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a: Thay x=-4 vào B, ta được:
\(B=\dfrac{-4+3}{-4}=\dfrac{-1}{-4}=\dfrac{1}{4}\)
b: \(P=A\cdot B=\dfrac{x^2-3x+2x-9+3x+9}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}\)
\(=\dfrac{x^2+2x}{\left(x-3\right)}\cdot\dfrac{1}{x}=\dfrac{x+2}{x-3}\)
c: Để P nguyên thì \(x-3\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{4;2;8;-2\right\}\)
a)B = \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{7x+3}{9-x^2}\left(ĐK:x\ne\pm3\right)\)
= \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{7x+3}{x^2-9}\)
= \(\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-7x-3}{\left(x+3\right)\left(x-3\right)}\)
= \(\dfrac{3x^2-9x}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x}{x+3}\)
b) \(\left|2x+1\right|=7< =>\left[{}\begin{matrix}2x+1=7< =>x=3\left(L\right)\\2x+1=-7< =>x=-4\left(C\right)\end{matrix}\right.\)
Thay x = -4 vào B, ta có:
B = \(\dfrac{-4.3}{-4+3}=12\)
c) Để B = \(\dfrac{-3}{5}\)
<=> \(\dfrac{3x}{x+3}=\dfrac{-3}{5}< =>\dfrac{3x}{x+3}+\dfrac{3}{5}=0\)
<=> \(\dfrac{15x+3x+9}{5\left(x+3\right)}=0< =>x=\dfrac{-1}{2}\left(TM\right)\)
d) Để B nguyên <=> \(\dfrac{3x}{x+3}\) nguyên
<=> \(3-\dfrac{9}{x+3}\) nguyên <=> \(9⋮x+3\)
| x+3 | -9 | -3 | -1 | 1 | 3 | 9 |
| x | -12(C) | -6(C) | -4(C) | -2(C) | 0(C) | 6(C) |
\(a,P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\left(x\ne\pm3;x\ne-2\right)\\ P=\dfrac{2x^2-7x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x+2}\\ P=\dfrac{3x^2+6x}{\left(x-3\right)\left(x+2\right)}=\dfrac{3x\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}=\dfrac{3x}{x-3}\\ b,x^2-7x+12=0\\ \Leftrightarrow\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow x=4\left(x\ne3\right)\\ \Leftrightarrow A=\dfrac{3\cdot4}{4-3}=12\\ c,P=\dfrac{3\left(x-3\right)+9}{x-3}=3+\dfrac{9}{x-3}\in Z\\ \Leftrightarrow x-3\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;4;6;12\right\}\)
a) ĐKXĐ: \(x\notin\left\{3;-3;-2\right\}\)
Ta có: \(P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\)
\(=\left(\dfrac{\left(2x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3-10x}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{x+2}{x-3}\)
\(=\dfrac{2x^2-6x-x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+2}{x-3}\)
\(=\dfrac{3x^2+6x}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+2}{x-3}\)
\(=\dfrac{3x\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x+2}\)
\(=\dfrac{3x}{x+3}\)
b) Ta có: \(x^2-7x+12=0\)
\(\Leftrightarrow x^2-3x-4x+12=0\)
\(\Leftrightarrow x\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
Thay x=4 vào biểu thức \(P=\dfrac{3x}{x+3}\), ta được:
\(P=\dfrac{3\cdot4}{4+3}=\dfrac{12}{7}\)
Vậy: Khi \(x^2-7x+12=0\) thì \(P=\dfrac{12}{7}\)
a) Ta có: \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)
\(=\dfrac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2-6x+x^2+4x+3+11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x}{x-3}\)
b)
ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)
Ta có: P=AB
\(=\dfrac{3x}{x-3}\cdot\dfrac{x-3}{x+1}\)
\(=\dfrac{3x}{x+1}\)
Để \(P=\dfrac{9}{2}\) thì \(\dfrac{3x}{x+1}=\dfrac{9}{2}\)
\(\Leftrightarrow9\left(x+1\right)=6x\)
\(\Leftrightarrow9x-6x=-9\)
\(\Leftrightarrow3x=-9\)
hay x=-3(loại)
Vậy: Không có giá trị nào của x để \(P=\dfrac{9}{2}\)
a: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)
\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{6\left(x+1\right)\cdot x\left(x+2\right)}{3\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x}{x-2}\)
a: Thay x=0 vào A, ta được:
\(A=\dfrac{0+1}{0-3}=\dfrac{1}{-3}=-\dfrac{1}{3}\)
b: \(B=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)
\(=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)+11x-3}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{2x^2-6x+x^2+4x+3+11x-3}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x^2+9x}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x}{x-3}\)
c: \(P=\dfrac{B}{A}=\dfrac{3x}{x-3}:\dfrac{x+1}{x-3}=\dfrac{3x}{x+1}\)
Để P nguyên thì \(3x⋮x+1\)
=>\(3x+3-3⋮x+1\)
=>\(-3⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
`#3107.101107`
`a.`
Tại `x = 0:`
`A = (0 + 1)/(0 - 3) = 1/(-3) = -1/3`
Vậy, `A = -1/3` tại `x = 0`
`b.`
`B= (2x)/(x+3) + (x+1)/(x-3) + (3 - 11x)/(9-x^2)`
\(=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{\left(3-x\right)\left(3+x\right)}\)
\(=\dfrac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2-6x+x^2+4x+3-3+11x}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{\left(2x^2+x^2\right)+\left(11x-6x+4x\right)+\left(3-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{3x^2+9x}{\left(x+3\right)\left(x-3\right)}\\ =\dfrac{3x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\\ =\dfrac{3x}{x-3}\)
`c.`
`P = B/A`
`=> P =` \(\dfrac{3x}{x-3}\div\dfrac{x+1}{x-3}\)
`P = (3x)/(x - 3) * (x - 3)/(x + 1)`
`P = (3x)/(x + 1)`
Ta có: `3x \vdots (x + 1)`
`=> (3x + 3 - 3) \vdots (x + 1)`
`=> -3 \vdots (x + 1)`
`=> x + 1 \in \text{Ư(-3)} = {+-1; +-3}`
`=>`\(\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=3\\x+1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=2\\x=-4\end{matrix}\right.\)