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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 )531 + ( 218 - x ) = 735
218 - x = 735 - 531
218 - x = 204
x = 218 - 204
x = 14
a ) 541 + ( 218 - x ) = 735
218 - x = 735 - 541
218 - x = 194
x = 218 - 194
x = 24
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 \(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 3\)
\(3\frac{1}{5}x+125x=178\)
=> 3,2x + 125x = 178
=> 128,2x = 178
\(\Rightarrow x=\frac{890}{641}\)
\(3\frac{1}{5}x+125x=178 \)
\(\frac{16}{5}x+125x=178\)
\(x.\left(\frac{16}{5}+125\right)=178\)
\(x.\frac{641}{5}=178\)
\(x=178:\frac{641}{5}\)
\(x=\frac{890}{641}\)
vậy \(x=\frac{890}{641}\)