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\(B=\left(1,6\cdot0,4\right)^2-\dfrac{57}{4}+\dfrac{7}{5}-\dfrac{7}{4}=\dfrac{256}{625}-16+\dfrac{7}{5}=-\dfrac{8869}{625}\)
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
(7 + 3 1/4 - 3/5) + (0,4 - 5) - (4 1/4 - 1)
(7 + 13/4 - 3/5) + (2/5 - 5) - (17/4 - 1)
= 7 + 13/4 - 3/5 + 2/5 - 5 - 17/4 + 1
= (7 - 5 + 1) + (13/4 - 17/4) + (-3/5 + 2/5)
= 3 - 1 - 1/5
= 2 - 1/5
= 9/5
(7 + 3\(\dfrac{1}{4}\) - \(\dfrac{3}{5}\)) + (0,4 - 5) - ( 4\(\dfrac{1}{4}\) - 1)
= 7 + 3 + \(\dfrac{1}{4}\) - 0,6 - 4,6 - 4 - \(\dfrac{1}{4}\) + 1
= ( 7 + 3 - 4 + 1) + (\(\dfrac{1}{4}\) - \(\dfrac{1}{4}\)) - ( 0,6 + 4,6)
= 7 - 5,2
= 1,8
a: \(0.4\cdot\sqrt{0.25-\sqrt{\dfrac{1}{4}}}=0.4\cdot\sqrt{0.25-0.5}\)(đề này sai rồi bạn)
b: \(\dfrac{3}{2}+2\left(x-1\right)=-5\dfrac{1}{2}\)
\(\Leftrightarrow2\left(x-1\right)=\dfrac{-11}{2}-\dfrac{3}{2}=-7\)
\(\Leftrightarrow x-1=\dfrac{-7}{2}\)
hay \(x=-\dfrac{5}{2}\)
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
<=> \(\left[\begin{matrix}2y-1=6\\2y-1=-6\end{matrix}\right.\)
<=>\(\left[\begin{matrix}y=3,5\\y=-2,5\end{matrix}\right.\)
= \(\dfrac{15}{2}\)
Bn có thể trình bày đầy đủ giúp mk đc ko ạ?