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0,75 : 4,5 = \(\dfrac{1}{15}\) : (2x)
\(\Rightarrow\) \(\dfrac{1}{6}\) = \(\dfrac{1}{30}\) : x
\(\Rightarrow\) \(\dfrac{1}{30}\) : \(\dfrac{1}{6}\) = x
\(\Rightarrow\) x = \(\dfrac{1}{5}\)
0,75 : 4,5 = 1/15 : (2x)
1/6 = 1/15 : (2x)
(2x) = 1/15 : 1/6
2x = 2/5
x = 2/5 : 2
x = 1/5
Vậy x = 1/5
chúc bn hc tốt môn toán
đó là đáp án đúng đó ạk
a) 0,75 : 4,5 = \(\frac{1}{15}\) : 2x
=> \(\frac{0,75}{4,5}\) = \(\frac{\frac{1}{15}}{2x}\)
=> 0,75 . 2x = \(\frac{1}{15}\) . 4,5
=> 0,75 . 2x = 0,3
=> 2x = 0,3 : 0,75
=> 2x = 0,4
=> x = 0,4 : 2
=>x = 0,2
Ta có: 0,75:4,5= \(\frac{1}{15}\):(2x)
suy ra \(\frac{1}{6}\)=\(\frac{1}{15}\):(2x)
suy ra 2x = \(\frac{1}{15}:\frac{1}{6}\)
suy ra 2x= \(\frac{2}{5}\)
suy ra x= \(\frac{1}{5}\)
Vậy x=...
a/ 2x - 10 - [3x - 14 - (4 - 5x) - 2x] = 2
=> 2x - 10 - (3x - 14 - 4 + 5x - 2x) = 2
=> 2x - 10 - 3x + 14 + 4 - 5x + 2x = 2
=> -4x + 6 = 0
=> -4x = -6
=> x = 3/2
b/ \(\left(\frac{1}{4}x-1\right)+\left(\frac{5}{6}x-2\right)-\left(\frac{3}{8}x+1\right)=4,5\)
\(\Rightarrow\frac{1}{4}x-1+\frac{5}{6}x-2-\frac{3}{8}x-1-\frac{9}{2}=0\)
\(\Rightarrow\frac{17}{24}x-\frac{17}{2}=0\)
\(\Rightarrow\frac{17}{24}x=\frac{17}{2}\)
\(\Rightarrow x=12\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
- Ta chứng minh bất đẳng thức phụ dưới đây: \(\frac{1}{\sqrt{x}\left(x+1\right)}=\frac{\sqrt{x}}{x\left(x+1\right)}=\sqrt{x}\left(\frac{1}{x}-\frac{1}{x+1}\right)=\sqrt{x}\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x+1}}\right)\)\(=\left(1+\frac{\sqrt{x}}{\sqrt{x+1}}\right)\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)< 2\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)\)
Áp dụng : \(\frac{1}{\sqrt{1}.2}< 2.\left(1-\frac{1}{\sqrt{2}}\right)\)
\(\frac{1}{\sqrt{2}.3}< 2.\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)\)
...................................
\(\frac{1}{\sqrt{2015}.2016}< 2.\left(\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\right)\)
Cộng các BĐT trên với nhau được : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}}< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\right)=2\left(1-\frac{1}{\sqrt{2016}}\right)< 2\left(1-\frac{1}{\sqrt{2025}}\right)=\frac{88}{45}\)
Từ đó suy ra đpcm
Cái ............... là gì vậy bn
Giải:
\(0,75:4,5=\dfrac{1}{15}:2x\)
\(\Leftrightarrow0,75.2x=\dfrac{1}{15}.4,5\)
\(\Leftrightarrow\dfrac{3}{2}x=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{3}{10}:\dfrac{3}{2}\)
\(\Leftrightarrow x=\dfrac{1}{5}\)
Vậy \(x=\dfrac{1}{5}\).
Chúc bạn học tốt!
a)\(2\left|2x-3\right|=\frac{1}{2}\)
\(\Leftrightarrow\left|2x-3\right|=\frac{1}{4}\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=\frac{1}{4}\\2x-3=-\frac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{8}\\x=\frac{11}{8}\end{matrix}\right.\)
Vậy....
b)\(7,5-3\left|5-2x\right|=-4,5\)
\(\Leftrightarrow\left|5-2x\right|=4\)
\(\Rightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{9}{2}\end{matrix}\right.\)
VẬy...
c)\(\left|3x-4\right|+\left|5-2x\right|=0\)
Có: \(\left|3x-4\right|\ge0với\forall x\\ \left|5-2x\right|\ge0với\forall x\)
\(\Rightarrow\left[{}\begin{matrix}3x-4=0\\5-2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow x\in\varnothing\)
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\Leftrightarrow\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{1}{93}\)
\(\Leftrightarrow2x+3=93\)
\(\Leftrightarrow2x=90\)
\(\Leftrightarrow x=45\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Rightarrow\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(\Rightarrow2x=90\)
\(\Rightarrow x=45\)
Vậy x = 45.
0,75:4,5=\(\frac{1}{15}\):(2x)
\(\frac{1}{6}\) = \(\frac{1}{15}\):(2x)
\(\frac{1}{15}\):(2x)=\(\frac{1}{6}\)
2x= \(\frac{1}{15}\):\(\frac{1}{6}\)
2x= 0,4 \(\Rightarrow\)x=0,4:2 \(\Rightarrow\)x=0,2
K giùm mình nha
0.75/4.5=1/15/2x (hãy viết dưới dạng phân số)
khi đó: 0.75.2x=4.5.1
0.75.2x=4.5
2x=4.5/0.75
2x=6
x=6/2
x=3