Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

=100

Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)

\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)

\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)

\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)

\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)

\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)

\(=7-\dfrac{26}{5}\)

\(=\dfrac{9}{5}\)

\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)

\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)

\(=\dfrac{2}{3}+\dfrac{21}{8}\)

\(=\dfrac{79}{24}\)

\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)

\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)

\(=\dfrac{31}{4}:\dfrac{49}{8}\)

\(=\dfrac{62}{49}\)

\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)

\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)

\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)

-_-

a) = (\(-\dfrac{141}{20}\)- \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\dfrac{1}{15}\)

    = \(-\dfrac{73}{10}\) : - 5

    = \(\dfrac{73}{50}\)

b) = \(\left(\dfrac{3}{25}-\dfrac{28}{25}\right)\). \(\dfrac{7}{3}\) : \(\left(\dfrac{7}{2}-\dfrac{11}{3}.14\right)\)

    = \(-\dfrac{7}{3}\) . \(-\dfrac{6}{287}\)

    = \(\dfrac{2}{41}\)

A= 4/7.

Biết có cái

1, \(x\left(x+\dfrac{2}{3}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)

2, a, \(\left|x+\dfrac{4}{6}\right|\ge0\)

Để \(\left|x+\dfrac{4}{6}\right|\) đạt GTNN thì \(\left|x+\dfrac{4}{6}\right|=0\)

\(\Leftrightarrow x+\dfrac{4}{6}=0\Rightarrow x=\dfrac{-2}{3}\)

Vậy, ...

b, \(\left|x-\dfrac{1}{3}\right|\ge0\)

Để \(\left|x-\dfrac{1}{3}\right|\) đạt GTLN thì \(\left|x-\dfrac{1}{3}\right|=0\)

\(\Leftrightarrow x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)

Vậy, ...

1)

a)

\(x\cdot\left(x+\dfrac{2}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)

2)

a)

\(\left|x+\dfrac{4}{6}\right|\ge0\)

Dấu \("="\) xảy ra khi \(x+\dfrac{4}{6}=0\Leftrightarrow x=\dfrac{-4}{6}\Leftrightarrow x=\dfrac{-2}{3}\)

Vậy \(Min_{\left|x+\dfrac{4}{6}\right|}=0\text{ khi }x=\dfrac{-2}{3}\)

b)

\(\left|x-\dfrac{1}{3}\right|\ge0\)

Dấu \("="\) xảy ra khi \(x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)

Vậy \(Min_{\left|x-\dfrac{1}{3}\right|}=0\text{ khi }x=\dfrac{1}{3}\)

Ai lm đc câu nào thì giúp mk với , cảm ơn !!

\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)

a: \(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{5}\)

\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)

\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)

a) `A=a. 1/3 + a. 1/4 - a.1/6 = a. (1/3+1/4 -1/6)=a. 5/12`

Thay `a=-3/5: A=-3/5 . 5/12 =-1/4`

b) `B=b. 5/6+ b. 3/4-b. 1/2=b.(5/6+3/4-1/2)=b. 13/12`

Thay `b=12/13: B=12/13 . 13/12=1`.

a) Ta có: \(A=a\cdot\dfrac{1}{3}+a\cdot\dfrac{1}{4}-a\cdot\dfrac{1}{6}\)

\(=a\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{6}\right)\)

\(=a\cdot\left(\dfrac{4}{12}+\dfrac{3}{12}-\dfrac{2}{12}\right)\)

\(=a\cdot\dfrac{5}{12}\)

\(=\dfrac{-3}{5}\cdot\dfrac{5}{12}=\dfrac{-1}{4}\)

b) Ta có: \(B=b\cdot\dfrac{5}{6}+b\cdot\dfrac{3}{4}-b\cdot\dfrac{1}{2}\)

\(=b\left(\dfrac{5}{6}+\dfrac{3}{4}-\dfrac{1}{2}\right)\)

\(=b\cdot\left(\dfrac{10}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\)

\(=b\cdot\dfrac{5}{4}\)

\(=\dfrac{12}{13}\cdot\dfrac{5}{4}=\dfrac{60}{52}=\dfrac{15}{13}\)

a) \(A=a\cdot\dfrac{1}{3}+a\cdot\dfrac{1}{4}-a\cdot\dfrac{1}{6}\\ A=a\cdot\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{6}\right)\\ A=a\cdot\dfrac{-5}{12}\)

Khi \(a=\dfrac{-3}{5}\), ta có:

\(A=\dfrac{-3}{5}\cdot\dfrac{-5}{12}\\ A=\dfrac{1}{4}\)

Vậy khi \(a=\dfrac{-3}{5}\) thì \(A=\dfrac{1}{4}\)

b. \(B=b\cdot\dfrac{5}{6}+b\cdot\dfrac{3}{4}-b\cdot\dfrac{1}{2}\\ B=b\cdot\left(\dfrac{5}{6}+\dfrac{3}{4}-\dfrac{1}{2}\right)\\ B=b\cdot\dfrac{13}{12}\)

Khi \(a=\dfrac{12}{13}\), ta có:

\(B=\dfrac{12}{13}\cdot\dfrac{13}{12}\\ B=1\)

Vậy khi \(a=\dfrac{-3}{5}\) thì B = 1