(1/3+12/67+13/41)-(79/67-28/41)
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12 tháng 3 2016
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)+\frac{1}{3}\)
\(=\left(-1\right)+1+\frac{1}{3}\)
\(=\frac{1}{3}\)
NP
1
SS
18 tháng 10 2020
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
=\(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(=\frac{1}{3}-1+1\)
\(=\frac{1}{3}\)
NH
0
TP
1
9 tháng 8 2017
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\frac{79}{67}+\frac{28}{41}\) \(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\) \(=\frac{1}{3}+\left(-1\right)+1=\frac{1}{3}\)
Chúc bạn hc giỏi ,Trần Phúc Đông❣ღ!
TJ
4
24 tháng 7 2018
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
=\(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
=\(\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
=\(\frac{1}{3}+\left(\frac{-67}{67}\right)+\left(\frac{41}{41}\right)\)
=\(\frac{1}{3}-1+1\)
=\(\frac{1}{3}\)
24 tháng 7 2018
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
=\(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
=\(\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
=\(\frac{1}{3}+\left(\frac{-67}{67}\right)+\left(\frac{41}{41}\right)\)
=\(\frac{1}{3}-1+1\)
=\(\frac{1}{3}\)
11 tháng 3 2018
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}-\frac{28}{41}\)
\(=\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)+\frac{1}{3}\)
\(=\left(-1\right)+1+\frac{1}{3}\)
\(=0+\frac{1}{3}\)
\(=\frac{1}{3}\)
18 tháng 3 2020
=1/3+12/67+13/41-79/67+28/41
=1/3+(12/67-79/67)+(13/41+28/41)
=1/3+(-1)+1
=1/3+0
=1/3
S
24 tháng 12 2022
\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}+\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\left(-1\right)+1=\dfrac{1}{3}+0=\dfrac{1}{3}\)
\(\left(\dfrac{15}{4}-5x\right).\left(9x^2-4\right)=0\)
\(\left[{}\begin{matrix}\dfrac{15}{4}-5x=0\\9x^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}5x=\dfrac{15}{4}\\9x^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\\ =\left(\dfrac{103}{201}+\dfrac{13}{41}\right)-\dfrac{1363}{2747}\\ =\dfrac{6836}{8241}-\dfrac{1363}{2747}\\ =\dfrac{1}{3}\)
\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\\ =\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\\ =\dfrac{1}{3}+\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)\\ =\dfrac{1}{3}-\dfrac{67}{67}+\dfrac{41}{41}\\ =\dfrac{1}{3}-1+1\\ =\dfrac{1}{3}\)