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a: \(P=\dfrac{x^2+6x+9-x^2+6x-9-4}{\left(x-3\right)\left(x+3\right)}:\dfrac{3x-1}{x-3}\)
\(=\dfrac{4\left(3x-1\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{3x-1}=\dfrac{4}{x+3}\)
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Cái biểu thức A ban ghi rõ thì mình mới giải được chứ , ghi như thế ai hiểu mà giải.
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a) ĐK: \(x-5\ne0\Leftrightarrow x\ne5\)
b)
ĐK: \(\left(\dfrac{1}{2}x+4\right)\ne0\Leftrightarrow\dfrac{1}{2}x\ne-4\\ \Leftrightarrow x\ne-8\)
c)ĐK:
\(-2x-10\ne0\\ \Leftrightarrow-2x\ne10\\ \Leftrightarrow x\ne-5\)
a) ĐKXĐ: \(x\ne5\)
b) ĐKXĐ: \(x\ne-8\)
c) ĐKXĐ: \(x\ne-5\)
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a, ĐKXĐ:\(\left\{{}\begin{matrix}x+3\ne0\\x^2+x-6\ne0\\2-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x^2+x-6\ne0\\x\ne2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)
b, \(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+3\right)}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x-4}{x-2}\)
\(c,A=\dfrac{-3}{4}\\ \Leftrightarrow\dfrac{x-4}{x-2}=\dfrac{-3}{4}\\ \Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\\ \Leftrightarrow4x-16x=-3x+6\\ \Leftrightarrow4x-16x+3x-6=0\\ \Leftrightarrow7x-22=0\\ \Leftrightarrow x=\dfrac{22}{7}\)
d, \(A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)
Để \(A\in Z\Rightarrow\dfrac{2}{x-2}\in Z\Rightarrow x-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Ta có bảng:
x-2 | -2 | -1 | 1 | 2 |
x | 0 | 1 | 3 | 4 |
Vậy \(x\in\left\{0;1;3;4\right\}\)
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a: Để C là số nguyên thì \(3x^3+6x^2+3x+x^2+2x+1-2⋮x^2+2x+1\)
=>\(x^2+2x+1\in\left\{1;-1;2;-2\right\}\)
=>(x+1)^2=1 hoặc (x+1)^2=2
=>\(x\in\left\{0;-2;\sqrt{2}-1;-\sqrt{2}-1\right\}\)
b: Để D là số nguyên thì \(x^4+x^2+x^3+x-29⋮x^2+1\)
=>\(x^2+1\in\left\{1;-1;29;-29\right\}\)
=>x^2+1=1 hoặc x^2+1=29
=>\(x\in\left\{0;2\sqrt{7};-2\sqrt{7}\right\}\)