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a,
A=1−3−5−7−9−...−97−99a)A=1−3−5−7−9−...−97−99
=1−(3+5+7+...+99)=1−(3+5+7+...+99)
=1−(99+3).[(99−3):2+1]2=1−(99+3).[(99−3):2+1]2
=1−2499=−2498=1−2499=−2498
b)B=1+3−5−7+9+...+97−99b)B=1+3−5−7+9+...+97−99
=(−8)+(−8)+(−8)+...+(−8)+97−99=(−8)+(−8)+(−8)+...+(−8)+97−99
=(−8).12+(−2)=−98=(−8).12+(−2)=−98
c)C=1−3−5+7+9−11−13+15+...+97−99c)C=1−3−5+7+9−11−13+15+...+97−99
=0+0+0+0+0+...+0−99=0+0+0+0+0+...+0−99
=−99
a) 5/9 + 4/9 . 3/7 + 4/9 . 4/7
= 5/9 + 4/9 . (3/7 + 4/7)
= 5/9 + 4/9 . 1
= 5/9 + 4/9
= 1
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
a) \(\left(\left(\frac{-12}{16}\right)+\frac{7}{14}\right)-\left(\frac{1}{13}-\frac{3}{13}\right)\) \(=\left(\left(\frac{-3}{4}\right)+\frac{1}{2}\right)-\left(\frac{-2}{13}\right)\) \(=\left(\frac{-2}{8}\right)-\left(\frac{-2}{13}\right)\) \(=\left(\frac{-10}{104}\right)\) \(=\left(\frac{-5}{72}\right)\) | b) \(\frac{10}{11}+\frac{4}{11}:4-\frac{1}{8}\) \(=\frac{10}{11}+\frac{4}{11}:\frac{4}{1}-\frac{1}{8}\) \(=\frac{10}{11}+\frac{4}{11}\cdot\frac{1}{4}-\frac{1}{8}\) \(=\frac{10}{11}+\frac{1}{11}-\frac{1}{8}\) \(=\frac{11}{11}-\frac{1}{8}\) \(=1-\frac{1}{8}\) \(=\frac{7}{8}\) |
HT
\(a.\frac{8}{7}+\frac{4}{7}\times\left(-\frac{6}{11}\right)-\frac{4}{7}\times\frac{5}{11}\)
\(=\frac{8}{7}+\frac{4}{7}\left(-\frac{6}{11}-\frac{5}{11}\right)\)
\(=\frac{8}{7}+\frac{4}{7}.\left(-1\right)\)
\(=\frac{8}{7}-\frac{4}{7}\)
\(=\frac{4}{7}\)
bài 1 :
a)\(2,5.16,27.4+7,3=\left(2,5.4\right).16,27+7,3=10.16,27+7,3\)
\(=162,7+7,3=170\)
b) \(\frac{2}{3}+\frac{3}{4}-\frac{5}{6}=\left(\frac{2}{3}-\frac{5}{6}\right)+\frac{3}{4}=\frac{-1}{6}+\frac{3}{4}=\frac{7}{12}\)
c) \(17,6-5,3+16,8-7,6+15,3-6,8\)
\(=\left(17,6-7,6\right)+\left(-5,3+15,3\right)+\left(16,8-6,8\right)\)
\(=10+10+10=30\)
d)\(\frac{38}{11}+\left(13,16+\frac{6}{11}\right)=\frac{38}{11}+13,16+\frac{6}{11}\)
\(=\left(\frac{38}{11}+\frac{6}{11}\right)+13,16=4+13,16=17,16\)
\(2,45.46+8.0,75+54.2,45+0,5.8\\ =\left[2,45.46+54.2,45\right]\)
\(+\left[8.0,75+0,5.8\right]=\left[2,45.\left(46+54\right)\right]+\left[8.\left(0,75+0,5\right)\right]\)
\(=\left(2,45.100\right)+\left(8.1,25\right)=245+10=255\)
bài 2 :
a) .....
b) \(1-\left(12,5+x-4,25\right):21,75=0\)
\(\Rightarrow1-\left(12,5+x-4,25\right)=21,75\)
\(\Rightarrow12,5+x-4,25=-20,75\\ \Rightarrow12,5+x=-16,5\)
\(\Rightarrow x=-29\)
cậu có thể tham khảo bài làm trên đây ạ, mn ai thấy đúng thì cho mk xin 1 t.i.c.k đúng ạ ,thank nhiều
Dễ 1+1=2