tim x thuoc N, biet [x-1] [x+3]=6
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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{x\left(x+1\right):2}=\frac{2001}{2003}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{2003}:2=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}=\frac{1}{2003}\)
=> x+1 = 2003
=> x = 2003 - 1
=> x = 2002
\(\left(x-1\right)\left(2y+3\right)=1\cdot26=26\cdot1=2\cdot13=13\cdot2\)
Ta co bang sau:
x-1 | 1 | 26 | 2 | 13 | ||||
2y+3 | 26 | 1 | 13 | 2 | ||||
x | 2 | 27 | 3 | 14 | ||||
y | 11.5 (loai) | -1(loai) | 5 | -0.5(loai) |
Ta có : \(\left(x-2\right)^{2016}\)dương
\(\Rightarrow x-3=0\Rightarrow x=3\)
Thay x ta thử :
\(\left(3-2\right)^{2016}+\left(3-3\right)=1+0=1\)thỏa đề
Vậy \(x=3\)
a) 6x / 2 - 84 / 2 - 72 =201
3x - 42 -72 = 201
3x - 114 =201
3x = 315
x = 105
b)3x -3^4 =6^5 / 6^3
3x - 3^4 = 36
3x - 3 * 3^3 = 36
3*(x-27) = 36
x - 27 = 36 / 3
x - 27 = 12
x = 39
\(\left(x-1\right)\left(x+3\right)=6\)
\(\Rightarrow6⋮\left(x-1\right),\left(x+3\right)\)
\(\Rightarrow\left(x-1\right),\left(x+3\right)\inƯ\left(6\right)\)
\(\RightarrowƯ\left(6\right)=\left\{1;2;3;6\right\}\)
Ta có bảng :