Tìm x
a) \(1+\frac{-1}{60}\)\(+\frac{19}{120}\)<\(\frac{x}{36}\)\(+\frac{-1}{60}\)<\(\frac{58}{90}\)\(+\frac{59}{72}\)\(+\frac{-1}{60}\)
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\(1+\frac{-1}{60}+\frac{19}{120}< \frac{x}{36}< \frac{58}{90}+\frac{59}{72}+\frac{-1}{60}\)
=> \(\frac{137}{120}< \frac{x}{36}< \frac{521}{360}\)
=> \(\frac{411}{360}< \frac{10x}{360}< \frac{521}{360}\)
=> 411 < 10x < 521
=> x \(\in\){ 42,43,44,...,52}
Bài 2,
a, \(\frac{1}{9+1}+\frac{1}{9\left(9+1\right)}\)
\(=\frac{1}{10}+\frac{1}{9.10}\)
\(=\frac{9+1}{9.10}\)
\(=\frac{1}{9}\)
\(\Leftrightarrow\frac{1}{9}=\frac{1}{9+1}+\frac{1}{9\left(9+1\right)}\)(đpcm)
b, \(\frac{1}{m+1}+\frac{a\left(m+1\right)-b}{b\left(m+1\right)}\)
\(=\frac{b+a\left(m+1\right)-b}{b\left(m+1\right)}\)
\(=\frac{a}{b}\)
\(\Leftrightarrow\frac{a}{b}=\frac{1}{m+1}+\frac{a\left(m+1\right)-b}{b\left(m+1\right)}\)(đpcm)
\(1\frac{13}{15}\cdot\left(0,5\right)^2\cdot3-\left(\frac{8}{15}+1\frac{19}{60}\right):1\frac{19}{60}\)
\(=\frac{28}{15}\cdot\frac{1}{4}\cdot\frac{3}{1}-\frac{37}{20}\cdot\frac{60}{79}\)
\(=\frac{7}{5}-\frac{111}{79}\)
\(=\frac{-2}{395}\)
=28/15.1/4.3-8/15-79/60.60/79
=7.4/15.1/4.3-8/15-1
=7/15.3-8/15-1
=21/15-8/15-1
=13/15-1
=-2/15