bài 1: Tính nhanh:
A= \(\frac{1}{3}\)- \(\frac{3}{4}\)- (\(\frac{-3}{5}\))+ \(\frac{1}{72}\)- \(\frac{2}{9}\)- \(\frac{1}{36}\)+ \(\frac{1}{15}\)
B= (3+ \(\frac{1}{2}\)- \(\frac{2}{3}\)) - (2- \(\frac{2}{3}\)+ \(\frac{5}{2}\))- (5- \(\frac{5}{2}\)+ \(\frac{4}{3}\))
bài 2: tìm các số nguyên x,y; biết:
\(\frac{x}{6}\)- \(\frac{1}{y}\)= \(\frac{1}{2}\)
bài 3:tính :
S1= \(\frac{3}{1.4}\)+ \(\frac{6}{4.10}\)+ \(\frac{9}{10.19}\)+ \(\frac{12}{19.31}\)+ \(\frac{15}{31.46}\)
S2= \(\frac{3}{1.5}\)+ \(\frac{3}{5.9}\)+ \(\frac{3}{9.13}\)+....+ \(\frac{3}{89.93}\)
MỌI NGƯỜI GIẢI NHANH GIÚP MÌNH VỚI. MÌNH ĐANG CẦN GẤP TRONG TRƯA NAY
Bài 1 :
\(A=\frac{1}{3}-\frac{3}{4}-\frac{\left(-3\right)}{5}+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow A=\frac{3}{9}-\frac{3}{4}+\frac{9}{15}+\frac{1}{72}-\frac{2}{9}-\frac{2}{72}+\frac{1}{15}\)
\(\Rightarrow A=\left(\frac{3}{9}-\frac{2}{9}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)+\left(\frac{1}{72}+\frac{-2}{72}\right)-\frac{3}{4}\)
\(\Rightarrow A=\frac{1}{9}+\frac{2}{3}+\frac{-1}{72}-\frac{3}{4}=\frac{8}{72}+\frac{48}{72}+\frac{-1}{72}-\frac{54}{72}\)
\(\Rightarrow A=\frac{1}{72}\)
Vậy : \(A=\frac{1}{72}\)
Bài 2:
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\(\frac{x}{6}-\frac{1}{y}=\frac{1}{2}\Leftrightarrow\frac{x}{6}-\frac{1}{2}=\frac{1}{y}\Leftrightarrow\frac{x-3}{6}=\frac{1}{y}\Leftrightarrow\left(x-3\right)y=6\) ta co bang
sau
x | y | x-3 |
-3 | -1 | -6 |
0 | -2 | -3 |
2 | -6 | -1 |
1 | -3 | -2 |
4 | 6 | 1 |
5 | 3 | 2 |
6 | 2 | 3 |
9 | 1 | 6 |
3:
\(\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\)
\(S_1=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+....-\frac{1}{46}=1-\frac{1}{46}=\frac{45}{46}\)
\(S_2=\frac{3}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+.....+\frac{4}{89.93}\right)=\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-.....-\frac{1}{93}\right)=\frac{3}{4}.\frac{92}{93}=\frac{23}{31}\)
Bài 1
b) \(B=\left(3+\frac{1}{2}-\frac{2}{3}\right)-\left(2-\frac{2}{3}+\frac{5}{2}\right)-\left(5-\frac{5}{2}+\frac{4}{3}\right)\)
\(\Rightarrow B=3+\frac{1}{2}-\frac{2}{3}-2+\frac{2}{3}-\frac{5}{2}-5+\frac{5}{2}-\frac{4}{3}\)
\(\Rightarrow B=\left(3-2-5\right)+\left(\frac{2}{3}-\frac{2}{3}\right)+\left(\frac{5}{2}-\frac{5}{2}\right)+\left(\frac{1}{2}-\frac{4}{3}\right)\)
\(\Rightarrow B=-4+\frac{-5}{6}=-\frac{29}{6}\)
Vậy : \(B=-\frac{29}{6}\)