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a, 3/2 + 3/6 + 3/12 + . . . + 3/90
= 3/1*2 + 3/2*3 + 3/3*4 + . . . + 3/9*10
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + . . . + 1/9 - 1/10
= 1/1 - 1/10 = 9/10
Vậy a = 9/10
ko chắc chắn lắm
Theo đề bài :
\(S=\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}\)
S có tất cả 10 hạng tử, do đó :
\(S\) > \(\left(\dfrac{1}{15}+\dfrac{1}{15}+\dfrac{1}{15}+\dfrac{1}{15}+\dfrac{1}{15}\right)+\left(\dfrac{1}{20}+\dfrac{1}{20}+\dfrac{1}{20}+\dfrac{1}{20}+\dfrac{1}{20}\right)\)
\(S\) > \(5\times\dfrac{1}{15}+5\times\dfrac{1}{20}=\dfrac{7}{12}\)
Vậy \(S>\dfrac{7}{12}\)
a,8/3x + 26/3= 10/3
8/3x = 10/3- 26/3 = -16/3
=>x = -16/3 : 8/3 = -2
b, (2/3-1/2)x = 5/12
1/6x = 5/12
=>x = 5/2
xong rùi đó
nhớ tk nha
Ta có: \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-1-\dfrac{1}{2}-...-\dfrac{1}{10}\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\)
Vậy \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\)
\(|x-3|-12=|-5|\)
\(\Leftrightarrow|x-3|-12=5\)
\(\Leftrightarrow|x-3|=17\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=17\\x-3=-17\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=20\\x=-17\end{cases}}\)
Vậy \(x=\left\{20;-17\right\}\)
\(|x-3|-12=|-5|\)\(\Leftrightarrow|x-3|-12=5\)
\(\Leftrightarrow|x-3|-12=5\)\(\Leftrightarrow|x-3|=17\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=-17\\x-3=17\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-14\\x=20\end{cases}}\)
Vậy \(x=-14\)hoặc \(x=20\)
\(\dfrac{423134\cdot846267-423133}{846267\cdot423133+423134}\)
\(=\dfrac{(423133+1)\cdot846267-423133}{846267\cdot423133+423134}\)
\(=\dfrac{423133\cdot846267+846267-423133}{846267\cdot423133+423134}\)
\(=\dfrac{423133\cdot846267+423134}{864267\cdot423133+423134}=1\)

\(2\sqrt3\)
\(\sqrt{12}\) = \(\sqrt{4.3}\) = 2\(\sqrt3\)
2√3