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\(-\left(-23\right)+\left(-36\right)+\left|-57\right|-\left(-29\right)-35=23-36+57+29-3=38\)
\(\frac{29}{9}.\frac{109}{7}-\frac{29}{9}.\frac{57}{7}+\frac{29}{9}.\frac{12}{7}-\frac{29}{9}.\frac{1}{7}\)
= \(\frac{29}{9}.\left(\frac{109}{7}-\frac{57}{7}+\frac{12}{7}-\frac{1}{7}\right)\)
= \(\frac{29}{9}.\frac{63}{7}\)
=\(\frac{29}{9}.9\)
=\(\frac{29.9}{9}\)
=\(\frac{261}{9}\)
= \(29\)
hứ 7 nộp rùi giải giúp mình à
Được cập nhật 10 giờ trước (09:01)
Toán lớp 6
Chau Nguyen Van 9 giờ trước (09:40)
Báo cáo sai phạm
299 .1097 −299 .577 +299 .127 −299 .17
= 299 .(1097 −577 +127 −17 )
= 299 .637
=299 .9
=29.99
=2619
=
vi UCLN(a,b)=57
\(\Rightarrow a⋮57,b⋮57\)
\(\Rightarrow a=57m\)va \(b=57n\)
ma a+b=228\(\Rightarrow56m+56n=228\)
=>m+n=4
Vay: Ta co bang:
m | 1 | 2 | 3 |
n | 3 | 2 | 1 |
Vay:\(\hept{\begin{cases}a=1.57=57\\b=3.57=171\end{cases}}\)hoac\(\hept{\begin{cases}a=2.57=114\\b=2.57=114\end{cases}}\)hoac\(\hept{\begin{cases}a=3.57=171\\b=1.57=57\end{cases}}\)
a) \(56+\left(-29\right)+\left(-7\right)+28+13+\left(-35\right)\)
\(=27+\left(-7\right)+28+13+\left(-35\right)\)
\(=20+28+13+\left(-35\right)\)
\(=48+13+\left(-35\right)\)
\(=61+\left(-35\right)\)
\(=26\)
b) \(\left(-213\right)+186+\left(-14\right)+217+54+\left(-49\right)\)
\(=\left(-27\right)+\left(-14\right)+217+54+\left(-49\right)\)
\(=\left(-41\right)+217+54+\left(-49\right)\)
\(=176+54+\left(-49\right)\)
\(=230+\left(-49\right)\)
\(=181\)
c) \(435+\left(-43\right)+\left(-483\right)+\left(-57\right)+383+\left(-415\right)\)
\(=392+\left(-483\right)+\left(-57\right)+383+\left(-415\right)\)
\(=\left(-91\right)+\left(-57\right)+383+\left(-415\right)\)
\(=\left(-148\right)+383+\left(-415\right)\)
\(=235+\left(-415\right)\)
\(=-180\)
a,
\(\dfrac{13}{17}=1-\dfrac{4}{17}\\ \dfrac{25}{29}=1-\dfrac{4}{29}\\ \dfrac{4}{17}>\dfrac{4}{29}\Rightarrow1-\dfrac{4}{17}< 1-\dfrac{4}{29}\Leftrightarrow\dfrac{13}{17}< \dfrac{25}{29}\)
Vậy \(\dfrac{13}{17}< \dfrac{25}{29}\)
b,
\(\dfrac{59}{101}>\dfrac{56}{101}>\dfrac{56}{105}\\ \Rightarrow\dfrac{59}{101}>\dfrac{56}{105}\)
Vậy \(\dfrac{59}{101}>\dfrac{56}{105}\)
c,
\(\dfrac{14}{55}>\dfrac{14}{56}=\dfrac{1}{4}=\dfrac{20}{80}>\dfrac{20}{83}\)
Vậy \(\dfrac{14}{55}>\dfrac{20}{83}\)
vi ca hai ko co uoc chung vi thê UCLN la1