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a) \(\frac{3}{-4}=\frac{-3}{4};\frac{-1}{-4}=\frac{1}{4}\)
Vì - 3 < 1 nên \(\frac{-3}{4}< \frac{1}{4}\)
hay \(\frac{3}{-4}< \frac{-1}{-4}\)
Quy đồng mẫu ta được:
15/17=15.27/17.27=405/459
25/27=25.17/27.27=425/459
⇒405/459<425/459⇒15/17<25/27
1)
A = \(\frac{3}{8}+\frac{4}{9}+\frac{1}{3}\)
A = \(\frac{27}{72}+\frac{32}{72}+\frac{24}{72}\)
A = \(\frac{83}{72}\)
Vì \(\frac{83}{72}>1\)nên A > 1
B = \(\frac{4}{15}+\frac{4}{13}+\frac{1}{3}\)
B = \(\frac{52}{195}+\frac{60}{195}+\frac{65}{195}\)
B = \(\frac{177}{195}\)
Vì \(\frac{177}{195}< 1\)nên B < 1
a, Ta có : 3/8 > 3/9 = 1/3
4/9 > 3/9 = 1/3
=> A > 1/3 + 1/3 + 1/3 = 1
b, Ta có : 4/15 < 5/15 = 1/3
4/13 < 4/12 = 1/3
=> B < 1/3 + 1/3 + 1/3 = 1
Tk mk nha
Vì 15.27 < 17.25 nên \(\frac{15}{17}< \frac{25}{27}\)
\(a,\frac{5}{7}+\frac{2}{7}:x=\frac{-11}{14}\)
\(\Rightarrow\frac{2}{7}:x=\frac{-11}{14}-\frac{5}{7}\)
\(\Rightarrow\frac{2}{7}:x=\frac{-11}{14}-\frac{10}{14}\)
\(\Rightarrow\frac{2}{7}:x=\frac{-21}{14}\)
\(\Rightarrow\frac{2}{7}:x=\frac{-3}{2}\)
Tự tìm x được rồi :v
\(b,\frac{2}{3}x-\frac{1}{2}x=\frac{5}{12}\)
\(\Rightarrow\left[\frac{2}{3}-\frac{1}{2}\right]x=\frac{5}{12}\)
\(\Rightarrow\left[\frac{4}{6}-\frac{3}{6}\right]x=\frac{5}{12}\)
\(\Rightarrow\frac{1}{6}x=\frac{5}{12}\)
\(\Rightarrow x=\frac{5}{12}:\frac{1}{6}=\frac{5}{12}\cdot\frac{6}{1}=\frac{5}{2}\cdot\frac{1}{1}=\frac{5}{2}\)
\(c,\left[x+\frac{1}{2}\right]^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left[x+\frac{1}{2}\right]^2=\frac{26}{25}-\frac{17}{25}\)
\(\Rightarrow\left[x+\frac{1}{2}\right]^2=\frac{9}{25}\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{3}{5}\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{2}=\frac{3}{5}\\x+\frac{1}{2}=-\frac{3}{5}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{10}\\x=-\frac{11}{10}\end{cases}}\)
d, Tự làm
P/S : Thông cảm cho mk nhé vì máy tính mk ko bấm dấu ngoặc tròn được :v
\(-3\frac{3}{5}.x=\frac{4}{15}\)
\(\frac{-18}{5}.x=\frac{4}{15}\)
\(x=\frac{4}{15}:\frac{-18}{5}\)
\(x=\frac{-2}{27}\)
\(\frac{\frac{4}{17}-\frac{4}{45}+\frac{4}{156}}{\frac{3}{17}-\frac{3}{45}+\frac{3}{156}}=\frac{4.\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}{3.\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}=\frac{4}{3}\)
câu 2:đặt B=1/1*2+1/2*3+...+1/2007*2008
ta có:\(A=3\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2008^2}\right)\)
\(\frac{A}{3}=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2008^2}
câu 2:đặt B=1/1*2+1/2*3+...+1/2007*2008
\(A=3\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2008^2}\right)\)
\(\frac{A}{3}=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2008^2}\)\( (1)
mà \(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2007.2008}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2007}-\frac{1}{2008}\)
\(=1-\frac{1}{2008}\)<1 (2)
mà 1<3 (3)
từ (1),(2) và (3)=> đpcm
Ta có : \(\frac{3}{-4}=\frac{-3}{4}< 0\)
\(\frac{-1}{-4}=\frac{1}{4}>0\)
\(\Rightarrow\frac{3}{-4}< \frac{-1}{-4}\) hay \(\frac{-3}{4}< \frac{1}{4}\)
Vậy : ....
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