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\(b,\left(\sqrt{1\frac{9}{16}-\sqrt{\frac{9}{16}}}\right):5\)
\(=\left(\sqrt{\frac{25}{16}-\frac{3}{4}}\right):5\)
\(=\sqrt{\frac{13}{16}}:5\)
\(=\frac{\sqrt{13}}{4}:5\)
\(=\frac{\sqrt{13}}{20}\)
\(a)=\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-\frac{17}{13}\)
\(=1-1-\frac{5}{2}\)
\(=-\frac{5}{2}\)
a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)
Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)
\(a,\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{1}{18}\)
\(b,2^8:2^5+3^3.2-12\)
\(=2^3+9.2-12\)
\(=8+18-12\)
\(=26-12\)
\(=14\)
Câu c,d em chưa học nên không biết làm ạ, mong mọi người thông cảm!!!
Sửa lại câu b
\(=2^3+27.2-12\)
\(=8+54-12\)
\(=62-12\)
\(=50\)
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
Bài 1:
a: \(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\cdots+\frac{1}{19\cdot21}\)
\(=\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{19\cdot21}\right)\)
\(=\frac12\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{19}-\frac{1}{21}\right)=\frac12\left(1-\frac{1}{21}\right)=\frac12\cdot\frac{10}{21}=\frac{5}{21}\)
b: \(\frac12\cdot\sqrt{64}-\sqrt{\frac{4}{25}}+1^{2012}\)
\(=\frac12\cdot8-\frac25+1\)
\(=4+1-\frac25=5-\frac25=\frac{23}{5}\)
Bài 2:
a: \(-\frac59\left(\frac{3}{10}-\frac25\right)\)
\(=-\frac59\left(\frac{3}{10}-\frac{4}{10}\right)=-\frac59\cdot\frac{-1}{10}=\frac{5}{18}\)
b: \(\frac29-\left(\frac12\right)^2+\frac{5}{18}\)
\(=\frac{4}{18}+\frac{5}{18}-\frac14\)
\(=\frac{9}{18}-\frac14=\frac12-\frac14=\frac14\)
c: \(-\frac{5}{17}\cdot\frac{31}{33}+\frac{-5}{17}\cdot\frac{2}{33}+2\frac{5}{17}\)
\(=-\frac{5}{17}\left(\frac{31}{33}+\frac{2}{33}\right)+2+\frac{5}{17}\)
\(=-\frac{5}{17}+2+\frac{5}{17}=2\)
a) \(S = \sum_{k = 0}^{9} \frac{1}{\left(\right. 2 k + 1 \left.\right) \left(\right. 2 k + 3 \left.\right)}\).
Ta dùng đẳng thức \(\frac{1}{a \left(\right. a + 2 \left.\right)} = \frac{1}{2} \left(\right. \frac{1}{a} - \frac{1}{a + 2} \left.\right)\).
Do đó
\(S = \frac{1}{2} \left(\right. 1 - \frac{1}{21} \left.\right) = \frac{1}{2} \cdot \frac{20}{21} = \frac{10}{21} .\)
b) (theo giả thiết) \(\frac{1}{2} \sqrt{64} - \sqrt{\frac{4}{25}} + 1^{2012}\).
Tính: \(\sqrt{64} = 8 , \&\text{nbsp}; \sqrt{\frac{4}{25}} = \frac{2}{5} , \&\text{nbsp}; 1^{2012} = 1.\)
\(\frac{1}{2} \cdot 8 - \frac{2}{5} + 1 = 4 - \frac{2}{5} + 1 = 5 - \frac{2}{5} = \frac{25 - 2}{5} = \frac{23}{5} .\)
Bài 2
a) \(\frac{- 5}{9} \cdot \left(\right. \frac{3}{10} - \frac{2}{5} \left.\right)\).
Tính trong ngoặc: \(\frac{3}{10} - \frac{2}{5} = \frac{3}{10} - \frac{4}{10} = - \frac{1}{10} .\)
\(\frac{- 5}{9} \cdot \left(\right. - \frac{1}{10} \left.\right) = \frac{5}{90} = \frac{1}{18} .\)
b) (theo giả thiết: \(\frac{2}{9} - \left(\right. \frac{1}{2} \left.\right)^{2} + \frac{5}{18}\))
Tính: \(\left(\right. \frac{1}{2} \left.\right)^{2} = \frac{1}{4}\). Quy đồng mẫu 36:
\(\frac{2}{9} - \frac{1}{4} + \frac{5}{18} = \frac{8 - 9 + 10}{36} = \frac{9}{36} = \frac{1}{4} .\)
c) (theo giả thiết: \(\frac{- 5}{17} \cdot \frac{31}{33} + \frac{- 5}{17} \cdot \frac{2}{33} + 2 \frac{5}{17}\))
Gộp hai thừa số đầu: \(\frac{31}{33} + \frac{2}{33} = \frac{33}{33} = 1\).
Do đó hai hạng tử đầu bằng \(\frac{- 5}{17}\).
Hạng tử thứ ba: \(2 \frac{5}{17} = \frac{39}{17}\).
\(\frac{39}{17} - \frac{5}{17} = \frac{34}{17} = 2.\)
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