tìm x \(\inℤ\)biết 2x+9 \(⋮\) 2x-3
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xy - 2x - 3y = 1
=> (xy - 2x) - 3y + 6 = 1 + 6
=> x(y - 2) - 3(y - 2) = 7
=> (x-3)(y-2) = 7
=> x - 3; y - 2 thuộc Ư(7) = {-1; 1; -7; 7}
ta có bảng :
x-3 | -1 | 1 | -7 | 7 |
y-2 | -7 | 7 | -1 | 1 |
x | 2 | 4 | -4 | 10 |
y | -5 | 9 | 1 | 3 |
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\frac{2x}{2x-6}=\frac{2x-6+6}{2x-6}=1+\frac{3}{x-3}\)
Để M nguyên thì \(3⋮x-3\)
\(\Rightarrow x-3\in\left\{1,3,-1,-3\right\}\)
\(\Rightarrow x\in\left\{4,6,2,0\right\}\)
Ta có : 2x \(⋮\)2x - 6
\(\Leftrightarrow\)2x - 6 + 6 \(⋮\)2x - 6
Để M đạt giá trị nguyên
\(\Leftrightarrow\)2x - 6 \(\in\)Ư( 6 ) = { \(\pm\)1 ; \(\pm\)2 ; \(\pm\)3 ; \(\pm\)6 }
Ta lập bảng :
2x - 6 | 1 | - 1 | 2 | - 2 | 3 | - 3 | 6 | - 6 |
x | 7 / 2 | 5 / 2 | 4 | 2 | 9 / 2 | 3 / 2 | 6 | 0 |
Vì x\(\in\)Z nên ta chọn : x \(\in\){ 0 ; 2 ; 4 ; 6 }
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{1-x}{x+4}=\frac{5-4-x}{x+4}=\frac{5}{x+4}-1\inℤ\Leftrightarrow\frac{5}{x+4}\inℤ\)
mà \(x\inℤ\Rightarrow x+4\inƯ\left(5\right)=\left\{-5,-1,1,5\right\}\)
\(\Leftrightarrow x\in\left\{-9,-5,-3,1\right\}\)
b) \(\frac{11-2x}{x-5}=\frac{1+10-2x}{x-5}=\frac{1}{x-5}-2\inℤ\Leftrightarrow\frac{1}{x-5}\inℤ\)
mà \(x\inℤ\Rightarrow x-5\inƯ\left(1\right)=\left\{-1,1\right\}\Leftrightarrow x\in\left\{4,6\right\}\)
c) \(\frac{x+1}{2x+1}\inℤ\Rightarrow\frac{2\left(x+1\right)}{2x+1}=\frac{2x+1+1}{2x+1}=1+\frac{1}{2x+1}\inℤ\Leftrightarrow\frac{1}{2x+1}\inℤ\)
mà \(x\inℤ\Rightarrow2x+1\inƯ\left(1\right)=\left\{-1,1\right\}\Leftrightarrow x\in\left\{-1,0\right\}\).
Thử lại đều thỏa mãn.
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\(\frac{x-2}{4}=\frac{-9}{2-x}\)
\(\Rightarrow\frac{x-2}{4}=\frac{9}{x-2}\)
\(\Rightarrow\left(x-2\right)^2=36\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}}\)
\(\frac{3}{x+2}=\frac{5}{2x+1}\)
\(\Rightarrow3\left(2x+1\right)=\left(x+2\right)5\)
\(\Rightarrow6x+3=5x+10\)
\(\Rightarrow6x-5x=10-3\)
\(\Rightarrow x=7\)
c;giống câu trên :v
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:3(x+1)+24=16-(12-2x)
3x+3+24=16-12+2x
3x+27=4+2x
3x+27-2x=4
3x-2x+27=4
x+27=4
x=-23 Vậy x=-23
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Áp dụng BĐT giá trị tuyệt đối ta có:
\(\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=\left|4\right|=4\) (1)
Mặt khác:\(\left(y-5\right)^2\ge0\Rightarrow2\left(y-5\right)^2\ge0\Rightarrow2\left(y-5\right)^2+2\ge2\)
\(\Rightarrow\frac{8}{2\left(y-5\right)^2+2}\le\frac{8}{2}=4\) (2)
Từ (1) và (2) \(\Rightarrow\left|2x+3\right|+\left|2x-1\right|=\frac{8}{2\left(y-5\right)^2+2}\) khi \(\hept{\begin{cases}y=5\\\left(2x+3\right)\left(1-2x\right)\ge0\end{cases}}\)
Với \(\hept{\begin{cases}2x+3\ge0\\1-2x\ge0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x\ge-\frac{3}{2}\\x\le\frac{1}{2}\end{cases}}\)\(\Rightarrow-\frac{3}{2}\le x\le\frac{1}{2}\)
Với \(\hept{\begin{cases}2x+3\le0\\1-2x\le0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x\le-\frac{3}{2}\\x\ge\frac{1}{2}\end{cases}}\)(loại)
Vậy \(\frac{-3}{2}\le x\le\frac{1}{2};y=5\) thỏa mãn
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{5}{2}x+\frac{1}{2}x=x+400\%\)
\(\Rightarrow\left(\frac{5}{2}+\frac{1}{2}\right)x=x+4\)
\(\Rightarrow\frac{6}{2}x=x+4\)
\(\Rightarrow3x=x+4\)
\(\Rightarrow3x-x=4\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
Chúc bạn học tốt !!!