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23 tháng 8 2017
B=\(6\frac{4}{9}-4\frac{4}{9}+3\frac{7}{11}\)
B=\(2+3\frac{7}{11}\)
B=\(5\frac{7}{11}\)
23 tháng 8 2017
B = \(5\frac{7}{11}=\frac{62}{11}\)
C = 1
D = \(\frac{5}{2}=2\frac{1}{2}\)
MV
23 tháng 4 2017
9) \(\dfrac{x}{4}=\dfrac{9}{x}\)
Theo định nghĩa về hai phân số bằng nhau, ta có:
\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
8)
\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)
7)
\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)
6)
\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)
5)
\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)
4)
\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
3)
\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)
2)
\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)
1)
\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)
24 tháng 4 2016
\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
= \(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{2}{6}-\frac{2}{8}+\frac{2}{10}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
= \(\frac{2\times\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\times\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2\times\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\times\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
= \(\frac{2}{7}-\frac{2}{7}\)
= 0
29 tháng 4 2018
\(B=\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\right)\cdot\dfrac{1842009}{1842010}\)
\(B=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right)\cdot\dfrac{1842009}{1842010}\)
\(B=\left[\dfrac{2\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{1\cdot\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}{\dfrac{7}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}\right]\cdot\dfrac{1842009}{1842010}\)
\(B=\left(\dfrac{2}{7}-1:\dfrac{7}{2}\right)\cdot\dfrac{1842009}{1842010}\)
\(B=\left(\dfrac{2}{7}-\dfrac{2}{7}\right)\cdot\dfrac{1842009}{1842010}\)
\(B=0\cdot\dfrac{1842009}{1842010}=0\)
4 tháng 8 2018
1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)
\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)
\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)
\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)
Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
N
20 tháng 7 2017
\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
\(=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{2}{6}-\frac{2}{8}+\frac{2}{10}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{2.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(=\frac{2}{7}-\frac{2}{7}\)
\(=0\)
1 tháng 4 2019
\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\frac{2}{7}-\frac{1}{\frac{7}{2}}\)
\(=\frac{2}{7}-\frac{2}{7}=0\)
Study well ! >_<
\(\frac25-\frac{x}{7}\) = 0,25 + \(\frac{2}{-9}\)
\(\frac15-\frac{x}{7}\) = \(\frac14\) - \(\frac29\)
\(\frac15-\frac{x}{7}\) = \(\frac{9}{36}-\) \(\frac{8}{36}\)
\(\frac15-\frac{x}{7}\) = \(\frac{1}{36}\)
\(\frac{x}{7}\) = \(\frac25-\frac{1}{36}\)
\(\frac{x}{7}\) = \(\frac{72}{180}\) - \(\frac{5}{180}\)
\(\frac{x}{7}\) = \(\frac{67}{180}\)
\(x=\frac{67}{180}\times7\)
\(x\) = \(\frac{469}{180}\)
Vậy \(x=\frac{469}{180}\)
Ta có: \(\dfrac{2}{5}-\dfrac{x}{7}=0,25+\dfrac{2}{-9}\)
=>\(\dfrac{2}{5}-\dfrac{x}{7}=\dfrac{1}{4}-\dfrac{2}{9}=\dfrac{9}{36}-\dfrac{8}{36}=\dfrac{1}{36}\)
=>\(\dfrac{x}{7}=\dfrac{2}{5}-\dfrac{1}{36}=\dfrac{72}{180}-\dfrac{5}{180}=\dfrac{67}{180}\)
=>\(x=\dfrac{67}{180}\cdot7=\dfrac{469}{180}\)