so sánh \(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}\)với 3
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a) A=\(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}\)
ta có :
\(A=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)
\(\Rightarrow A=\left(1+1+1\right)-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)
\(\Rightarrow A=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)
Vậy \(A< 3\)
a. Ta có :
\(\frac{178}{179}< 1\left(\frac{1}{179}\right)\)
\(\frac{179}{180}< 1\left(\frac{1}{180}\right)\)
\(\frac{183}{181}>1\left(\frac{3}{181}\right)\left(1\right)\)
Mà \(\frac{3}{181}>\frac{1}{179}+\frac{1}{180}\left(=\frac{359}{32220}< \frac{3}{181}\right)\left(2\right)\)
Từ \(\left(1\right)\&\left(2\right)\Rightarrow\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 1+1+1\)
Vậy \(A< 3\)
\(A=\frac{4\cdot0,125\cdot20,2\cdot800\cdot0,25}{1,01\cdot75+0,26\cdot101-1,01}\)
\(=\frac{4\cdot0,25\cdot0,125\cdot800\cdot20,2}{1,01\cdot75+0,26\cdot100\cdot1,01-1,01}\)
\(=\frac{1\cdot100\cdot20,2}{1,01\cdot\left(75+26-1\right)}\)
\(=\frac{100\cdot20,2}{100\cdot1,01}\)
= 20
\(B=\left(\frac{178}{179}+\frac{179}{180}+\frac{180}{181}\right)\cdot\left(\frac{80}{56}-\frac{15}{12}:\frac{7}{8}\right)\)
\(=\left(\frac{178}{179}+\frac{179}{180}+\frac{180}{181}\right)\cdot\left(\frac{10}{7}-\frac{5}{4}\cdot\frac{8}{7}\right)\)
\(=\left(\frac{178}{179}+\frac{179}{180}+\frac{180}{181}\right)\cdot\left(\frac{10}{7}-\frac{10}{7}\right)\)
\(=\left(\frac{178}{179}+\frac{179}{180}+\frac{180}{181}\right)\cdot0\)
= 0
Ta có: 179/197 < 1 (Vì 179 < 197)
971/917 > 1 (Vì 971 > 917)
Vì 179/197 < 1 < 971/917
Nên 179/197 < 971/917
Chúc Bạn Thi Tốt
Lời giải:
$\frac{179}{197}< 1< \frac{971}{917}$
$\frac{183}{184}> 0> \frac{-184}{183}$
1/Vì 179/197<1 ; 971/917>1
=>179/197<971/917
2/Vì 183/184<1 ; 184/183>1
=>183/184<184/183
3/Ta có : -3/31=-3*101/31*101=-303/3131
Vì -303>-789 =>-303/3131>-789/3131 =>-3/31>-789/3131
\(\dfrac{-178}{179}>-1>\dfrac{-191}{189}\\ \dfrac{127}{129}=1-\dfrac{2}{129};\dfrac{871}{873}=1-\dfrac{2}{873}\\ \dfrac{2}{129}>\dfrac{2}{873}\left(129< 873\right)\Leftrightarrow1-\dfrac{2}{129}< 1-\dfrac{2}{873}\Leftrightarrow\dfrac{127}{129}< \dfrac{871}{873}\)
a) 179/197 là số bé
971/917 là số lớn
=>179/197<971/917
b)183/184 là số bé
-184/-183=184/183 là số lớn
=> 183/184<184/183
a) Do 179/197 < 1
971/917 > 1
=> 179/197 < 971/917
b) Do 183/184 < 1
-184/-183 = 184/183 > 1
=> 183/184 < -184/-183
Ta có \(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)
\(=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)
Ta thấy \(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}>0\)suy ra \(3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)
Khi đó \(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 3\)