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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 ) .
\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)
\(M=1-\frac{1}{7}=\frac{6}{7}\)
Mình làm câu 1 thoi nha!
1.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
a) Ta có:
\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)
\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)
\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)
\(=x+2x+-3+1-21\)
\(=3x-23\)
=> \(3x-23=2020\)
\(3x=2020+23=2043\)
=> \(x=2043:3=681\)
Nhầm
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)
\(=\frac{1}{100}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{100}\right)\)
Đặt : \(A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}\)
\(A=\frac{1.2.3.4.....99}{2.3.4.5.....100}\)
\(A=\frac{1}{100}\)
Vậy : \(A=\frac{1}{100}\)
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdot\cdot\cdot\left(1-\frac{1}{n^2}\right)\)
\(\Rightarrow A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\cdot\cdot\cdot\left(1-\frac{1}{n^2}\right)\)
\(\Rightarrow A=\frac{3}{4}\cdot\frac{8}{9}\cdot\cdot\cdot\frac{n^2-1}{n^2}\)
\(\Rightarrow A=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\cdot\cdot\frac{\left(n-1\right)\left(n+1\right)}{n\cdot n}\)
\(\Rightarrow A=\frac{\left(1\cdot3\right)\cdot\left(2\cdot4\right)\cdot\cdot\cdot\left[\left(n-1\right)\left(n+1\right)\right]}{\left(2\cdot2\right)\cdot\left(3\cdot3\right)\cdot\cdot\cdot\left(n\cdot n\right)}\)
\(\Rightarrow A=\frac{\left[1\cdot2\cdot\cdot\cdot\cdot\cdot\left(n-1\right)\right]\cdot\left[3\cdot4\cdot\cdot\cdot\cdot\cdot\left(n+1\right)\right]}{\left(2\cdot3\cdot\cdot\cdot\cdot\cdot n\right)\cdot\left(2\cdot3\cdot\cdot\cdot\cdot\cdot n\right)}\)
\(\Rightarrow A=\frac{1\cdot\left(n+1\right)}{n\cdot2}\)
\(\Rightarrow A=\frac{n+1}{2n}\)
A=(1-1/2^2)(1-1/3^2).....(1-1/n^2)
A=1(1/2^2-1/3^2-...-1/n^2)
......
xin lỗi bạn nha mình phải tắt máy rồi bạn cố gắng suy nghĩ tiếp nha
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
Câu 1:
Sai bét choét ...
Câu 2:
Đúng ròi