Tìm x: \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2x-1\right)\left(2x+1\right)}=\frac{49}{99}\)
tìm x nguyên biết
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right)\left(2x+1\right)}=\frac{49}{99}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{\left(2x-1\right)\left(2x+1\right)}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{\left(2x-1\right)\left(2x+1\right)}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{\left(2x-1\right)}-\frac{1}{\left(2x+1\right)}\)
\(2A=1-\frac{1}{2x+1}=\frac{2x}{2x+1}\)
\(A=\frac{x}{2x+1}\)
Mà \(A=\frac{49}{99}\) \(\Leftrightarrow\frac{x}{2x+1}=\frac{49}{99}\Leftrightarrow x=49\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right)\left(2x-1\right)}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}+\frac{1}{2x+1}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{2x+1}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2x+1}=\frac{49}{99}\)
\(\Rightarrow99x=49\left(2x+1\right)\)
\(\Rightarrow99x=98x+49\)
\(\Rightarrow x=49\)
Vậy : \(x=49\)
BT1: Tìm x\(\in\)N * :
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\cdot\left(2x+3\right)}\)=\(\frac{49}{99}\)
\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left\{\left(2x+1\right).\left(2x+3\right)\right\}}\right)=\frac{49}{99}\)
\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{49}{99}\)
\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2x+3}\right)=\frac{49}{99}\)
\(\frac{1}{2}.\cdot\left(\frac{2x+3}{2x+3}-\frac{1}{2x+3}\right)=\frac{49}{99}\)
\(\frac{1}{2}.\frac{2x+2}{2x+3}=\frac{49}{99}\)
\(\frac{2x+2}{2x+3}=\frac{49}{99}:\frac{1}{2}\)
\(\frac{2x+2}{2x+3}=\frac{98}{99}\)
=) \(2x+2=98\)và \(2x+3=99\)
TH1 : \(2x+2=98\)
\(2x=98-2\)
\(2x=96\)
\(x=96:2\)
\(x=48\)( THỎa mãn )
TH2 :
\(2x+3=99\)
\(2x=99-3\)
\(2x=96\)
\(x=96:2\)
\(x=48\)( THỎa mãn )
Vậy x = 48
Tìm x,y
a, \(\left|3x-1\right|+\left(y-2\right)^2=0\)
b \(\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{x.\left(x+1\right)}=\frac{19}{100}\)
c. \(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{x.\left(x+2\right)}=\frac{49}{99}\)
d. \(\left(2x+1\right).\left(y-2\right)=6\)
e \(^{x^2-3xy+3y-x=1}\)
b \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{19}{100}\)
=>\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{100}\)
=>\(\frac{1}{5}-\frac{1}{x+1}\)\(=\frac{19}{100}\)
=>\(\frac{1}{x+1}=\frac{1}{5}-\frac{19}{100}\)
=>\(\frac{1}{x+1}=\frac{1}{100}\)
=> x+1 =100
=>x=99
b) \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{19}{100}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{100}\)
\(\Rightarrow x+1=100\)
\(\Rightarrow x=99\)
c) \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{49}{99}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{49}{99}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{49}{99}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{49}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{50}{99}\)
\(\Rightarrow50.\left(x+2\right)=99\)
\(\Rightarrow x+2=\frac{99}{50}\)
\(\Rightarrow x=-\frac{1}{99}\)
d) Ta có : 6 = 1.6 = 2.3 = (-2) . (-3)
Lâp bảng xét 6 trường hợp:
\(2x+1\) | \(1\) | \(6\) | \(2\) | \(3\) | \(-2\) | \(-3\) |
\(y-2\) | \(6\) | \(1\) | \(3\) | \(2\) | \(-3\) | \(-2\) |
\(x\) | \(0\) | \(\frac{5}{2}\) | \(\frac{1}{2}\) | \(1\) | \(-\frac{3}{2}\) | \(-2\) |
\(y\) | \(8\) | \(3\) | \(5\) | \(4\) | \(-1\) | \(0\) |
Vậy các cặp (x,y) \(\inℤ\)thỏa mãn là : (0;4) ; (1; 4) ; (-2 ; 0)
e) \(x^2-3xy+3y-x=1\)
\(\Rightarrow x\left(x-3y\right)+3y-x=1\)
\(\Rightarrow x\left(x-3y\right)-\left(x-3y\right)=1\)
\(\Rightarrow\left(x-3y\right)\left(x-1\right)=1\)
Lại có : 1 = 1.1 = (-1) . (-1)
Lập bảng xét các trường hợp :
\(x-1\) | \(1\) | \(-1\) |
\(x-3y\) | \(1\) | \(-1\) |
\(x\) | \(2\) | \(0\) |
\(y\) | \(\frac{1}{3}\) | \(\frac{1}{3}\) |
Vậy các cặp(x,y) thỏa mãn là : \(\left(2;\frac{1}{3}\right);\left(0;\frac{1}{3}\right)\)
Tìm \(x\in Z\) biết :
\(a.\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)
\(b.\)
\(1-3+3^2-3^3+...+\left(-3^x\right)=\frac{1-9^{1006}}{4}\)
\(a.\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right).\left(2x+1\right)}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}-\frac{1}{2x+1}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{2x+1}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2x+1}=\frac{49}{99}\)
\(\Rightarrow99x=49.\left(2x+1\right)\)
\(\Rightarrow99x=98x+49\)
\(\Rightarrow x=49\)
Vậy : \(x=49\)
\(b.\)
\(1-3+3^2-3^3+...+\left(-3^x\right)=\frac{1-9^{1006}}{4}\)
Đặt \(A=1-3+3^2-3^3+...+\left(-3^x\right)\)
\(\Rightarrow3A=3-3^2+3^3-3^4+...+\left(-3^{x+1}\right)\)
\(\Rightarrow3A+A=1+\left(-3^{x+1}\right)\)
\(\Rightarrow4A=1+\left(-3^{x+1}\right)\)
\(\Rightarrow A=\frac{1+\left(-3^{x+1}\right)}{4}\)
\(\Rightarrow\frac{1+\left(-3^{x+1}\right)}{4}=\frac{1-9^{1006}}{4}\)
\(\Rightarrow-3^{x+1}=-9^{1006}\)
\(\Rightarrow-3^{x+1}=-3^{2012}\)
\(\Rightarrow x+1=2012\)
\(\Rightarrow x=2012-1\)
\(\Rightarrow x=2011\)
Vậy : \(x=2011\)
Tìm x biết :\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{1}{2}\left[\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\right]=\dfrac{49}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Leftrightarrow2x+1=99\Leftrightarrow x=49\)
Tìm x, biết \(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{x\left(x+2\right)}\right)=\frac{4016}{2007}\)
Tìm x
\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+...+\left|x+\frac{1}{97.99}\right|\)
Xinloi, t ghi thiếu đề
\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+...+\left|x+\frac{1}{97.99}\right|=50x\)
\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+...+\left|x+\frac{1}{97.99}\right|=50x\)
Vì \(\left|x+\frac{1}{1.3}\right|\ge0\forall x\)
\(\left|x+\frac{1}{3.5}\right|\ge0\forall x\)
................
\(\left|x+\frac{1}{97.99}\right|\ge0\forall x\)
(VT: Vế trái; VP: Vế phải)
\(\Rightarrow VT\ge0\Rightarrow VP=50x\ge0\)mà \(50>0\)
\(\Rightarrow x>0\)
\(\Rightarrow x+\frac{1}{1.3}>0\forall x\)
..............
\(x+\frac{1}{97.99}>0\forall x\)(1)
(1) \(\Leftrightarrow x+\frac{1}{1.3}+x+\frac{1}{3.5}+...+x+\frac{1}{97.99}=50x\)
\(\Leftrightarrow49x+\left(\frac{1}{1.3}+...+\frac{1}{97.99}\right)=50x\)
\(\Leftrightarrow50x-49x=\frac{1}{2}\left(\frac{2}{1.3}+...+\frac{2}{97.99}\right)\)
\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(\Leftrightarrow x=\frac{1}{2}\cdot\frac{98}{99}=\frac{49}{99}\)
Vậy....
P/s: Làm bừa :) Ko chắc đúng nhé
3. Cho \(A=\frac{3x-1}{x-1}\)và \(B=\frac{2x^2+x-1}{x+2}\)
a) Tìm \(x\inℤ\)để A; B là số nguyên
b) Tìm \(x\inℤ\)để A và B cùng là số nguyên
4. Thực hiện phép tính
\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2017.2019}\right)\)
\(S+\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{6^2}\right)...\left(1-\frac{1}{99^2}\right)\)
là S =... nhé, ko phải S +...
3. a) \(đk:x\ne1;x\ne-2\)
Ta có: \(A=\frac{3x-3+2}{x-1}=\frac{3\left(x-1\right)+2}{x-1}=3+\frac{2}{x-1}\)
Để A là số nguyên thì x là số nguyên và x-1 là ước của 2 . Ta có bảng:
x-1 | 1 | -1 | 2 | -2 |
x | 2 | 0 | 3 | -1 |
Lại có: \(B=\frac{2x^2+4x-3x-6+5}{x+2}=\frac{2x\left(x+2\right)-3\left(x+2\right)+5}{x+2}=2x-3+\frac{5}{x+2}\)
Để B là số nguyên thì x là số nguyên và x+2 là ước của 5. Ta có bảng:
x+2 | 1 | -1 | 5 | -5 |
x | -1 | -3 | 3 | -7 |
b) Để A và B cùng nguyên thì \(x\in\left\{-1;3\right\}\)
a)Tìm số nguyên dương n thỏa mãn:
\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)=\frac{2013}{2014}\)
b)tìm a sao cho
\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)=11.a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)