\(x.\left(4,6+\frac{3}{5}\right)=7,2-8,15\) \(\left(x-\frac{-7}{8}\right)+2,3=3,24\)
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\(=\dfrac{35.85\cdot14}{0.5+2.3}\cdot\dfrac{6}{17}\cdot\dfrac{16.8}{259,2}-\left(4.625-\dfrac{13}{6}:\dfrac{26}{3}\right):\left(3.25:2.25\right)\)
\(=\dfrac{1673}{408}-\dfrac{315}{104}=\dfrac{1421}{1326}\)
b) Ta có: \(|x-3,5|\ge0;\forall x\)
\(\Rightarrow|x-3,5|+2,3\ge2,3;\forall x\)
\(\Rightarrow\frac{4,6}{|x-3,5|+2,3}\le\frac{4,6}{2,3};\forall x\)
Hay \(I\le2;\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow|x-3,5|=0\)
\(\Leftrightarrow x=3,5\)
Vậy MAX I =2 \(\Leftrightarrow x=3,5\)
a) Ta có: \(\hept{\begin{cases}|x+2,1|\ge0;\forall x\\|y-4,6-2015|\ge0;\forall y\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}-|x+2,1|\le0;\forall x\\-|y-2019,6|\le0;\forall x\end{cases}}\)
\(\Rightarrow-|x+2,1|-|y-2019,6|\le0;\forall x,y\)
Hay \(G\le0;\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}|x+2,1|=0\\|y-2019,6|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-2,1\\y=2019,6\end{cases}}\)
Vậy MAX G=0 \(\Leftrightarrow\hept{\begin{cases}x=-2,1\\y=2019,6\end{cases}}\)
\(\frac{1}{x+2}-\frac{1}{x+5}+...+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\frac{1}{x+2}-\frac{1}{x+7}=\frac{x}{\left(x+2\right)\left(x+7\right)}\)
\(\Rightarrow x=1\)
\(x:\frac{13}{10}+\frac{42}{5}.\frac{6}{7}.\left(6-\frac{\left(2,3+0,8\right).7}{0,1+6,9}\right)=\frac{39}{7}:\frac{15}{14}=\frac{37}{7}.\frac{14}{15}=\frac{74}{15}\)
\(x.\frac{10}{13}+\frac{36}{7}\left(6-\frac{3,1.7}{7}\right)=\frac{74}{15}\)
\(x.\frac{10}{13}+\frac{36}{7}\left(6-3,1\right)=\frac{74}{15}\)
\(x.\frac{10}{13}+\frac{36}{7}.\frac{29}{10}=\frac{74}{15}\)
\(x.\frac{10}{13}+\frac{522}{35}=\frac{74}{15}\)
\(x.\frac{10}{13}=\frac{74}{15}-\frac{522}{35}=-9\frac{103}{105}\)
\(x=-9\frac{103}{105}:\frac{10}{13}=-12\frac{512}{525}\)