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HH
0
PL
14 tháng 3 2018
a) \(\frac{31}{23}-\left(\frac{7}{23}+\frac{8}{23}\right)\)
\(=\frac{31}{23}-\frac{15}{23}\)
\(=\frac{16}{23}\)
b) \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{-67}{67}+\frac{41}{41}\)
\(=\frac{1}{3}-1+1\)
\(=\frac{1}{3}\)
c) \(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{52}-\frac{3}{11}\right)\)
\(=\frac{38}{45}-\frac{8}{45}+\frac{17}{52}+\frac{3}{11}\)
\(=\frac{30}{45}+\frac{17}{52}+\frac{3}{11}\)
\(=\frac{2}{3}+\frac{17}{52}+\frac{3}{11}\)
\(=\frac{104+51}{156}+\frac{3}{11}\)
\(=\frac{155}{156}+\frac{3}{11}\)
\(=\frac{156}{156}-\frac{1}{156}+\frac{3}{11}\)
\(=1-\frac{1}{156}+\frac{3}{11}\)
\(=1-\left(\frac{11-468}{1716}\right)\)
\(=1-\frac{-457}{1716}\)
\(=1+\frac{457}{1716}\)
\(=\frac{2173}{1716}\)
14 tháng 3 2018
a)31/23-(7/32+8/23)=31/23-7/32-8/23=(31/23-8/23)-7/32=1-7/32=25/32
17 tháng 4 2018
a) 7/15-(2/15-12/18)
=7/15-2/15+2/3
=5/15+2/3
=1/3+2/3
=1
b)(7/41- 4/9) - (3/19 + 7/41) + (4/9 - 16/19)
=7/41 - 4/9 - 3/19 - 7/41 + 4/9 + 16/19
=(7/41 - 7/41) - (4/9 - 4/9) - (3/19 + 16/19)
= -1
LT
17 tháng 4 2018
b) 7/41 - 4/9 - 3/19 - 7/41 + 4/9 + 16/19
= (7/41 - 7/41 ) - (4/9 - 4/9 ) + ( 3/19 + 16/19 )
= 0 - 0 + 1
= 1
YN
22 tháng 2 2020
\(A=\frac{5}{13}+\frac{-5}{7}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)
\(\Leftrightarrow A=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{41}\right)+\frac{-5}{7}\)
\(\Leftrightarrow A=1+\left(-1\right)+\frac{-5}{7}\)
\(\Leftrightarrow A=0+\frac{-5}{7}=\frac{-5}{7}\)
Vậy A = \(\frac{-5}{7}\)
B= \(\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{7}{15}\)
\(\Leftrightarrow B=\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{-4}{9}+\frac{7}{15}\)
\(\Leftrightarrow B=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)
\(\Leftrightarrow B=-1+1+\frac{-2}{11}\)
\(\Leftrightarrow B=0+\frac{-2}{11}\)
\(\Leftrightarrow\) \(B=\frac{-2}{11}\)
Vậy \(B=\frac{-2}{11}\)
@@ Học tốt
Chiyuki Fujito
K cần tk nhá
G
17 tháng 5 2018
Giải:
a) \(A=\dfrac{5}{13}.\dfrac{5}{7}+\dfrac{-20}{41}+\dfrac{5}{13}+\dfrac{-21}{41}\)
\(\Leftrightarrow A=\dfrac{5}{13}.\dfrac{5}{7}+\dfrac{5}{13}+\dfrac{-21}{41}+\dfrac{-20}{41}\)
\(\Leftrightarrow A=\dfrac{5}{13}\left(\dfrac{5}{7}+1\right)+\dfrac{-41}{41}\)
\(\Leftrightarrow A=\dfrac{5}{13}.\dfrac{12}{7}+\left(-1\right)\)
\(\Leftrightarrow A=\dfrac{60}{91}+\left(-1\right)=-\dfrac{31}{91}\)
Vậy ...
b) \(B=\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{12}{11}-\dfrac{5}{7}.\dfrac{7}{11}\)
\(\Leftrightarrow B=\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{12}{11}-\dfrac{7}{11}\right)\)
\(\Leftrightarrow B=\dfrac{5}{7}.\dfrac{7}{11}\)
\(\Leftrightarrow B=\dfrac{5}{11}\)
Vậy ...
c) \(C=\dfrac{-2}{3}+\dfrac{-5}{7}+\dfrac{2}{3}+\dfrac{-2}{7}\)
\(\Leftrightarrow C=\left(\dfrac{-2}{3}+\dfrac{2}{3}\right)+\left(\dfrac{-2}{7}+\dfrac{-5}{7}\right)\)
\(\Leftrightarrow C=0+\left(-1\right)=-1\)
Vậy ...
11 tháng 3 2017
d)
Ta có: \(\dfrac{1}{51}>\dfrac{1}{100}\)
\(\dfrac{1}{52}>\dfrac{1}{100}\)
...
\(\dfrac{1}{99}>\dfrac{1}{100}\)
\(\dfrac{1}{100}=\dfrac{1}{100}\)
\(\Rightarrow S=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{99}+\dfrac{1}{100}>\dfrac{1}{100}.50=\dfrac{1}{2}\)\(\Rightarrow S>\dfrac{1}{2}\)
20 tháng 3 2017
d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B
26 tháng 3 2017
cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A
Suy ra B>A(chuc ban hoc goi nhe)
a=4100, b=−4059
a=73a+27a+100b
=100a+100b
=100(a+b)
\(=100\cdot41=4100\)