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b.-3/2007
Cách làm:
=(-3/4).(-4/5).(-5/6).....(-2006/2007)
=-3/2007
a: \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)\)
\(=\dfrac{37}{3}-3-4-\dfrac{3}{2}\)
\(=\dfrac{74-9}{6}-7=\dfrac{65}{6}-7=\dfrac{65-42}{7}=\dfrac{23}{7}\)
b: \(3\dfrac{5}{6}+2\dfrac{1}{6}\cdot6\)
\(=3+\dfrac{5}{6}+\dfrac{13}{6}\cdot6\)
\(=16+\dfrac{5}{6}=\dfrac{101}{6}\)
c: \(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)
\(=3+\dfrac{1}{2}+4+\dfrac{5}{7}-5-\dfrac{5}{14}\)
\(=2+\dfrac{7+10-5}{14}=2+\dfrac{12}{14}=2+\dfrac{6}{7}=\dfrac{20}{7}\)
d: \(=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}=\dfrac{9}{2}+\dfrac{1}{11}=\dfrac{99+2}{22}=\dfrac{101}{22}\)
\(G=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)....\left(1-\frac{1}{780}\right)\)
\(=\left(1-\frac{1}{\frac{2.3}{2}}\right)\left(1-\frac{1}{\frac{3.4}{2}}\right)\left(1-\frac{1}{\frac{4.5}{2}}\right).....\left(1-\frac{1}{\frac{39.40}{2}}\right)\)
Ta có : \(1-\frac{1}{\frac{n\left(n+1\right)}{2}}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Áp dụng ta được :
\(G=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.......\frac{38.41}{39.40}\)
\(=\frac{\left(2.3....38\right)\left(4.5.6.....41\right)}{\left(2.3.4....39\right)\left(3.4.5....40\right)}=\frac{41}{39.3}=\frac{41}{117}\)
Bài 1 :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+............+\dfrac{1}{7^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..................
\(\dfrac{1}{7^2}< \dfrac{1}{6.7}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{6.7}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+........+\dfrac{1}{6}-\dfrac{1}{7}\)
\(\Leftrightarrow A< 1-\dfrac{1}{7}< 1\)
\(\Leftrightarrow A< 1\rightarrowđpcm\)
Câu 2 có vấn đề nặng nề :v
\(1)\frac{1}{5}+\frac{2}{11}< \frac{x}{55}< \frac{2}{5}+\frac{1}{55}\)
\(\Rightarrow\frac{11}{55}+\frac{10}{55}< \frac{x}{55}< \frac{22}{55}+\frac{1}{55}\)
\(\Rightarrow\frac{21}{55}< \frac{x}{55}< \frac{23}{55}\)
\(\Rightarrow21< x< 23\)
\(\Rightarrow x=22\)
\(2)\frac{11}{3}+\frac{-19}{6}+\frac{-15}{2}\le x\le\frac{19}{12}+\frac{-5}{4}+\frac{-10}{3}\)
\(\Rightarrow\frac{22}{6}+\frac{-19}{6}+\frac{-45}{6}\le x\le\frac{19}{12}+\frac{-15}{12}+\frac{-40}{12}\)
\(\Rightarrow\frac{22+\left[-19\right]+\left[-45\right]}{6}\le x\le\frac{19+\left[-15\right]+\left[-40\right]}{12}\)
\(=\frac{-42}{6}\le x\le\frac{-36}{12}\)
\(\Rightarrow-7\le x\le-3\)
\(\Rightarrow x\in\left\{-7;-6;-5;-4;-3\right\}\)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}<\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=2.\frac{1}{2}+2.\frac{1}{4}+3.\frac{1}{6}=2\)
\(-\frac{2}{3}x+\frac{1}{5}=\frac{1}{3}\times\left(\frac{-1}{6}\right)\)
\(-\frac{2}{3}x+\frac{1}{5}=-\frac{1}{18}\)
\(-\frac{2}{3}x=-\frac{1}{18}-\frac{1}{5}\)
\(\frac{-2}{3}x=\frac{-5}{90}-\frac{18}{90}=-\frac{23}{90}\)
\(x=-\frac{23}{90}:\frac{-2}{3}\)
\(x=-\frac{23}{90}.\frac{3}{-2}=\frac{23}{60}\)
Vậy \(x=\frac{23}{60}\)
Làm như zậy bạn nhé ^_^"
Đặt :
\(A=\left(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\right)\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{5.6}\)( mik lười viết bạn thông cảm nhé !!! )
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)
\(A=\frac{1}{1}-\frac{1}{6}\)
\(A=\frac{5}{6}\)
Thay A vào biểu thức :
Ta có : \(\frac{5}{6}.10-x=0\)
=> \(\frac{25}{3}=x+0\)
=> \(x=\frac{25}{3}\)
(1/1*1/2+1/2*1/3+1/3*1/4+1/4*1/5+1/5*1/6)*10-x=0
=> (1/1.2+1/2.3+1/3.4+1/4.5+1/5.6)*10-x=0
=> (1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6)*10-x=0
=> (1-1/6)*10-x=0
=> 5/6*10-x=0
=> 25/3-x =0
=> x=0+25/3
=> x=25/3
Vậy x=25/3