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1/
a) ta có \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33x}{2009}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33}{2009}.x\Rightarrow x=\dfrac{33}{100}:\dfrac{0,33}{2009}=2009\)
b,1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x+1)=1 1991/1993
2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 3984/1993
2.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x+1) = 3984/1993
2.(1 − 1/2 + 1/2 − 1/3 + ... + 1/x − 1/x+1)=3984/1993
2.(1 − 1/x+1) = 3984/1993
1 − 1/x + 1= 3984/1993 :2
1 − 1/x+1 = 1992/1993
1/x+1 = 1 − 1992/1993
1/x+1=1/1993
<=>x+1 = 1993
<=>x+1=1993
<=> x+1=1993
<=> x = 1993-1
<=> x = 1992
Lời giải:
$6A=\frac{6^{2021}+6}{6^{2021}+1}=1+\frac{5}{6^{2021}+1}>1+\frac{5}{6^{2022}+1}$
$=\frac{6^{2022}+6}{6^{2022}+1}=6.\frac{6^{2021}+1}{6^{2022}+1}=6B$
$\Rightarrow A>B$
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...........+\dfrac{1}{x\left(x+3\right)}=\dfrac{6}{19}\)
\(\Rightarrow\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...........+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{6}{19}\)
\(\Rightarrow\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+............+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{6}{19}\)
\(\Rightarrow\dfrac{1}{3}\left(1-\dfrac{1}{x+3}\right)=\dfrac{6}{19}\)
\(\Rightarrow1-\dfrac{1}{x+3}=\dfrac{6}{19}:\dfrac{1}{3}\)
\(\Rightarrow1-\dfrac{1}{x+3}=\dfrac{18}{19}\)
\(\Rightarrow\dfrac{1}{x+3}=1-\dfrac{18}{19}\)
\(\Rightarrow\dfrac{1}{x+3}=\dfrac{1}{19}\)
\(\Rightarrow x+3=19\)
\(\Rightarrow x=19-3\)
\(\Rightarrow x=16\)
Vậy \(x=16\) laf giá trị cần tìm
5\(\dfrac{8}{17}\):x + (-\(\dfrac{1}{17}\)) : x + 3\(\dfrac{1}{17}\) : 17\(\dfrac{1}{3}\)= \(\dfrac{4}{17}\)
\(\dfrac{93}{17}\).\(\dfrac{1}{x}\) + (-\(\dfrac{1}{17}\)) .\(\dfrac{1}{x}\) +\(\dfrac{3}{17}\)= \(\dfrac{4}{17}\)
\(\dfrac{1}{x}\).\(\dfrac{92}{17}\)=\(\dfrac{1}{17}\)
\(\dfrac{1}{1.4}\)+\(\dfrac{1}{4.7}\)+\(\dfrac{1}{7.10}\)+...+\(\dfrac{1}{x.\left(x+3\right)}\)=\(\dfrac{6}{19}\)
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{x(x+3)}=\dfrac{6}{19}\)
\(\Rightarrow\)\(\dfrac{1}{3}.(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{x(x+3)})=\dfrac{6}{19}\)
\(\Rightarrow\)\(\dfrac{1}{3}.(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3})=\dfrac{6}{19}\)
\(\Rightarrow\)\(\dfrac{1}{3}.(\dfrac{1}{1}-\dfrac{1}{x+3})=\dfrac{6}{19}\)
\(\Rightarrow\) \(\dfrac{1}{1}-\dfrac{1}{x+3}=\dfrac{6}{19}:\dfrac{1}{3}\)
\(\Rightarrow\) \(\dfrac{1}{1}-\dfrac{1}{x+3}=\dfrac{18}{19}\)
\(\Rightarrow\) \(\dfrac{1}{x+3}=\dfrac{1}{1}-\dfrac{18}{19}\)
\(\Rightarrow\) \(\dfrac{1}{x+3}=\dfrac{1}{19}\)
\(\Rightarrow\) \(x+3=19\)
\(x=19-3\)
\(x=16\)
Vậy \(x=16\)
Ta chi can tach ra , xong ta luoc bot,neu co so bi thua ra thi ta tinh tong.....
\(\dfrac{-11}{-32}>\dfrac{16}{49}\)
\(\dfrac{-2020}{-2021}>\dfrac{-2021}{2022}\)
1.
E = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.10}\) + \(\dfrac{3}{10.13}\) + \(\dfrac{3}{13.16}\) + \(\dfrac{3}{16.19}\) + \(\dfrac{3}{19.22}\)
E = 1 - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ... +\(\dfrac{1}{19}\) - \(\dfrac{1}{22}\)
E = 1 - \(\dfrac{1}{22}\)
E = \(\dfrac{21}{22}\)
2.
(x - 4)(x - 5) = 0
TH1:
x - 4 = 0 => x = 4
TH2:
x - 5 = 0 => x = 5
Vậy: x = 4 hoặc x = 5
c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)
\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)
\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)
\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)
\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)
\(\Rightarrow x=-36\)
Vậy ...
a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)
\(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\)
\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)
Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)
Chúc bạn học tốt!!!