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\(.1.\)
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)
Ta có : \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)
....................
\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)
____________
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}.100=\sqrt{100}=10\)
Vậy : \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>0\)
Bài 2: Ta thấy:\(\left\{\begin{matrix}\left|2x-6\right|\ge0\\\left|3y+9\right|\ge0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}-\left|2x-6\right|\le0\\-\left|3y+9\right|\le0\end{matrix}\right.\)
\(\Rightarrow-\left|2x-6\right|-\left|3y+9\right|\le0\)
\(\Rightarrow-18-\left|2x-6\right|-\left|3y+9\right|\le-18\)
\(\Rightarrow C\le-18\)
Dấu "=" xảy ra khi \(\left\{\begin{matrix}-\left|2x-6\right|=0\\-\left|3y+9\right|=0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x=3\\y=-3\end{matrix}\right.\)
Vậy với \(\left\{\begin{matrix}x=3\\y=-3\end{matrix}\right.\) thì C đạt GTLN là -18
a: \(=0.5\cdot10-\dfrac{1}{7}+15=20-\dfrac{1}{7}=\dfrac{139}{7}\)
b: \(=6\cdot\dfrac{-2}{3}+12\cdot\dfrac{4}{9}+18\cdot\dfrac{-8}{27}\)
\(=-4+\dfrac{16}{3}-\dfrac{16}{3}=-4\)
c: \(=\left(\dfrac{5}{2}+\dfrac{3}{8}-\dfrac{5}{8}+\dfrac{2}{3}\right):\left(\dfrac{17}{2}+\dfrac{49}{4}-\dfrac{17}{8}+\dfrac{34}{15}\right)\)
\(=\dfrac{35}{12}:\dfrac{2507}{120}=\dfrac{350}{2507}\)
Bài 1:
a: \(\Leftrightarrow2-3\sqrt{x}+5\sqrt{x}=8\)
=>2 căn x=6
=>căn x=3
=>x=9
b: \(\Leftrightarrow\dfrac{1}{\sqrt{x}}\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}}=\dfrac{2}{3}:\dfrac{2}{3}=1\)
=>x=1
B1
a. = 7/3. ( 37/5 - 32/5)
= 7/3 . 1
= 7/3
Phần b có gì đó sai sao lại có 3:+
c. = 4 + 6 - 3 + 5
= 12
d. = -5/21 : -19/21 : 4/5
= 25/76
B2
a. 1/4 : x =1/2 - 3/4
x = -1/4
x = 1/4 : -1/4
x = -1
b. 2 . | 2x - 3 | = 4 - (-8)
2 . | 2x - 3| = 12
| 2x - 3 | = 12:2
| 2x - 3 | = 6
| x - 3 | = 6:2
| x - 3 | = 3
=> x - 3 = +- 3
* x - 3 = 3
x = 6
* x - 3 = -3
x = 0
Chúc bạn vui vẻ 
a: \(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)
\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)
\(=\dfrac{5}{3}\cdot12=20\)
b: \(=\dfrac{3}{5}:\left(\dfrac{-2-5}{30}\right)+\dfrac{3}{5}:\left(\dfrac{-1}{3}-\dfrac{16}{15}\right)\)
\(=\dfrac{3}{5}:\dfrac{-7}{30}+\dfrac{3}{5}:\dfrac{-21}{15}\)
\(=\dfrac{3}{5}\left(\dfrac{-30}{7}-\dfrac{15}{21}\right)=\dfrac{3}{5}\cdot\left(\dfrac{-30}{7}-\dfrac{5}{7}\right)=\dfrac{3}{5}\cdot\left(-5\right)=-3\)
c: \(=5.7\left(-6.5-3.5\right)\)
\(=5.7\cdot\left(-10\right)=-57\)
d: \(=10\cdot0.1\cdot\dfrac{4}{3}+3\cdot7-\dfrac{1}{6}\cdot2\)
\(=\dfrac{4}{3}+21-\dfrac{1}{3}=22\)
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
1)
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}=\dfrac{100}{\sqrt{100}}=10\left(đpcm\right)\)
2)
\(C=-18-\left|2x-6\right|-\left|3y+9\right|\le-18\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\)