BT1: Tìm x, biết:
3) \(\dfrac{0,3.x+2,5}{x}-3=17\)
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Giải:
\(0,28-0,3:\left(50\%x-1\dfrac{1}{3}\right)=-1\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{7}{25}-\dfrac{3}{10}:\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)=-\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3}{10}:\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)=\dfrac{7}{25}-\left(-\dfrac{5}{3}\right)\)
\(\Leftrightarrow\dfrac{3}{10}:\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)=\dfrac{146}{75}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{4}{3}=\dfrac{3}{10}:\dfrac{146}{75}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{4}{3}=\dfrac{45}{292}\)
\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{45}{292}-\dfrac{4}{3}\)
\(\Leftrightarrow\dfrac{1}{2}x=-\dfrac{1033}{876}\)
\(\Leftrightarrow x=-\dfrac{1033}{876}:\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1033}{438}\)
Vậy \(x=-\dfrac{1033}{438}\).
Chúc bạn học tốt!
\(\dfrac{1}{3}-\left(1\dfrac{1}{2}-x\right)=0,3\\ \dfrac{1}{3}-\left(\dfrac{3}{2}-x\right)=\dfrac{3}{10}\\ \dfrac{3}{2}-x=\dfrac{1}{3}-\dfrac{3}{10}\\ \dfrac{3}{2}-x=\dfrac{1}{30}\\ x=\dfrac{3}{2}-\dfrac{1}{30}=\dfrac{44}{30}\)
16 : (0,3 * x +2,5) : x = 17 + 3= 20
-> \(\frac{0,3\times x+2,5}{x}=\frac{16}{20}=\frac{4}{5}\)
\(\frac{0,3\times x}{x}+\frac{2,5}{x}=\frac{4}{5}\)
\(0,3+\frac{2,5}{x}=\frac{4}{5}\)
\(\frac{2,5}{x}=\frac{4}{5}-0,3\)
\(\frac{2,5}{x}=\frac{1}{2}\)
\(x=2,5:\frac{1}{2}=5\)
Mình thay x là a nha:
16 : (0,3 x a + 2,5) : a - 3 = 17
(16 : 0,3 x 16 :a + 26 : 2,5) : a = 17 + 3
(\(\frac{160}{3}\) x \(\frac{16}{a}\) + \(\frac{52}{5}\)) : a = 20
\(\frac{160}{3}\) : a x \(\frac{16}{a}\) :a + \(\frac{52}{5}:a\)= 20
\(\frac{160}{3a}\times\frac{16}{a\times a}+\frac{52}{5a}=20\)
...
\(\dfrac{-8}{13}+\dfrac{-7}{17}+\dfrac{21}{13}\le x\le\dfrac{-9}{14}+3+\dfrac{5}{-14}\)
=> \(\dfrac{10}{17}\le x\le2\)
=> \(\dfrac{10}{17}\le\dfrac{17x}{17}\le\dfrac{34}{17}\)
=> 10 \(\le17x\le34\)
=> x = 1; 2 (thỏa mãn)
@Khánh Linh
16 : (0,3 x x +2,5):x-3=17
(0,3 x x +2,5):x-3=16:17=16/17
(0,3 x x +2,5):x=16/17+3=67/17
=>0,3 x x +2,5=67x/17
0,3 x x=67x/17-2,5=67x-42.5/17
=>x=67x-42.5/17:0,3=670x-425/51
=>51x x=670x-425
=>670xx-51xx=425
=>619xx=425
=>x=425:619
Chắc mình sai
0,3 : x + 2,5 : x - 3 = 17
0,3 : x + 2,5 : x = 17 + 3
0,3 : x + 2,5 : x = 20
( 0,3 + 2,5 ) : x = 20
2,8 : x = 20
x = 2,8 : 20
x = 0,14
cho mk sửa lại
tacó:
\(\dfrac{-64}{125}=\left(\dfrac{-4}{5}\right)^3\)
suy ra\(\dfrac{2}{3}-\dfrac{3}{5}x=\dfrac{-4}{5}\)
\(\dfrac{3}{5}x=\dfrac{2}{3}-\dfrac{-4}{5}\)
\(\dfrac{3}{5}x=\dfrac{22}{15}\)
\(x=\dfrac{22}{15}:\dfrac{3}{5}\)
\(x=\dfrac{22}{9}\)
ta có:
\(\dfrac{-64}{125}=\left(\dfrac{-16}{5}\right)^3\)
suy ra \(\dfrac{2}{3}-\dfrac{3}{5}x=\dfrac{-16}{5}\)
\(\dfrac{3}{5}x=\dfrac{2}{3}-\dfrac{-16}{5}\)
\(\dfrac{3}{5}x=\dfrac{58}{15}\)
\(x=\dfrac{58}{15}:\dfrac{3}{5}\)
\(x=\dfrac{58}{9}\)
\(\dfrac{x+1}{2017}+\dfrac{x+2}{2016}=\dfrac{x+3}{2015}-1\)
\(\Leftrightarrow\dfrac{x+1}{2017}+1+\dfrac{x+2}{2016}+1=\dfrac{x+3}{2015}-1+2\)
\(\Leftrightarrow\dfrac{x+100}{2017}+\dfrac{x+100}{2016}=\dfrac{x+100}{2015}\)
\(\Leftrightarrow\dfrac{x+100}{2017}+\dfrac{x+100}{2016}+\dfrac{x+100}{2015}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\right)=0\)
Do \(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\ne0\) nên \(x+100=0\)
\(\Leftrightarrow x=\left(-100\right)\)
Vậy \(x=\left(-100\right)\)
\(\dfrac{0,3.x+2,5}{x}-3=17\)
<=> \(\dfrac{0,3.x+2,5}{x}=20\)
<=> 0,3.x + 2,5 = 20x
<=> 2,5 = 20x - 0,3.x
<=> 2,5 = 19,7.x
<=> x = \(\dfrac{25}{197}\)
@Khánh Linh