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a) \(5\frac{3}{7}+\frac{7}{29}-0,7-\frac{3}{17}+\frac{22}{19}-2,3\)
\(=\frac{38}{7}+\frac{7}{29}-\frac{7}{10}-\frac{3}{17}+\frac{22}{19}-\frac{23}{10}\)
\(=\left(-\frac{7}{10}-\frac{23}{10}\right)+\frac{38}{7}+\frac{7}{29}-\frac{3}{17}+\frac{22}{19}\)
\(=\left(-3\right)+\frac{38}{7}+\frac{7}{29}-\frac{3}{17}+\frac{22}{19}\)
\(=\frac{17}{7}+\frac{7}{29}-\frac{3}{17}+\frac{22}{19}\)
\(=\frac{8605}{3451}+\frac{22}{19}\)
\(\approx4.\)
d) \(\sqrt{0,36}-\sqrt{0,81}+\sqrt{0,49}\)
\(=0,6-0,9+0,7\)
\(=\left(-0,3\right)+0,7\)
\(=0,4.\)
Chúc bạn học tốt!
mk làm bài 1 thui,bài 2 chỉ qui đồng ms
3a/6 = 3b/4 => 3(a-b)/ (6-4) = 3.4,5/2= 13,5/2 =k
a = 2k=13,5
b = 4k/3 =9
a,(1/3/7-2/1/4) . 3/1/3
= -23/28 .3/1/3
= -115/42
b,(2/1/3+3/1/2):(-4/1/6+3/1/7)+7/1/2
= 35/6 : -43/42 +7/1/2
= -245/43 +7/1/2
= 155/86
\(\left(1\frac{3}{7}-2\frac{1}{4}\right).3\frac{1}{3}\)
\(=\left(\frac{10}{7}-\frac{9}{4}\right).\frac{10}{3}\)
\(=-\frac{23}{28}.\frac{10}{3}\)
\(=\frac{-115}{42}\)
Ta có: \(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...............+\frac{2}{2007}+\frac{1}{2008}\)
\(B=\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+........+\left(1+\frac{1}{2008}\right)+1\)
\(B=\frac{2009}{2}+\frac{2009}{3}+..............+\frac{2009}{2008}+\frac{2009}{2009}\)
\(B=2009\left(\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{2009}\right)\)
Khi đó: \(\text{}\text{}\text{}\frac{A}{B}=\frac{1}{2009}\)
Chuc bạn học tốt!!
Ta có: \(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(=2008+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(=\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{2}{2007}\right)+\left(1+\frac{1}{2008}\right)\)
\(=\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}\)
\(=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}\right)\)
Ta có: \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}}{2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}\right)}\)
hay \(\frac{A}{B}=\frac{1}{2009}\)
Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)
\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
\(A=\frac{B}{6}=\frac{100}{2}=50\)
Vậy \(A=50\)
1)\(\frac{-8}{5}+\frac{207207}{201201}\)
=\(\frac{-8}{5}+\frac{207}{201}\)
=\(\frac{-8}{5}+\frac{69}{67}\)
=\(\frac{-191}{335}\)
Bài 1 : Thực hiện phép tính :
a, \(\frac{4}{5}+1\frac{1}{6}\cdot\frac{3}{4}\)
= \(\frac{4}{5}+\frac{7}{6}\cdot\frac{3}{4}\)
= \(\frac{4}{5}+\frac{7}{8}\)
= \(\frac{32+35}{40}=\frac{67}{40}\)
b, \(\frac{2}{3}:\left(\frac{3}{4}\cdot\frac{4}{3}\right)+2\)
\(=\frac{2}{3}:1+2\)
\(=\frac{2}{3}+2=\frac{2+6}{3}=\frac{8}{3}\)
c, \(\frac{1}{2}\times\left(\frac{2}{3}+\frac{3}{5}\cdot\frac{5}{7}\right)+1\frac{1}{3}\)
\(=\frac{1}{2}\cdot\left(\frac{2}{3}+\frac{9}{35}\right)+\frac{4}{3}\)
\(=\frac{1}{2}\cdot\frac{97}{105}+\frac{4}{3}\)
\(=\frac{97}{210}+\frac{4}{3}=\frac{377}{210}\)
Bài 2 : Tìm \(x\inℤ\), biết :
a, \(\frac{2}{3}< \frac{x}{6}\le\frac{10}{3}\)
\(\Leftrightarrow\frac{4}{6}< \frac{x}{6}\le\frac{20}{6}\)
mà \(x\inℤ\Rightarrow\text{x}\in\) {\(5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20\)}
b, \(\frac{1}{3}+x=1\frac{1}{2}\)
\(\frac{1}{3}+x=\frac{3}{2}\)
\(x=\frac{3}{2}+\frac{\left(-1\right)}{3}\)
\(x=\frac{7}{6}\) (loại vì \(x\notinℤ\))
\(\Rightarrow x\in\varnothing\)
c, \(\frac{1}{7}+x=\frac{25}{14}+\frac{5}{14}\)
\(\frac{1}{7}+x=\frac{15}{7}\)
\(x=\frac{15}{7}+\frac{(-1)}{7}\)
\(x=\frac{14}{7}=2\).
=09999389648
Really?