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VK
12 tháng 7 2019
\(a.\) \(\frac{x}{3.15}\) \(=\) \(\frac{0.15}{7.2}\)
<=> \(x=\) \(\frac{0.15\cdot3.15}{7.2}\)
<=>\(x=0.065625\)
\(b.\) \(\frac{-2.6}{x-11}=\frac{-12}{42}\)
<=>\(x-11=\frac{-2.6\cdot-12}{42}\)
<=>\(x-11=0.742857142\)
<=>\(x=11.74285714\)
14 tháng 7 2019
\(3\frac{4}{5}:2x=0,25:2\frac{2}{3}\)
\(\Leftrightarrow\frac{19}{5}:2x=0,25:\frac{8}{3}\)
\(\Leftrightarrow\frac{19}{5}:2x=0,25\cdot\frac{3}{8}\)
\(\Leftrightarrow\frac{19}{5}:2x=\frac{3}{32}\)
\(\Leftrightarrow2x=\frac{19}{5}:\frac{3}{32}=\frac{19}{5}\cdot\frac{32}{3}=\frac{608}{15}\)
\(\Leftrightarrow x=\frac{608}{15}:2=\frac{608}{30}=\frac{304}{15}\)
TN
1
14 tháng 10 2015
a)\(\) Ta có : x / 3,15 = 0,15 / 7,2
(=) x = 0,15 . 3,15 / 7,2
= 21 / 320
b) Ta có : -2,6 / x = -12 / 42
(=) x = -2,6 . 42 / -12
= 91 / 10
18 tháng 8 2015
\(\frac{-2,6}{12}=\frac{x}{-42}\Leftrightarrow x=\frac{\left(-2,6\right)\left(-42\right)}{12}=9,1\)
\(\frac{1,4}{-5}=\frac{2,1}{x}\Leftrightarrow x=\frac{2,1.\left(-5\right)}{1,4}=-7,5\)
\(\frac{3}{x-1}=\frac{21}{16}\Leftrightarrow x-1=\frac{3.16}{21}=\frac{16}{7};x=\frac{23}{7}\)
\(\frac{1,2}{30}=\frac{3x+4}{50}\Leftrightarrow3x+4=\frac{50.1,2}{30}=2;3x=-1;x=-\frac{1}{3}\)
18 tháng 8 2015
a) \(\frac{-2,6}{12}=\frac{x}{42}\)
\(x=\frac{-2,6.\left(-42\right)}{12}=9,1\)
b) \(\frac{1.4}{-5}=\frac{2,1}{x}\)
\(x=\frac{-5.2,1}{1,4}=-7,5\)
c) \(\frac{3}{x-1}=\frac{21}{16}\)
\(x-1=\frac{3.16}{21}=\frac{16}{7}\)
\(x=\frac{16}{7}+1=\frac{23}{7}\)
d) \(\frac{1,2}{30}=\frac{3x+4}{50}\)
\(3x+4=\frac{12.50}{30}=20\)
\(3x=20-4=16\)
\(x=16:3=\frac{16}{3}\)
14 tháng 8 2020
\(\left|x-1,5\right|=2\\ \Rightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
Vậy \(x\in\left\{3,5;-0,5\right\}\)
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\(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\\ \Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\\ \Rightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-\frac{5}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};-\frac{5}{4}\right\}\)
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\(\left|x-2\right|=x\left(ĐK:x\ge0\right)\\ \Rightarrow\left[{}\begin{matrix}x-2=x\\x-2=-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-x=2\\x+x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=2\left(\text{vô lý}\right)\\2x=2\end{matrix}\right.\\ \Rightarrow x=1\left(tmđk\right)\)
Vậy \(x=1\)
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\(\left|x-3,4\right|+\left|2,6-x\right|=0\\ \Rightarrow\left|x-3,4\right|=-\left|2,6-x\right|\)
Mà \(\left|2,6-x\right|\ge0\forall x\Rightarrow-\left|2,6-x\right|\le0\forall x\)
\(\Rightarrow\left|x-3,4\right|\le0\forall x\left(\text{vô lý}\right)\)
Vậy \(x\in\varnothing\)
TG
14 tháng 8 2020
a/ \(\left|x-1,5\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2+1,5=3,5\\x=-2+1,5=-0,5\end{matrix}\right.\)
b/ \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=0+\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}-\frac{3}{4}=\frac{2}{4}-\frac{3}{4}=-\frac{1}{4}\\x=-\frac{1}{2}-\frac{3}{4}=\left(-\frac{2}{4}\right)+\left(-\frac{3}{4}\right)=-\frac{5}{4}\end{matrix}\right.\)
c/ \(\left|x-2\right|=x\)
\(\Rightarrow\left[{}\begin{matrix}x-2=x\\x-2=-x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-x=2\\x+x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0=2\left(vô-lý\right)\\2x=2\end{matrix}\right.\)
=> 2x = 2
=> x = 2 : 2 = 1
d/ \(\left|x-3,4\right|+\left|2,6-x\right|=0\)
Ta có: \(\left\{{}\begin{matrix}\left|x-3,4\right|\ge0\\\left|2,6-x\right|\ge0\end{matrix}\right.\)
=> Để \(\left|x-3,4\right|+\left|2,6-x\right|=0\) thì \(\left\{{}\begin{matrix}\left|x-3,4\right|=0\\\left|2,6-x\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-3,4=0\\2,6-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0+3,4=3,4\\x=2,6-0=2,6\end{matrix}\right.\)
XO
24 tháng 6 2019
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
= \(\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\frac{8}{9}\)
= \(0\)
VN
24 tháng 1 2017
Nhiều thế :( Làm 1,2 câu thôi nhé
a) \(\frac{1}{3}+\frac{1}{4}=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}\) (bị mất nét nhưng vẫn nhìn ra là số 12 nhỉ?)
b) \(\frac{-2}{5}+\frac{7}{21}=\frac{-42}{105}+\frac{35}{105}=\frac{-7}{105}=\frac{-1}{15}\)
9 tháng 10 2018
a, Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{a-1}{2}=\frac{b-2}{3}=\frac{c-3}{4}=\frac{2b-4}{6}=\frac{3c-9}{12}=\frac{\left(a-1\right)-\left(2b-4\right)+\left(3c-9\right)}{2-6+12}=\frac{8}{8}=1\)
\(\Rightarrow\hept{\begin{cases}a-1=2\\b-2=3\\c-3=4\end{cases}\Rightarrow}\hept{\begin{cases}a=3\\b=5\\c=7\end{cases}}\)
b, \(xy=-30\Rightarrow x=\frac{-30}{y}\)
\(yz=42\Rightarrow z=\frac{42}{y}\)
\(z-x=-12\)
\(\Rightarrow\frac{42}{y}-\frac{-30}{y}=-12\Rightarrow\frac{72}{y}=-12\Rightarrow y=-6\)
Ta có: \(x=\frac{-30}{y}=\frac{-30}{-6}=5\)
\(z=\frac{42}{y}=\frac{42}{-6}=-7\)
Chúc bạn học tốt.
\(\frac{-2,6}{x}=\frac{-12}{42}\Rightarrow x=-2,6:\frac{-12}{42}=\frac{13}{5}:\frac{2}{7}=\frac{91}{10}=9,1\)
Ta có: \(\frac{-2,6}{x}=\frac{-12}{42}\)
\(\Rightarrow x=42.\left(-2,6\right):\left(-12\right)\)
\(\Rightarrow x=9,1\)
Vậy x = 9,1