1-2+3-4+5-6+7-8+...-98+99
Giải hộ mk nha mk dg cần gấp
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1 + 2 + 3 + ... + 199 + 200
Ta có : 1 + 2 + 3 + ... + 199 + 200 ( có 200 số )
= (200 + 1) x 200 : 2 = 20100
3 + 5 + 7 + ... + 97 + 99
Ta có : 3 + 5 + 7 + ... + 97 + 99 (có 49 số )
= (99 + 3) . 49 : 2 = 2499
\(-\frac{1}{10}< =x< =\frac{3}{5}\)
\(\frac{-4}{9}< x< =\frac{2}{3}\)
\(S=\dfrac{2}{4\cdot7}+\dfrac{2}{7\cdot10}-\dfrac{3}{5\cdot9}-\dfrac{3}{9\cdot13}\)
\(=\dfrac{2}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}\right)-\dfrac{3}{4}\left(\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}\right)-\dfrac{3}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}\right)\)
\(=\dfrac{2}{5}\cdot\dfrac{3}{20}-\dfrac{3}{4}\cdot\dfrac{8}{65}=\dfrac{-21}{650}\)
a) \(x\cdot3\dfrac{1}{4}+\left(-\dfrac{7}{6}\right)\cdot x-1\dfrac{2}{3}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{3}{4}x-\dfrac{7}{6}x-\dfrac{2}{3}=\dfrac{5}{12}\)
\(\Leftrightarrow9x-14x-8=5\)
\(\Leftrightarrow-5x-8=5\)
\(\Leftrightarrow-5x=5+8\)
\(\Leftrightarrow-5x=13\)
\(\Rightarrow x=-\dfrac{13}{5}\)
Vậy \(x=-\dfrac{13}{5}\)
b) \(5\dfrac{8}{17}:x+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)
\(\Rightarrow5\dfrac{8}{17}:x+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\left(đk:x\ne0\right)\)
\(\Leftrightarrow\dfrac{93}{17}\cdot\dfrac{1}{x}+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{93}{17x}+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{93}{17x}+2x-\dfrac{3}{4}=-\dfrac{7}{4}\left(đk:2x-\dfrac{3}{4}\ge0\right)\\\dfrac{93}{17x}-\left(2x-\dfrac{3}{4}\right)=-\dfrac{7}{4}\left(đk:2x-\dfrac{3}{4}< 0\right)\end{matrix}\right.\)
đến đây bạn giải tiếp nhé
c) \(\left(x+\dfrac{1}{2}\right)\cdot\left(\dfrac{2}{3}-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0-\dfrac{1}{2}\\2x=0+\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{2}{3}:2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{1}{2};x_2=\dfrac{1}{3}\)
\(x+\frac{2}{15}=\frac{1}{3}\)
\(x=\frac{1}{3}-\frac{2}{15}\)
\(x=\frac{1}{5}\)
h, \(h,\frac{1}{3}-\frac{2}{3}:x=\frac{1}{4}\)
\(\frac{2}{3}:x\)= \(\frac{1}{3}-\frac{1}{4}\)
\(\frac{2}{3}:x=\frac{1}{12}\)
\(x=\frac{2}{3}:\frac{1}{12}\)
\(x=8\)
1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 97 + 98 - 99 - 100 ( có 100 số; 100 : 4 = 25)
= (1 + 2 - 3 - 4) + (5 + 6 - 7 - 8) + ... + (97 + 98 - 99 - 100) ( có 25 nhóm)
= -4 + (-4) + ... + (-4) ( có 25 số)
= -4 × 25
= -100
1 + 2 - 3 - 4 + 5 - 6 - 7 - 8 + ... + 97 + 98 - 99 - 100
= 1 + ( 2 - 3 - 4 + 5 ) + ( 6 - 7 - 8 + 9 ) + ( 98 - 99 - 100 )
= 1 + 0 + 0 + .... + 0 + ( -101 )
= 1 + ( -101 )
= 100