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a) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)
\(A=\frac{2x-12}{x^2-5x+6}-\frac{x+3}{x-2}+\frac{2x}{x-3}\)
\(\Leftrightarrow A=\frac{2x-12-x^2+9+2x^2-4x}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{x^2-2x-3}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{\left(x-3\right)\left(x+1\right)}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{x+1}{x-2}\)
b) Thay \(x=5\)vào A ta được :
\(A=\frac{5+1}{5-2}=2\)
c) Để \(A\inℤ\)
\(\Leftrightarrow x+1⋮x-2\)
\(\Leftrightarrow3⋮x-2\)
\(\Leftrightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow x\in\left\{1;3;-1;5\right\}\)
Vì \(x\ne3\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{1;-1;5\right\}\)
a, ĐKXĐ: \(x\ne\pm3\)
\(A=\frac{x\left(x-3\right)+2x\left(x+3\right)-3x^2-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(=\frac{3x-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}=\frac{3x-12}{3x+9}\)
b, \(x=-4\Rightarrow A=\frac{3.\left(-4\right)-12}{3.\left(-4\right)+9}=8\)
c, \(A\in Z\Rightarrow3x-12⋮\left(3x+9\right)\Rightarrow3x+9-21⋮\left(3x+9\right)\Rightarrow21⋮\left(3x+9\right)\)
\(\Rightarrow3x+9\inƯ\left(21\right)=\left\{\pm1;\pm3;\pm7;\pm21\right\}\)
Mà \(3x+9⋮3\Rightarrow3x+9\in\left\{-21;-3;3;21\right\}\Rightarrow x\in\left\{-10;-4;-2;4\right\}\) (thỏa mãn điều kiện)
a, ĐỂ A xác định :
\(\Rightarrow\hept{\begin{cases}x+3\ne0\\x-3\ne0\\x^2-9\ne0\end{cases}}\Rightarrow x\ne\pm3.\)
\(A=\left(\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x+3\right)\left(x-3\right)}\right):\frac{3}{x-3}\)
\(A=\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}:\frac{3}{x-3}\)
\(A=\frac{x^2-3x+2x^2+6x-3x^2+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(A=\frac{3x+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(A=\frac{x-4}{x+3}\)
b
a: \(B=\dfrac{3x\left(2x-3\right)-4\left(2x+3\right)-4x^2+23x+12}{\left(2x-3\right)\left(2x+3\right)}\cdot\dfrac{2x+3}{x+3}\)
\(=\dfrac{6x^2-9x-8x-12-4x^2+23x+12}{2x-3}\cdot\dfrac{1}{x+3}\)
\(=\dfrac{2x^2+6x}{\left(2x-3\right)}\cdot\dfrac{1}{x+3}=\dfrac{2x}{2x-3}\)
b: 2x^2+7x+3=0
=>(2x+3)(x+2)=0
=>x=-3/2(loại) hoặc x=-2(nhận)
Khi x=-2 thì \(A=\dfrac{2\cdot\left(-2\right)}{-2-3}=\dfrac{-4}{-7}=\dfrac{4}{7}\)
d: |B|<1
=>B>-1 và B<1
=>B+1>0 và B-1<0
=>\(\left\{{}\begin{matrix}\dfrac{2x+2x-3}{2x-3}>0\\\dfrac{2x-2x+3}{2x-3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3< 0\\\dfrac{4x-3}{2x-3}>0\end{matrix}\right.\Leftrightarrow x< \dfrac{3}{4}\)
a) ĐKXĐ: \(x\notin\left\{0;3;1\right\}\)
Sửa đề: \(A=\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right):\dfrac{2x-2}{x}\)
Ta có: \(A=\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right):\dfrac{2x-2}{x}\)
\(=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-6x+18}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-6\left(x-3\right)}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-3}{x-1}\)
b) Để A nguyên thì \(-3⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;-2;4\right\}\)
\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}\)
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) Điều kiện : \(x\ne2;x\ne3\)
\(B=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2x+4}{x-3}\)
\(=\frac{2x-9-\left(x-3\right)\left(x+3\right)+2\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x+4}{x-3}\)
b) Điều kiện \(x\in Z;x\ne2;x\ne3\)
Có \(B=\frac{x+4}{x-3}\in Z\), mà x+4 và x-3 nguyên do x nguyên, nên
\(x+4⋮x-3\Leftrightarrow7⋮x-3\), do đó \(x-3\inƯ\left(7\right)=\left\{1;7;-1;-7\right\}\Rightarrow x\in\left\{4;10;2;-4\right\}\)
mà do x khác 2 (điều kiện) nên ta kết luận \(x\in\left\{4;10;-4\right\}\)