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\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)
\(A=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2x+4}{x-3}\)
\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x^2-9}{\left(x-2\right)\left(x-3\right)}+\frac{2x^2-8}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\frac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x+4}{x-3}\)
b) Ta có : \(A=\frac{x+4}{x-3}=\frac{x-3+7}{x-3}=1+\frac{7}{x-3}\)
Để A đạt giá trị nguyên thì \(\frac{7}{x-3}\)đạt giá trị nguyên
=> 7 ⋮ x - 3
=> x - 3 ∈ Ư(7) = { ±1 ; ±7 }
| x-3 | 1 | -1 | 7 | -7 |
| x | 4 | 2 | 10 | -4 |
So với ĐKXĐ ta thấy x = 4 , x = 10 , x = -4 thỏa mãn
Vậy với x ∈ { ±4 ; 10 } thì A đạt giá trị nguyên
(....) dùng để nhìn được chữ số ở phân số cuối cùng thôi, ko dùng để làm gì.
( ác ) là từ ( các )
(gia strij) là từ ( giá trị )
a: \(M=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
ĐKXĐ: \(x\ne\pm1;-2\)
\(P=\left(\frac{x+1}{x-1}+\frac{2}{x^2-1}-\frac{x}{x+1}\right).\frac{x-1}{x+2}\)
\(=\left(\frac{\left(x+1\right)^2}{\left(x-1\right).\left(x+1\right)}+\frac{2}{\left(x-1\right).\left(x+1\right)}-\frac{x\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}\right).\frac{x-1}{x+2}\)
\(=\left(\frac{x^2+2x+1}{\left(x-1\right).\left(x+1\right)}+\frac{2}{\left(x-1\right).\left(x+1\right)}-\frac{x^2-x}{\left(x-1\right).\left(x+1\right)}\right).\frac{x-1}{x+2}\)
\(=\left(\frac{x^2+2x+1+2-x^2+x}{\left(x-1\right).\left(x+1\right)}\right).\frac{x-1}{x+2}\)
\(=\frac{3x+3}{\left(x-1\right).\left(x+1\right)}.\frac{x-1}{x+2}=\frac{3.\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}.\frac{x-1}{x+2}=\frac{3}{x+2}\)
c. \(x^2-3x=0\Leftrightarrow x.\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Nếu x=0 thì: \(P=\frac{3}{x+2}=\frac{3}{0+2}=\frac{3}{2}\)
Nếu x=3 thì: \(P=\frac{3}{x+2}=\frac{3}{3+2}=\frac{3}{5}\)
d. Ta có: \(P=\frac{3}{x+2}\inℤ\)
Vì \(x\inℤ\Rightarrow x+2\inℤ\Rightarrow x+2\inƯ\left\{3\right\}\Rightarrow x+2\in\left\{\pm1;\pm3\right\}\Leftrightarrow x\in\left\{-3;-1;1;-5\right\}\)
Kết hợp ĐKXĐ \(\Rightarrow x\in\left\{-3;-5\right\}\)
a, ĐK : \(x\ne\pm3;\frac{1}{2}\)
\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)
\(=\left(\frac{\left(x-1\right)\left(x-3\right)+2\left(x+3\right)-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{2x-1-2x-1}{2x+1}\right)\)
\(=\frac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}:\left(-\frac{2}{2x+1}\right)\)
\(=\frac{-2x+6}{\left(x+3\right)\left(x-3\right)}.\frac{-\left(2x+1\right)}{2}=\frac{2x+1}{x+3}\)
b, Ta có : \(\left|x+1\right|=\frac{1}{2}\)
TH1 : \(x+1=\frac{1}{2}\Leftrightarrow x=-\frac{1}{2}\)
Thay vào biểu thức A ta được : \(\frac{-1+1}{-\frac{1}{2}+3}=0\)
TH2 : \(x+1=-\frac{1}{2}\Leftrightarrow x=-\frac{3}{2}\)
Thay vào biểu thức A ta được : \(\frac{-3+1}{-\frac{3}{2}+3}=\frac{-2}{\frac{3}{2}}=-\frac{4}{3}\)
c, Ta có : \(P=\frac{x}{2}\Rightarrow\frac{2x+1}{x+3}=\frac{x}{2}\Rightarrow4x+2=x^2+3x\)
\(\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)
b, Ta có : \(\frac{2x+1}{x+3}=\frac{2\left(x+3\right)-5}{x+3}=2-\frac{5}{x+3}\)
\(\Rightarrow x+3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
| x + 3 | 1 | -1 | 5 | -5 |
| x | -2 | -4 | 2 | -8 |
a.\(ĐKXĐ:\hept{\begin{cases}x^2-2x\ne0\\x-2\ne0\\x\left(x+1\right)\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\left(x-2\right)\ne0\\x-2\ne0\\x\left(x+1\right)\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-1\end{cases}}}\)
b.\(M=\left(\frac{1}{x^2-2x}+\frac{2}{x-2}\right)\div\frac{2x+1}{x\left(x+1\right)}\)
\(=\left(\frac{1}{x\left(x-2\right)}+\frac{2}{x-2}\right)\div\frac{2x+1}{x\left(x+1\right)}\)
\(=\left(\frac{1}{x\left(x-2\right)}+\frac{2x}{x\left(x-2\right)}\right)\div\frac{2x+1}{x\left(x+1\right)}\)
\(=\frac{2x+1}{x\left(x-2\right)}\div\frac{2x+1}{x\left(x+1\right)}\)
\(=\frac{2x+1}{x\left(x-2\right)}.\frac{x\left(x+1\right)}{2x+1}=\frac{x\left(2x+1\right)\left(x+1\right)}{x\left(x-2\right)\left(2x+1\right)}=\frac{x+1}{x-2}\)
c.Để \(M>1\)thì
\(\frac{x+1}{x-2}>1\)
c, Ta có : \(M>1\Rightarrow\frac{x+1}{x-2}>1\Leftrightarrow\frac{x+1}{x-2}-1>0\)
\(\Leftrightarrow\frac{x+1-x+2}{x-2}>0\Leftrightarrow\frac{3}{x-2}>0\)
\(\Rightarrow x-2>0\Leftrightarrow x>2\)vì 3 > 0
d, Để M nguyên khi \(x+1⋮x-2\Leftrightarrow x-2+3⋮x-2\)ĐK : \(x\ne2\)
\(\Leftrightarrow3⋮x-2\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
| x - 2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
a: ĐKXĐ: x<>2; x<>3
\(Q=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}=\dfrac{x+1}{x-3}\)
b: Để P<1 thì P-1<0
=>\(\dfrac{x+1-x+3}{x-3}< 0\)
=>x-3<0
=>x<3