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\(B=\left(1,6\cdot0,4\right)^2-\dfrac{57}{4}+\dfrac{7}{5}-\dfrac{7}{4}=\dfrac{256}{625}-16+\dfrac{7}{5}=-\dfrac{8869}{625}\)
Bài 3:
a: \(6\cdot15=2\cdot45\)
=>6/2=45/15; 6/45=2/15; 2/6=15/45; 45/6=15/2
b: \(-0.125\cdot16=0.4\cdot\left(-5\right)\)
\(\Leftrightarrow\dfrac{-0.125}{0.4}=\dfrac{-5}{16};\dfrac{-0.125}{-5}=\dfrac{0.4}{16};\dfrac{0.4}{-0.125}=\dfrac{16}{-5};\dfrac{-5}{-0.125}=\dfrac{16}{0.4}\)
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
<=> \(\left[\begin{matrix}2y-1=6\\2y-1=-6\end{matrix}\right.\)
<=>\(\left[\begin{matrix}y=3,5\\y=-2,5\end{matrix}\right.\)
a) \(-1\frac{5}{7}.15+\frac{2}{7}.\left(-15\right)+\left(-105\right).\left(\frac{2}{3}-\frac{4}{5}+\frac{1}{7}\right)\)
\(=1\frac{5}{7}.\left(-15\right)+\frac{2}{7}.\left(-15\right)+\left(-105\right).\left(\frac{70}{105}-\frac{84}{105}+\frac{15}{105}\right)\)
\(=\left(-15\right)\left(1+\frac{5}{7}+\frac{2}{7}\right)+\left(-105\right).\frac{1}{105}\)
\(=-30-1=-31\)
b) \(\frac{2}{3}+\frac{3}{4}.\left(-\frac{4}{9}\right)\)
= \(=\frac{2}{3}+\frac{3.\left(-4\right)}{4.9}=\frac{2}{3}+\frac{-1}{3}=\frac{1}{3}\)
c) \(\left(\frac{3}{4}-0,2\right).\left(0,4-\frac{4}{5}\right)\)
\(=\left(\frac{3}{4}-\frac{1}{5}\right).\left(\frac{2}{5}-\frac{4}{5}\right)=\frac{11}{20}.\left(\frac{-2}{5}\right)=\frac{11.\left(-1\right).2}{2.10.5}=\frac{-11}{50}\)
a) \(\frac{4}{5}\)✖\(\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
= \(\frac{4}{5}\)✖ \((\frac{15}{4})^2\)
= \(\frac{4}{5}\)✖ \(\frac{225}{16}\)
= \(\frac{1}{1}\times\frac{45}{4}\)
= \(\frac{45}{4}\)
a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-3}{2}+\dfrac{1}{2}+\dfrac{3}{4}\)
= (-1) + \(\dfrac{3}{4}\)
= \(\dfrac{-4}{4}+\dfrac{3}{4}\)
= \(\dfrac{-1}{4}\)
b; 0,5 + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)
= (\(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{5}{7}\)- \(\dfrac{4}{35}\)+ \(\dfrac{2}{5}\))
= ( \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{3}{5}\) + \(\dfrac{2}{5}\))
= 1 + 1
= 2
= \(\dfrac{15}{2}\)
Bn có thể trình bày đầy đủ giúp mk đc ko ạ?