4/3*y-3/4*y+y=0,5
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\(\frac{4}{3}y-\frac{3}{4}y+y=0,5=\frac{1}{2}\Leftrightarrow quydongMSC=12\) nếu bạn ghi đề sai dẫn đến sai
\(\Leftrightarrow\frac{16y-3y+12y}{12}=\frac{6}{12}\Rightarrow\frac{25y}{12}=\frac{6}{12}\Rightarrow25y=6\Rightarrow y=\frac{6}{25}\)
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\(\frac{4}{3}.y-\frac{3}{4}.y+y=0,5\)
\(y.\left(\frac{4}{3}-\frac{3}{4}+1\right)=\frac{1}{2}\)
\(y.\frac{19}{12}=\frac{1}{2}\)
\(y=\frac{6}{19}\)
Vậy \(y=\frac{6}{19}\)
a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)
y = \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)
y = \(\dfrac{4}{3}\)
b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)
y - 0,5 + 0,5 = \(\dfrac{3}{4}\)
y = \(\dfrac{3}{4}\)
c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2
0,8 - 0,4y = 0,2
0,4y = 0,8 - 0,2
0,4y = 0,6
y = 1,5
d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{14}{9}\)
y = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)
y = \(\dfrac{29}{36}\)
e, y : \(\dfrac{5}{4}\) = \(\dfrac{9}{5}\) + \(\dfrac{1}{2}\)
y : \(\dfrac{5}{4}\) = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{8}\)
f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\)) = \(\dfrac{4}{5}\)
2y = \(\dfrac{4}{5}\)
y = \(\dfrac{2}{5}\)
\(y\div\frac{2}{10}+y\div12,5\%+4\div\frac{1}{y}+y+y\div0,5=\frac{20}{3}\)
\(\Rightarrow y\times5+y\times8+4\times y+y+y=\frac{20}{3}\)
\(\Rightarrow y\times\left(5+8+4+1+1\right)=\frac{20}{3}\)
\(\Rightarrow y\times19=\frac{20}{3}\)
\(\Rightarrow y=\frac{20}{3}\div19\)
\(\Rightarrow y=\frac{20}{57}\)
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\(x\) = 4\(\dfrac{3}{7}\) = \(\dfrac{31}{7}\)
Thay \(x\) = \(\dfrac{31}{7}\) và y = 0,5 vào biểu thức: \(x^2\) . y2 ta có:
(\(\dfrac{31}{7}\))2 . (0,5)2 = (\(\dfrac{31}{7}\) . 0,5)2 = \(\left(\dfrac{31}{14}\right)^2\) = \(\dfrac{961}{196}\)
1.Thực hiện phép tính :
a) -4,3y - \(\frac{1}{2}y-\frac{3}{4}=-0,4\)
=> (-4,3- \(\frac{1}{2}\))y = -0,4 + \(\frac{3}{4}\)
=> \(\frac{-24}{5}\)y = \(\frac{7}{20}\)
=> y = \(\frac{7}{20}:\frac{-24}{5}\)
=> y = \(\frac{-7}{96}\)
b) 4\(\left(y-\frac{1}{3}\right)^3=0\)
=> y3 - \(\frac{1}{3}^3\) = 0
=> y3 - \(\frac{1}{27}=0\)
=> y3 = \(\frac{1}{27}\)
=> y = \(\frac{1}{3}\)
c) 13 -2 . | 1 - 2y | = 1
=> 2.| 1 - 2y | = 13 - 1
=> 2.| 1 - 2y | = 12
=> | 1 - 2y | = 6
=> \(\left[\begin{matrix}1-2y=6\\1-2y=-6\end{matrix}\right.\)
=> \(\left[\begin{matrix}2y=-5\\2y=7\end{matrix}\right.\)
=> \(\left[\begin{matrix}y=\frac{-5}{2}\\y=\frac{7}{2}\end{matrix}\right.\)
Bài 1:
a)\(\frac{-4}{3}.y.\frac{-1}{2}.y.\frac{-3}{4}=-0.4\)
\(\Leftrightarrow\frac{-4}{3}.\frac{-3}{4}.\frac{-1}{2}.y^2=\frac{-2}{5}\)
\(\Leftrightarrow\frac{-1}{2}y^2=\frac{-2}{5}\)
\(\Leftrightarrow y^2=\frac{4}{5}\)
\(\Leftrightarrow\left[\begin{matrix}y=\sqrt{\frac{4}{5}}\\y=-\sqrt{\frac{4}{5}}\end{matrix}\right.\)
b) \(4.\left(y-\frac{1}{3}\right)^3=0\)
\(\Leftrightarrow\left(y-\frac{1}{3}\right)^3=0\)
\(\Leftrightarrow y-\frac{1}{3}=0\)
\(\Leftrightarrow y=\frac{1}{3}\)
c) 13-2.|1-2y|=1
<=>2.|2y-1|=12
<=>|2y-1|=6
<=> \(\left[\begin{matrix}2y-1=6\\2y-1=-6\end{matrix}\right.\)
<=>\(\left[\begin{matrix}y=3,5\\y=-2,5\end{matrix}\right.\)
4/3*y-3/4*y+y=0,5
y*(4/3-3/4+1)=0,5
y*19/12=0,5
y=0,5/19/12
y=6/19