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a) \(P=\dfrac{2x+5}{x+3}\inℤ\left(x\inℤ;x\ne-3\right)\)
\(\Rightarrow2x+5⋮x+3\)
\(\Rightarrow2x+5-2\left(x+3\right)⋮x+3\)
\(\Rightarrow2x+5-2x-6⋮x+3\)
\(\Rightarrow-1⋮x+3\)
\(\Rightarrow x+3\in\left\{-1;1\right\}\)
\(\Rightarrow x\in\left\{-4;-2\right\}\)
b) \(P=\dfrac{3x+4}{x+1}\inℤ\left(x\inℤ;x\ne-1\right)\)
\(\Rightarrow3x+4⋮x+1\)
\(\Rightarrow3x+4-3\left(x+1\right)⋮x+1\)
\(\Rightarrow3x+4-3x-3⋮x+1\)
\(\Rightarrow1⋮x+1\)
\(\Rightarrow x+1\in\left\{-1;1\right\}\)
\(\Rightarrow x\in\left\{-2;0\right\}\)
c) \(P=\dfrac{4x-1}{2x+3}\inℤ\left(x\inℤ;x\ne-\dfrac{3}{2}\right)\)
\(\Rightarrow4x-1⋮2x+3\)
\(\Rightarrow4x-1-2\left(2x+3\right)⋮2x+3\)
\(\Rightarrow4x-1-4x-6⋮2x+3\)
\(\Rightarrow-7⋮2x+3\)
\(\Rightarrow2x+3\in\left\{-1;1;-7;7\right\}\)
\(\Rightarrow x\in\left\{-2;-1;-5;2\right\}\)
a) P=\(\dfrac{2x+5}{x+3}=\dfrac{2\left(x+3\right)-2}{x+3}=\dfrac{2\left(x+3\right)}{x+3}-\dfrac{2}{x+3}=2-\dfrac{2}{x+3}\)
để \(P\inℤ\) thì \(\dfrac{2}{x+3}\inℤ\) hay 2 ⋮ (x-3) ⇒x+3 ϵ Ư2= (2,-2,1,-1)
ta có bảng sau:
| x+3 | 2 | -2 | 1 | -1 |
| x | -1 | -5 | -2 | -4 |
Vậy x \(\in-1,-2,-5,-4\)
a) -4/5 + 5/2x = -3/10
5/2x = -3/10 + 4/5
5/2x = 1/5
5/2x = 1/2
x = 1/2 : 5/2
x = 1/5
b) 4/3 + 5/8 : x = 1/12
5/8x = 1/12 - 4/3
5/8x = -5/4
5 = -5/4.8x
5 = -10x
5/-10 = x
-1/2 = x
x = -1/2
c) (x - 1/3)(x - 2/5) = 0
x - 1/3 = 0 hoặc x - 2/5 = 0
x = 0 + 1/3 x = 0 + 2/5
x = 1/3 x = 2/5
Tìm số nguyên x để A;B;C là số nguyên
ta có: \(A=\frac{3x-1}{x+2}=\frac{3x+6-7}{x+2}=\frac{3.\left(x+2\right)-7}{x+2}=3-\frac{7}{x+2}\)
Để A nguyên
=>7/(x+2) nguyên
=> 7 chia het cho x +2
=> x + 2 thuộc U(7)={1;-1;7;-7}
...
bn tu lm tiep nha
a, Ta có :
\(\frac{2x-1}{x-2}\text{ nguyên khi }\left(2x-1\right)\text{ }⋮\text{ }x-2\)
\(\text{ }\frac{2x-1}{x-2}=\frac{2\left(x-2\right)+4-1}{x-2}=\frac{2\left(x-2\right)+3}{x-2}=\frac{2\left(x-2\right)}{x-2}+\frac{3}{x-2}\)
\(2x-1\text{ }⋮\text{ }x-2\text{ }\Rightarrow\text{ }3\text{ }⋮\text{ }x-2\text{ }\)
\(\Leftrightarrow\text{ }x-2\inƯ\left(3\right)\)
Ta có bảng :
\(\Rightarrow\text{ }x\in\text{ }\left\{1\text{ ; }3\text{ ; }-1\text{ ; }5\right\}\)
câu a biết làm rồi bn còn mỗi câu b,c thôi