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Ta có:5/20>5/25
5/21>5/25
5/22>5/25
5/23>5/25
5/24>5/25
=>S=5/20+5/21+5/22+5/23+5/24>5/25+5/25+5/25+5/25+5/25=1
=>5/20+5/21+5/22+5/23+5/24>1
DỄ
DO: 5/20 <1
5/21<1
5/22<1
5/23<1
5/24<1
=> 5/20+5/21+5/22+5/23+5/24<1
hay S<1 ( ĐPCM)
ĐÚNG NÈ ỦNG HỘ
2) Tinh nhanh:
a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)
= \(\dfrac{5}{598}\)
b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)
a: 7/30=21/90
8/45=16/90
11/90=11/90
b: -4/5=-168/210
1/6=35/210
-9/7=-270/210
c: -7/24=-21/72
11/12=66/72
-23/36=-46/72
d: 17/30=85/150
-22/75=-44/150
5=750/150
a: 2/9=4/18
1/3=6/18
5/18=5/18
b: 7/15=14/30
1/5=6/30
-5/6=-25/30
c: -21/56=-3/7
-3/16=-63/336
5/24=70/336
-21/56=-3/7=-144/336
d: \(\dfrac{-4}{7}=\dfrac{-36}{63}\)
8/9=56/63
\(-\dfrac{10}{21}=-\dfrac{30}{63}\)
e: 3/-20=-3/20=-9/60
-11/-30=11/30=22/60
7/15=28/60
a)\(\dfrac{5}{23}.\dfrac{17}{26}+\dfrac{5}{23}.\dfrac{10}{26}-\dfrac{5}{23}\)
\(=\dfrac{5}{23}\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
\(=\dfrac{5}{23}.\left(\dfrac{27}{26}-1\right)\)
\(=\dfrac{5}{23}.\dfrac{1}{26}\)
\(=\dfrac{5}{598}\)
b)\(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
\(=\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}.1=\dfrac{5}{9}\)
a)\(\dfrac{5}{23}.\dfrac{17}{26}+\dfrac{5}{23}.\dfrac{10}{26}-\dfrac{5}{23}\)
\(=\dfrac{5}{23}.\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
\(=\dfrac{5}{23}.\left(\dfrac{27}{26}-\dfrac{26}{26}\right)\)
=\(\dfrac{5}{23}.\dfrac{1}{26}\)
\(=\dfrac{5}{598}\)
b)\(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
\(=\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}.\left(\dfrac{7}{7}\right)\)
=\(\dfrac{5}{9}.1\)
\(=\dfrac{5}{9}\)
mình ghi nhầm nên các bạn cứ hết hai phân số là một câu nhé ví dụ như \(\dfrac{-5}{8}\):\(\dfrac{15}{4}\)
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 ...
Ta có:\(\dfrac{1}{20}>\dfrac{1}{21}>\dfrac{1}{22}>\dfrac{1}{23}>\dfrac{1}{24}>\dfrac{1}{25}\)
=>S=\(\dfrac{5}{20}+\dfrac{5}{21}+\dfrac{5}{22}+\dfrac{5}{23}+\dfrac{5}{24}=5\left(\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+\dfrac{1}{24}\right)>5\cdot\left(\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{25}\right)\)
=>S>\(5\cdot\dfrac{5}{25}\)
=>S>1(đpcm)
Cảm ơn bạn nhiều nha!!!
Giải:
Dễ thấy:
\(20< 25\Leftrightarrow\dfrac{5}{20}>\dfrac{5}{25}\)
\(21< 25\Leftrightarrow\dfrac{5}{21}>\dfrac{5}{25}\)
\(......................\)
\(24< 25\Leftrightarrow\dfrac{5}{24}>\dfrac{5}{25}\)
Cộng vế theo vế ta có:
\(S>\dfrac{5}{25}+\dfrac{5}{25}+...+\dfrac{5}{25}=\dfrac{5}{25}.5=1\)
Vậy \(S>1\) (Đpcm)
\(S=\dfrac{5}{20}+\dfrac{5}{21}+\dfrac{5}{22}+\dfrac{5}{23}+\dfrac{5}{24}>\dfrac{5}{25}+\dfrac{5}{25}+\dfrac{5}{25}+\dfrac{5}{25}+\dfrac{5}{25}=1\\ \Rightarrow S>1\)