(1/3+12/67+13/41)-(79/67-28/41)
giúp mình với
(1/3+12/67+13/41)-(79/67-28/41)
bằng 1/3 bạn nhé ..check mk đó.
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)+\frac{1}{3}\)
\(=\left(-1\right)+1+\frac{1}{3}\)
\(=\frac{1}{3}\)
(1/3+(-12)/13+13/41)-(79/67-28/41) = ?
tính bằng cách hợp lý
(1/3+12/67+13/41)-(79/67-28/41)
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
=\(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(=\frac{1}{3}-1+1\)
\(=\frac{1}{3}\)
(1/3+12/67+13/41)-(79/67-28/41). =1/3+12/613/41-79/67+28/41. =1/3+(12/67-79/67)+(13/41+28/41) =1/3+(-1)+1=1/3+0. =1/3
GIÚP MÌNH SO SÁNH NHAAAA
A=31/23-(7/32+8/2) và B=(1/3+12/67+13/41)-(79/67-28/41)\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\frac{79}{67}+\frac{28}{41}\)
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\frac{79}{67}+\frac{28}{41}\) \(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\) \(=\frac{1}{3}+\left(-1\right)+1=\frac{1}{3}\)
Chúc bạn hc giỏi ,Trần Phúc Đông❣ღ!
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
=\(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
=\(\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
=\(\frac{1}{3}+\left(\frac{-67}{67}\right)+\left(\frac{41}{41}\right)\)
=\(\frac{1}{3}-1+1\)
=\(\frac{1}{3}\)
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
=\(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
=\(\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
=\(\frac{1}{3}+\left(\frac{-67}{67}\right)+\left(\frac{41}{41}\right)\)
=\(\frac{1}{3}-1+1\)
=\(\frac{1}{3}\)
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(=\frac{1}{3}+\left(-1\right)+1\)
\(=\frac{1}{3}+0\)
\(=\frac{1}{3}\)
_Chúc bạn học tốt_
Tính
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}-\frac{28}{41}\)
\(=\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)+\frac{1}{3}\)
\(=\left(-1\right)+1+\frac{1}{3}\)
\(=0+\frac{1}{3}\)
\(=\frac{1}{3}\)
=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
2/15-2/65-4/39
32/13-(7/23+8/23)
(1/3+12/67+13/41) - (79/67 - 28/41)
38/45 - (8/45-17/51-3/11)
2/15-2/65-4/39=4/39-4/39=0
32/13-(7/23+8/23)=32/13-15/23=541/299
(1/3+12/67+13/41) - (79/67 - 28/41)=6836/8241-1363/2747=1/3
38/145 - (8/45-17/51-3/11)=38/145-(-212/495)=1982/2871
\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(\left(\dfrac{15}{4}-5x\right)\cdot\left(9x^2-4\right)=0\)
\(\sqrt{x-2}+\dfrac{1}{3}=1\)
\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}+\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\left(-1\right)+1=\dfrac{1}{3}+0=\dfrac{1}{3}\)
\(\left(\dfrac{15}{4}-5x\right).\left(9x^2-4\right)=0\)
\(\left[{}\begin{matrix}\dfrac{15}{4}-5x=0\\9x^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}5x=\dfrac{15}{4}\\9x^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)