\dfrac{-1}{24}-\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{x}{12}=-\dfrac{1}{6}\)
\(\Leftrightarrow6x=-12\)
\(\Leftrightarrow x=-2\)
Vậy ...
3. \(\dfrac{2}{5.7}+\dfrac{2}{7.9}+........+\dfrac{2}{19.21}\)
\(=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+......+\dfrac{1}{19}-\dfrac{1}{21}\)
\(=\dfrac{1}{5}-\dfrac{1}{21}\)
\(=\dfrac{16}{105}\)
\(-x-\frac{3}{4}=-\frac{8}{11}=>-x=-\frac{8}{11}+\frac{3}{4}=\frac{1}{44}=>x=-\frac{1}{44}\)
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