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\(B=\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}-\frac{9}{16}-\frac{7}{13}+\frac{2}{12}-\frac{5}{9}+\frac{3}{7}-\frac{1}{5}-\frac{1}{5}\)
\(B=\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{3}{7}-\frac{3}{7}\right)+\left(\frac{5}{9}-\frac{5}{9}\right)+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{2}{11}+\frac{2}{12}-\frac{9}{16}-\frac{1}{5}\)
\(B=0-0+0+0-\frac{2}{11}+\frac{2}{12}-\frac{9}{16}-\frac{1}{5}\)
\(B=\frac{-2}{11}+\frac{2}{12}-\frac{9}{16}-\frac{1}{5}\)
Đến đây chỉ còn cách quy đồng thôi
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\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)
Dễ thấy: \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\Rightarrow x+2004=0\Leftrightarrow x=-2014\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=16\frac{2}{7}:\left(\frac{-3}{5}\right)-28\frac{2}{7}:\left(\frac{-3}{5}\right)\)
\(B=\frac{114}{7}\cdot\frac{-5}{3}-\frac{198}{7}\cdot\frac{-5}{3}\)
\(B=\left(\frac{114}{7}-\frac{198}{7}\right)\cdot\frac{-5}{3}\)
\(B=\frac{-84}{7}\cdot\frac{-5}{3}\)
\(B=\frac{420}{21}\)
\(B=20\)
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a) \(\frac{x-3}{5}=\frac{4}{9}\)
=> ( x - 3) . 9 = 20
=> 9x - 27 = 20
=> 9x = 47
=> x = \(\frac{47}{9}\)
b) \(\left(x-\frac{1}{4}\right)^2:3^2=81\)
\(\Rightarrow\left(x-\frac{1}{3}\right)^2:9=81\)
\(\Rightarrow\left(x-\frac{1}{4}\right)^2=729\)
\(\Rightarrow x-\frac{1}{4}=27\) hoặc \(x-\frac{1}{4}=-27\)
+) \(x-\frac{1}{4}=27\Rightarrow x=\frac{109}{4}\)
+) \(x-\frac{1}{4}=-27\Rightarrow x=\frac{-107}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\)\(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\)\(\frac{2}{7}-\frac{1}{\frac{7}{2}}\)
\(=\)\(\frac{2}{7}-\frac{2}{7}\)
\(=\)\(0\)
Chúc bạn học tốt ~
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5/9 : (1/11 - 5/12) + 5/9 : (1/15 - 2/3)
= 5/9 : (-43/132) + 5/9 : (-3/5)
= 5/9 . (-132/43) + 5/9 . (-5/3)
= 5/9 . [-132/43 + (-5/3)]
= 5/9 . 445/129
= 2225/1161.
Ai k mình mình k lại.
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
\(\frac{-2}{3}+x+\frac{5}{8}=\frac{-7}{12}\)
\(x+\frac{-16}{24}+\frac{15}{24}=\frac{-14}{24}\)
Vậy x = \(\frac{-14}{24}-\frac{15}{24}-\frac{-16}{24}=\frac{13}{24}\)
sao mình ngu thế nhỉ ngu quá haha tao là hà dễ thương đẹp trai bụng bự