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3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
\(4\dfrac{1}{3}\cdot\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\\ \dfrac{13}{3}\cdot\dfrac{-1}{3}\le x\le\dfrac{2}{3}\cdot\dfrac{-11}{12}\\ \dfrac{-13}{9}\le x\le\dfrac{-11}{18}\\ \dfrac{-26}{18}\le x\le\dfrac{-11}{18}\\ \Rightarrow x=-1\)
\(6\dfrac{2}{9}.x+3\dfrac{10}{27}=22\dfrac{1}{7}\)
\(\dfrac{56}{9}.x+\dfrac{91}{27}=\dfrac{155}{7}\)
\(\left(\dfrac{56}{9}.x\right)\) \(=\dfrac{155}{7}-\dfrac{91}{27}\)
\(\left(\dfrac{56}{9}.x\right)\) \(=\dfrac{4185}{189}-\dfrac{637}{189}\)
\(\left(\dfrac{56}{9}.x\right)\) \(=\dfrac{3548}{189}\)
\(x\) \(=\dfrac{3548}{189}:\dfrac{56}{9}\)
\(x\) \(=\dfrac{3548}{189}.\dfrac{9}{56}\)
\(x\) \(=\dfrac{887}{294}\)
Vậy \(x\\\) \(=\dfrac{887}{294}\)
Chúc bạn học tốt
\(10A=\dfrac{10^{12}-10}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\)
\(10B=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)
Vì \(10^{12}-1>10^{11}+1\)
nên \(-\dfrac{9}{10^{12}-1}>-\dfrac{9}{10^{11}+1}\)
hay A>B
a, \(x+\dfrac{2}{3}=0,2\)
\(\Rightarrow x+\dfrac{2}{3}=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{-7}{10}\)
b, \(\dfrac{17}{7}-\dfrac{6}{5}x=\dfrac{17}{4}\)
\(\Rightarrow\dfrac{6}{5}x=\dfrac{17}{7}-\dfrac{17}{4}\)
\(\Rightarrow\dfrac{6}{5}x=\dfrac{-51}{28}\)
\(\Rightarrow x=\dfrac{-51}{28}:\dfrac{6}{5}\)
\(\Rightarrow x=\dfrac{-85}{56}\)
)
Gọi \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\)là \(S\)
\(S=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\\ S>\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{100\cdot101}\\ S>\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{100}-\dfrac{1}{101}\\ S>\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{5}\)
Vậy \(S>\dfrac{1}{5}\)(đpcm)