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a, \(\frac{-3}{5}+\frac{7}{21}+\frac{-4}{5}+\frac{7}{5}\)
\(=\left(\frac{-3}{5}+\frac{-4}{5}+\frac{7}{5}\right)+\frac{7}{21}\)
\(=0+\frac{7}{21}\)
\(=\frac{7}{21}\)
\(=\frac{1}{3}\)
b, \(\frac{8}{9}+\frac{1}{9}.\frac{7}{9}+\frac{1}{9}.\frac{2}{9}\)
\(=\frac{8}{9}+\frac{1}{9}.\left(\frac{7}{9}+\frac{2}{9}\right)\)
\(=\frac{8}{9}+\frac{1}{9}.1\)
\(=\frac{8}{9}+\frac{1}{9}\)
\(=1\)
a) \(\frac{-3}{5}\)+\(\frac{7}{21}\)+\(\frac{-4}{5}\)+\(\frac{7}{5}\)
=(\(\frac{-3}{5}\)+\(\frac{-4}{5}\)+\(\frac{7}{5}\)) +\(\frac{7}{21}\)
= 0+
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
có \(\frac{1}{2\cdot3}< \frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3\cdot4}< \frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(\frac{1}{4\cdot5}< \frac{1}{4^2}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{9\cdot10}< \frac{1}{9^2}< \frac{1}{8\cdot9}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}>A>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}>A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow1-\frac{1}{9}>A>\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow\frac{8}{9}>A>\frac{2}{5}\)
Bạn ơi, sai rồi, mình k nhầm
làm sao mà \(\frac{1}{2^2}< \frac{1}{1.2}\)được
Bài 2:
b: x+25%x=-1,25
=>1,25x=-1,25
hay x=-1
c: x-75%x=1/4
=>1/4x=1/4
hay x=1
Bài 2:
a: =3/2-11/4=6/4-11/4=-5/4
b: =-49/6-17/2=-49/6-51/6=-100/6=-50/3
a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
... . . . .
\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)
b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Suy ra \(\frac{2}{5}< S\) (1)
Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Từ đó suy ra S < 8/9
Từ (1) và (2) suy ra đpcm
Bài 1:
a) \(\frac{25}{4}+\frac{-5}{4}=\frac{25-5}{4}=\frac{20}{4}=5\)
b)\(\frac{-5}{9}+\left(\frac{-2}{7}\right)=\frac{-35}{63}+\left(\frac{-18}{63}\right)=\frac{-53}{63}\)
c) \(\frac{1}{4}+\frac{9}{11}+\frac{7}{4}+\left(\frac{-2}{11}\right)=\left(\frac{1}{4}+\frac{7}{4}\right)+\left(\frac{-2}{11}+\frac{9}{11}\right)=2+\frac{7}{11}=\frac{22+7}{11}=\frac{29}{11}\)
d) \(\frac{-5}{19}.\frac{8}{19}+\left(\frac{-14}{19}\right).\frac{11}{19}=\frac{-40}{361}-\frac{151}{361}=-\frac{191}{361}\)
Bài 2:
a) \(x+\frac{5}{9}=\frac{-8}{9}\) \(\Leftrightarrow x=\frac{-8}{9}-\frac{5}{9}\) \(\Leftrightarrow x=-\frac{13}{9}\)
b) \(\frac{-1}{8}-x=\frac{9}{20}\) \(\Leftrightarrow x=\frac{-1}{8}-\frac{9}{20}\) \(\Leftrightarrow x=\frac{-5}{40}-\frac{18}{40}\) \(\Leftrightarrow x=-\frac{23}{40}\)
c) (x + 5)3 - 12 = 15
\(\Leftrightarrow\)(x + 5)3 = 27
\(\Leftrightarrow\)x + 5 = 3
\(\Leftrightarrow\)x = -2
d) \(\left|x-3\right|-\frac{4}{15}=\frac{26}{15}\) \(\Leftrightarrow\left|x-3\right|=2\) \(\Leftrightarrow\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
a)\(=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)
\(=\left(\frac{-3}{7}+\frac{3}{7}\right)-\left(\frac{15}{26}+\frac{2}{13}\right)\)
\(=0-\frac{19}{26}\)
\(=-\frac{19}{26}\)
c)\(=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.2-\frac{1}{23}\)
\(=\frac{-22}{23}-\frac{1}{23}\)
\(=-1\)
Câu đầu tiên mẫu số \(9\)mũ là gì vậy
Ta có: \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)\(=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{9}=\dfrac{23}{36}< \dfrac{32}{36}=\dfrac{8}{9}\). (1)
Ta lại có: \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2^2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{19}{20}>\dfrac{8}{20}=\dfrac{2}{5}\). (2)
Từ (1) và (2) suy ra đpcm.
\(\frac{-8}{9}+\frac{1}{9}\times\frac{2}{9}+\frac{1}{9}\div\frac{7}{9}.\)
\(\Rightarrow\left(\frac{1}{9}+\frac{1}{9}\times\frac{2}{9}\right)+\frac{-8}{9}\div\frac{7}{9}\)
\(\Rightarrow1+\frac{-8}{7}\)
\(\Rightarrow\frac{-5}{7}\)