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
<=> \(\left[\begin{matrix}2y-1=6\\2y-1=-6\end{matrix}\right.\)
<=>\(\left[\begin{matrix}y=3,5\\y=-2,5\end{matrix}\right.\)
2 : Thực hiện phép tính :
a) \(\frac{22}{7}:\left(-\frac{10}{7}\right)-\frac{7}{3}\)
= \(\frac{22}{7}.\frac{-7}{10}-\frac{7}{3}\)
= \(\frac{-11}{5}-\frac{7}{3}\)
= \(\frac{-68}{15}\)
b) \(\frac{28}{7}:\frac{14}{3}-2.\left(\frac{-1}{13}\right)\)
= \(\frac{28}{7}.\frac{3}{14}-\frac{-2}{13}\)
= \(\frac{6}{7}+\frac{2}{13}\)
= \(\frac{92}{91}\)
c) \(\frac{31}{4}:0,5+\frac{7}{4}:\left(-0,5\right)\)
= \(\frac{31}{4}.2+\frac{7}{4}.\left(-2\right)\)
= \(\frac{31}{4}.2+\frac{-7}{4}.2\)
= \(\left(\frac{31}{4}.\frac{7}{4}\right).2\)
= \(\frac{19}{2}.2\)
= 19
