a) 4/5 + 1/3
b) 2/3 : 4
c) 13/7 - 7/7
d) 4/9 x 2/5
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a: \(=-\dfrac{5}{12}\cdot\dfrac{9}{20}\cdot\dfrac{7}{17}=-\dfrac{21}{272}\)
b: \(=\dfrac{13}{17}\left(-\dfrac{4}{5}-\dfrac{3}{4}\right)=\dfrac{-13}{17}\cdot\dfrac{31}{20}=-\dfrac{403}{340}\)
c: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{5}{7}=-\dfrac{5}{7}+\dfrac{5}{7}=0\)
d: \(=\dfrac{12}{5}-\dfrac{3}{10}=\dfrac{21}{10}\)
a: \(=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}=4-\dfrac{4}{7}=\dfrac{24}{7}\)
b: \(=\dfrac{11}{2}\cdot\dfrac{15}{4}=\dfrac{165}{8}\)
c: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}-6-\dfrac{2}{9}=6+\dfrac{3}{5}=\dfrac{33}{5}\)
d: \(=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}=5+\dfrac{7}{11}=\dfrac{62}{11}\)
Mình làm mẫu 1 bài rùi bạn tự giải những bài còn lại nha
1, 7A = 7+7^2+7^3+....+7^2008
6A = 7A - A = (7+7^2+7^3+....+7^2008)-(1+7+7^2+....+7^2007) = 7^2008-1
=> A = (7^2008-1)/6
Tk mk nha
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)
\(\Rightarrow6A=7^{2008}-1\)
\(\Rightarrow A=\frac{7^{2008}-1}{6}\)
a: =91/105+60/105-101/105
=50/105=10/21
c: \(\dfrac{3}{4}\cdot\dfrac{5}{2}\cdot\dfrac{7}{6}=\dfrac{3}{6}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{1}{2}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{35}{16}\)
d: =2-2/9
=18/9-2/9
=16/9
e: =24/36-9/36+8/36
=23/36
g: =5/2+1/2
=3
Bài 1: Tìm \(x\)
a; \(x-2\) + 7 = 1.3.(-9)
\(x\) - 2 + 7 = 3.(-9)
\(x\) - 2 + 7 = - 27
\(x\) = - 27 - 7 + 2
\(x\) = - 34 + 2
\(x\) = - 32
Vậy \(x=-32\)
Bài 1
c; - 2\(x\) + 5 = 7
- 2\(x\) = 7 - 5
- 2\(x\) = - 2
\(x\) = -2 : (-2)
\(x\) = - 1
Vậy \(x\) = - 1
a: \(\dfrac{7}{8}+3=\dfrac{3\cdot8+7}{8}=\dfrac{31}{8}\)
b: \(2-\dfrac{5}{4}=\dfrac{2\cdot4-5}{4}=\dfrac{3}{4}\)
\(4\cdot\dfrac{5}{9}=\dfrac{20}{9}\)
d: \(\dfrac{3}{8}:2=\dfrac{3}{16}\)
a) 12/5
b) 1/6
c) 6/7
d) 8/45
a) \(\dfrac{12}{5}\)
b) \(\dfrac{1}{6}\)
c) \(\dfrac{6}{7}\)
d) \(\dfrac{8}{45}\)