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Theo tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{5}\) = \(\dfrac{y}{4}\) = \(\dfrac{z}{3}\) = \(\dfrac{x+2y+3z}{5+8+9}\) = \(\dfrac{x+2y+3z}{22}\)
\(\dfrac{x}{5}\)= \(\dfrac{y}{4}\) = \(\dfrac{z}{3}\) = \(\dfrac{x-2y+3z}{5-8+9}\) = \(\dfrac{x-2y+3z}{6}\)
=> \(\dfrac{x+2y+3z}{22}\) = \(\dfrac{x-2y+3z}{6}\)
=> \(\dfrac{x+2y+3z}{x-2y+3z}\) = \(\dfrac{22}{6}\) =\(\dfrac{11}{3}\)
\(xy-3x-y=6\)
\(=>xy+3x-y-3=6-3\)
\(=>x\left(y+3\right)-\left(y+3\right)=3\)
\(=>\left(y+3\right)\left(x-1\right)=3\)
| y+3 | -1 | 3 | 1 | -3 | |
| x-1 | -3 | 1 | 3 | -1 |
| y+3 | -1 | 3 | -3 | 1 |
| y | -4 | -1 | -7 | -3 |
| x-1 | -3 | 1 | 3 | -1 |
| x | -2 | 2 | 4 | 0 |
3a)Vì A là số nguyên
=>\(3n+9⋮n-4=>3n-12+21⋮n-4=>3.\left(n-4\right)+21⋮n-4\)
Mà \(\text{3 . (n - 4)}⋮n-4\)
=>\(21⋮n-4=>n-4\inƯ\left(21\right)=\left\{-21;-7;-3;-1;1;3;7;21\right\}\)
(Vì n là số nguyên => n - 4 là 1 số nguyên)
=>\(n\in\left\{-17;-3;1;3;5;9;11;25\right\}\)
Ta có bảng sau:
| n | -17 | -3 | 1 | 3 | 5 | 9 | 11 | 25 |
| 3n + 9 | -42 | 0 | 12 | 18 | 24 | 36 | 42 | 84 |
| n - 4 | -21 | -7 | -3 | -1 | 1 | 3 | 7 | 21 |
| \(A=\dfrac{3n+9}{n-4}\) | 2 | 0 | -4 | -18 | 24 | 12 | 6 | 4 |
Vậy.....
b)Vì B là số nguyên
=>\(2n-1⋮n+5=>2n+10-11⋮n+5=>2\left(n+5\right)-11⋮n+5\)
Mà \(\text{2 ( n + 5)}⋮n+5\)
=>\(11⋮n+5=>n+5\in\left\{-11;-1;1;11\right\}\)
(Vì n là số nguyên=> n + 5 là số nguyên)
=> \(n\in\left\{-16;-6;-4;6\right\}\)
Ta có bảng sau:
| n | -16 | -6 | -4 | 6 |
| 2 n - 1 | -33 | -13 | -9 | 11 |
| n + 5 | -11 | -1 | 1 | 11 |
| \(B=\dfrac{2n-1}{n+5}\) | 3 | 13 | -9 |
1 |
Vậy.......
Câu 2:
\(\dfrac{x+2000}{x-2000}=\dfrac{y+2001}{y-2001}\)
\(\Leftrightarrow\left(x+2000\right)\left(y-2001\right)=\left(x-2000\right)\left(y+2001\right)\)
\(\Leftrightarrow xy-2001x+2000y-4002000=xy+2001x-2000y-4002000\)
=>-2001x+2000y=2001x-2000y
=>-4002x=-4000y
=>2001x=2000y
hay x/y=2000/2001
a)ta có 4+x/7+y=4/7
<=>7x+28=28+4y
<=> 7x=4y
lại có x+y=22
=>4/7y+y=22
<=>11/7y=22 <=> y=14
<=> x= 4/7*14=8
vậy x=8, y=14
b) Từ x/3=y/4 va y/5=z/6-->x/15=y/20=z/24 (1)
(1) = 2x/30=3y/60=4z/96=(2x+3y+4z)/186 (2) (t/c dãy tỉ số bằng nhau)
Ta lại có
(1) = 3x/45=4y/80=5z/120=(3x+4y+5z)/245 (3)(t/c dãy tỉ số bằng nhau)
Từ (2)(3) ta có(2x+3y+4z)/186=(3x+4y+5z)/245
Vậy M = (2x+3y+4z)/(3x+4y+5z)=186/245
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Rightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2z-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}\)
\(\Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{x}{2}=\dfrac{z}{4}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\\ \Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\\ =\dfrac{\left(12x-8y\right)+\left(6z-12x\right)+\left(8y-6z\right)}{16+9+4}=\dfrac{0}{29}=0\\ \Rightarrow3x=2y;2z=4x;4y=3z\\ \Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
a) Ta có: \(6x=4y=3z\Rightarrow\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{3z}{12}\Rightarrow\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}=\dfrac{x+2y-3z}{2+6-12}=\dfrac{-2}{-4}=\dfrac{1}{2}.\)
Với: \(\dfrac{x}{2}=\dfrac{1}{2}\Rightarrow x=1.\)
\(\dfrac{2y}{6}=\dfrac{y}{3}=\dfrac{1}{2}\Rightarrow y=\dfrac{1}{2}.3=\dfrac{3}{2}.\)
\(\dfrac{3z}{12}=\dfrac{z}{4}=\dfrac{1}{2}\Rightarrow z=\dfrac{1}{2}.4=\dfrac{4}{2}=2.\)
Vậy: \(x=1;y=\dfrac{3}{2};z=2.\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Leftrightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Theo tính chất của dãy tỉ số bằng nhau, có:
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8x+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}12x=8y\\6z=12x\\8y=6z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{y}{12}\\\dfrac{x}{6}=\dfrac{z}{12}\\\dfrac{y}{6}=\dfrac{z}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{x}{2}=\dfrac{z}{4}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)
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