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1, để \(\dfrac{2x+1}{x+3}\) là 1 số nguyên
= > 2x + 1 chia hết cho x + 3 ( x thuộc Z và x \(\ne3\) )
= > 2 ( x + 3 ) - 5 chia hết cho x + 3
=> -5 chia hết cho x + 3
hay x + 3 thuộc Ư(-5 ) \(\in\left\{\pm1;\pm5\right\}\)
Đến đây em tự tìm các giá trị của x
2, Tương tự câu 1, x - 1 chia hết cho x + 5 ( x thuộc Z và x khác - 5 )
= > - 6 chia hết cho x + 5
= > \(x+5\in\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
....
3, ( x - 1 ) ( y - 3 ) = 7
x,y thuộc Z = > x - 1 ; y - 3 thuộc Ư(7)
và ( x - 1 )( y - 3 ) = 7
( 1 ) \(\left\{{}\begin{matrix}x-1=1\\y-3=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=10\end{matrix}\right.\)
(2) \(\left\{{}\begin{matrix}x-1=7\\y-3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=4\end{matrix}\right.\)
( 3) \(\left\{{}\begin{matrix}x-1=-1\\y-3=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-4\end{matrix}\right.\)
( 4 ) \(\left\{{}\begin{matrix}x-1=-7\\y-3=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=2\end{matrix}\right.\)
Từ ( 1 ) , ( 2 ) , ( 3 ) , ( 4 ) các cặp giá trị ( x,y ) nguyên cần tìm là ....
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(x+1)+(x+3)+...+(x+99)=0
Tổng các số hạng là: (99+1):2=50 (số hạng)
=> (x+1)+(x+3)+...+(x+99)=0 <=> 50.x+(1+3+5+...+99) = 0
<=> 50.x+=0 <=> 50.x+2500=0 => x=-2500/50=-50
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a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
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Từ đầu bài suy ra:
\(\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(\Leftrightarrow x+y+y+z+z+x=\frac{13}{12}\)
\(\Leftrightarrow2x+2y+2z=\frac{13}{12}\)
\(\Leftrightarrow2\left(x+y+z\right)=\frac{13}{12}\)
\(\Rightarrow x+y+z=\frac{13}{12}:2=\frac{13}{24}\)
\(\Rightarrow x=\frac{13}{24}-\frac{1}{3}=\frac{5}{24}\)
\(y=\frac{13}{24}-\frac{1}{4}=\frac{7}{24}\)
\(z=\frac{13}{24}-\frac{1}{2}=\frac{1}{24}\)
Vậy...
x+y=1/2;y+z=1/3;z+x=1/4
=>2.(x+y+z)=1/2+1/3+1/4=13/12
x+y=1/2=>z=13/12-1/2=7/12
y+z=1/3=>x=13/12-1/3=3/4
z+x=1/4=>y=13/12-1/4=5/6
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Ta có :
\(x+y=\frac{1}{2};y+z=\frac{1}{3};z+x=\frac{1}{6}\)
\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow2x+2y+2z=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}\)
\(\Rightarrow2\left(x+y+z\right)=1\)
\(\Rightarrow x+y+z=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}\left(x+y+z\right)-\left(x+y\right)=\frac{1}{2}-\frac{1}{2}\Rightarrow z=0\\\left(x+y+z\right)-\left(y+z\right)=\frac{1}{2}-\frac{1}{3}\Rightarrow x=\frac{1}{6}\\\left(x+y+z\right)-\left(z+x\right)=\frac{1}{2}-\frac{1}{6}\Rightarrow y=\frac{1}{3}\end{cases}}\)
Vậy \(x=\frac{1}{6},y=\frac{1}{3};z=0\) .
\(x+y=\frac{1}{2};y+z=\frac{1}{3};z+x=\frac{1}{6}\)
Ta có:\(\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
\(\Leftrightarrow2\left(x+y+z\right)=1\)
\(\Leftrightarrow x+y+z=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}\left(x+y+z\right)-\left(x+y\right)=\frac{1}{2}-\frac{1}{2}=0\\\left(x+y+z\right)-\left(y+z\right)=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\\\left(x+y+z\right)-\left(z+x\right)=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\end{cases}}\)
Vậy....
X=-2;0