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\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)
\(M=1-\frac{1}{7}=\frac{6}{7}\)
Mình làm câu 1 thoi nha!
1.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
câu 1
A=-1
câu 2
\(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=8.2\)
\(\left(x+1\right).\left(x+1\right)=16\)
\(\left(x+1\right)^2=16\)
\(\Rightarrow x+1=4\)
vậy x=3
Bài 1:
1) \(\frac{11}{3}\): 3\(\frac{1}{3}\)- 3
= \(\frac{11}{3}\): \(\frac{10}{3}\)- 3
= \(\frac{11}{3}\). \(\frac{3}{10}\)- 3
= \(\frac{11}{10}\)- 3
= \(\frac{-19}{10}\)
2) \(\frac{5}{6}\): \(\frac{3}{52}\) - \(\frac{5}{6}\). 47\(\frac{1}{3}\)
= \(\frac{5}{6}\) . \(\frac{52}{3}\)- \(\frac{5}{6}\). 47\(\frac{1}{3}\)
= \(\frac{5}{6}\).(\(\frac{52}{3}\)- 47\(\frac{1}{3}\))
= \(\frac{5}{6}\).( -30)
= -25
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{3.7}+\frac{1}{4.7}+\frac{1}{4.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{2.3.7}+\frac{2}{2.4.7}+\frac{2}{2.4.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6}-\frac{2}{7}+\frac{2}{7}-\frac{2}{8}+....+\frac{2}{x}-\frac{2}{x+1}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6}-\frac{2}{x+1}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{6}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{1}{3}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{3}{9}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=17\)
câu a khó quá.Để nghĩ.
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{21\cdot2}+\frac{2}{28\cdot2}+\frac{2}{36\cdot2}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{x\left(x-1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{x-5}{6x+6}=\frac{1}{9}\)
\(\Rightarrow9\left(x-5\right)=6x+6\)
\(\Rightarrow9x-45=6x+6\)
\(\Rightarrow9x-6x=51\)
\(\Rightarrow3x=51\)
Tới đây bí:v
Câu 1 :
A = (2012+2) . [ ( 2012-2) : 3+1 ] : 2 = 2014 . 671 : 2 = 675697
B = \(\frac{1}{2}\). \(\frac{2}{3}\). \(\frac{3}{4}\)+...+ \(\frac{2010}{2011}\). \(\frac{2011}{2012}\)= \(\frac{1.2.3.....2010.2011}{2.3.4.....2011.2012}\)= \(\frac{1}{2012}\)
Câu 2 :
a) \(2x.\left(3y-2\right)+\left(3y-2\right)=-55\)
=> \(\left(3y-2\right).\left(2x+1\right)=-55\)
=> \(3y-2;2x+1\in\: UC\left(-55\right)\)
=> \(3y-2;2x+1=\left\{1;-1;5;-5;11;-11;55;-55\right\}\)
- Vậy ta có bảng
| \(2x+1\) | 1 | -1 | 5 | -5 | 11 | -11 | 55 | -55 |
| \(x\) | 0 | -1 | 2 | -3 | 5 | -6 | 27 | -28 |
| \(3y-2\) | -55 | 55 | -11 | 11 | -5 | 5 | -1 | 1 |
| \(3y\) | -53 | 57 | -9 | 13 | -3 | 7 | 1 | 3 |
| \(y\) | \(\frac{-53}{3}\)(loại) | 19(chọn) | -3(chọn) | \(\frac{13}{3}\)(loại) | -1(chọn) | \(\frac{7}{3}\)(loại) | \(\frac{1}{3}\)(loại) | 1(chọn) |
\(\Leftrightarrow\)Những cặp (x;y) tìm được là :
(-1;19) ; (2;-3) ; (5;-1) ; (-28;1)
b) Ta đặt vế đó là A
Ta xét A : \(\frac{1}{4^2}\)< \(\frac{1}{2.4}\)
\(\frac{1}{6^2}\)< \(\frac{1}{4.6}\)
\(\frac{1}{8^2}\)< \(\frac{1}{6.8}\)
...
\(\frac{1}{\left(2n\right)^2}\)< \(\frac{1}{\left(2n-2\right).2n}\)
\(\Leftrightarrow\)A < \(\frac{1}{2.4}\)+ \(\frac{1}{4.6}\)+...+ \(\frac{1}{\left(2n-2\right).2n}\)
\(\Leftrightarrow\)A < \(\frac{1}{2}\). ( \(\frac{2}{2.4}\)+ \(\frac{2}{4.6}\)+...+ \(\frac{2}{\left(2n-2\right).2n}\))
\(\Leftrightarrow\)A < \(\frac{1}{2}\). ( \(\frac{1}{2}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{6}\)+...+ \(\frac{1}{2n-2}\)- \(\frac{1}{2n}\))
\(\Leftrightarrow\)A < \(\frac{1}{2}\). ( \(\frac{1}{2}\)- \(\frac{1}{2n}\)) = \(\frac{1}{2}\). \(\frac{1}{2}\)- \(\frac{1}{2}\). \(\frac{1}{2n}\)
\(\Leftrightarrow\)A < \(\frac{1}{4}\)- \(\frac{1}{4n}\)< \(\frac{1}{4}\) ( Vì n \(\in\)N )
\(\Leftrightarrow\)A < \(\frac{1}{4}\)( đpcm ) .
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
Câu 2)
1)* Nếu : \(x^2-2\ge0;2-x^2\ge0=>x^2-2+2-x^2\)=28
=> \(x^2-x^2-2+2=28=>0x^2=28\) ( vô lý )
Vậy x không có giá trị
* Nếu : \(x^2-2< 0:2-x^2< 0\)
=> \(-\left(x^2-2\right)-\left(2-x^2\right)=28=>-x^2+2-2+x^2=28=>0x^2=28\left(l\right)\)
Vậy từ hai trường hợp trên x không có giá trị
2) 7762≡1(mod3)⇒776776≡1(mod3)7762≡1(mod3)⇒776776≡1(mod3)
777777≡0(mod3)777777≡0(mod3)
7782≡1(mod3)⇒778778≡1(mod3)7782≡1(mod3)⇒778778≡1(mod3)
⇒A≡2(mod3)⇒A≡2(mod3)