UU
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
27 tháng 7 2018
a) \(\frac{3}{7}x-\frac{1}{35}=\frac{3}{5}\)
\(\frac{3}{7}x=\frac{3}{5}+\frac{1}{35}\)
\(\frac{3}{7}x=\frac{22}{35}\)
\(x=\frac{49}{35}=1,4\)
27 tháng 7 2018
b) \(1,5-x:\frac{1}{2}=\frac{1}{4}\)
\(x:\frac{1}{2}=1,5-\frac{1}{4}\)
\(x:\frac{1}{2}=\frac{5}{4}\)
\(x=\frac{5}{4}.\frac{1}{2}\)
\(x=\frac{5}{8}\)
Vậy ..
15 tháng 6 2018
a) \(\left|3x-\frac{1}{2}\right|+\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
\(\Rightarrow\left|3x-\frac{1}{2}\right|=0\) \(\Rightarrow\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
\(\Rightarrow3x-\frac{1}{2}=0\) \(\Rightarrow\frac{1}{2}y+\frac{3}{5}=0\)
\(3x=\frac{1}{2}\) \(\frac{1}{2}y=\frac{-3}{5}\)
\(x=\frac{1}{2}:3\) \(y=\left(\frac{-3}{5}\right):\frac{1}{2}\)
\(x=\frac{1}{6}\) \(y=\frac{-6}{5}\)
KL: x = 1/6; y = -6/5
b) \(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\le0\)
mà \(\left|\frac{3}{2}x+\frac{1}{9}\right|>0;\left|\frac{1}{5}y-\frac{1}{2}\right|>0\)
\(\Rightarrow\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|>0\)
=> trường hợp |3/2x +1/9| + |1/5y -1/2| < 0 không thế xảy ra
\(\Rightarrow\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\)
rùi bn lm tương tự như phần a nhé!
27 tháng 6 2019
\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
HH
15 tháng 6 2018
Giải:
Vì:
\(\left\{{}\begin{matrix}\left|3x-\dfrac{1}{2}\right|\ge0\\\left|\dfrac{1}{2}y+\dfrac{3}{5}\right|\ge0\end{matrix}\right.\)
Nên dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\left|3x-\dfrac{1}{2}\right|=0\\\left|\dfrac{1}{2}y+\dfrac{3}{5}\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{2}y+\dfrac{3}{5}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{1}{2}\\\dfrac{1}{2}y=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{6}{5}\end{matrix}\right.\)
Vậy ...
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|\le0\)
Vì:
\(\left\{{}\begin{matrix}\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|\ge0\\\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|\ge0\end{matrix}\right.\)
Dấu "=" xảy ra, khi và chỉ khi:
\(\left\{{}\begin{matrix}\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|=0\\\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{1}{5}y-\dfrac{1}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x=-\dfrac{1}{9}\\\dfrac{1}{5}y=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{5}{2}\end{matrix}\right.\)
Vậy ...
15 tháng 6 2018
\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=\frac{2}{5}\)
\(\Leftrightarrow x=\frac{2}{5}\div\frac{11}{15}=\frac{2.15}{5.11}=\frac{6}{11}\)
Vậy x = 6/11
15 tháng 6 2018
a) \(\frac{1}{3}.x+\frac{2}{5}.\left(x-1\right)=0\)
\(\frac{1}{3}.x+\frac{2}{5}.x-\frac{2}{5}=0\)
\(x.\left(\frac{1}{3}+\frac{2}{5}\right)-\frac{2}{5}=0\)
\(x.\frac{11}{15}-\frac{2}{5}=0\)
\(x.\frac{11}{15}=\frac{2}{5}\)
\(x=\frac{2}{5}:\frac{11}{15}\)
\(x=\frac{6}{11}\)
b) \(3.\left(x-\frac{1}{2}\right)-5.\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)
\(3x-\frac{3}{2}-5x-3=x+\frac{1}{5}\)
\(3x-5x-\left(\frac{3}{2}+3\right)=x+\frac{1}{5}\)
\(-2x-\frac{9}{2}=x+\frac{1}{5}\)
\(\Rightarrow-2x-x=\frac{1}{5}+\frac{9}{2}\)
\(-3x=\frac{47}{10}\)
\(x=\frac{47}{10}:\left(-3\right)\)
\(x=\frac{-47}{30}\)
\(\Rightarrow\) 1/5-3x = 2/8 : 7/5 = 5/28
\(\Rightarrow\) 3x = 1/5 - 5/28 = 3/140
\(\Rightarrow\) x = 3/140 : 3 = 1/140