Leysa |x|/(x+4)=kx^2 | Microsoft stærðfræði leysa

Leystu fyrir k

\left\{\begin{matrix}k=\frac{|x|}{\left(x+4\right)x^{2}}\text{, }&x\neq -4\text{ and }x\neq 0\\k\in \mathrm{R}\text{, }&x=0\end{matrix}\right.

Leystu fyrir x

\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x=\sqrt{4+\frac{1}{k}}-2\text{, }&k>0\\x=\frac{\sqrt{4k^{2}-k}}{k}-2\text{, }&k<0\text{ or }k\geq \frac{1}{4}\\x=-\sqrt{4-\frac{1}{k}}-2\text{, }&k\geq \frac{1}{4}\end{matrix}\right.

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Spurningakeppni

Algebra

5 vandamál svipuð og:

\frac { | x | } { ( x + 4 ) } = k x ^ { 2 }

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How do you simplify \displaystyle\frac{{{x}^{{2}}+{2}{x}+{1}}}{{{x}+{1}}} ?

https://socratic.org/questions/how-do-you-simplify-x-2-2x-1-x-1

Konstantinos Michailidis Sep 25, 2015 It is \displaystyle\frac{{{x}^{{2}}+{2}{x}+{1}}}{{{x}+{1}}}=\frac{{\left({x}+{1}\right)}^{{2}}}{{{x}+{1}}}={\left({x}+{1}\right)}

How do you evaluate the limit \displaystyle\frac{{{x}^{{2}}+{4}{x}+{2}}}{{{x}^{{2}}-{3}}} as x approaches \displaystyle-{3} ?

https://socratic.org/questions/how-do-you-evaluate-the-limit-x-2-4x-2-x-2-3-as-x-approaches-3

By substitution. Explanation: Since \displaystyle\lim_{{{x}\rightarrow-{3}}}{\left({x}^{{2}}-{3}\right)}\ne{0} we can simply find the quotient of the limits, \displaystyle\lim_{{{x}\rightarrow-{3}}}\frac{{{x}^{{2}}+{4}{x}+{2}}}{{{x}^{{2}}-{3}}}=\frac{{\lim_{{{x}\rightarrow-{3}}}{\left({x}^{{2}}+{4}{x}+{2}\right)}}}{{\lim_{{{x}\rightarrow-{3}}}{\left({x}^{{2}}-{3}\right)}}} ...

Write f(x) = (3x^3+|x|)/x as a piecewise function (rational function)

https://math.stackexchange.com/q/2616812

f(x)=\frac{3x^3+x}{x}=3x^2+1 for x>0 and f(x)=\frac{3x^3-x}{x}=3x^2-1 for x<0

How do you find the Vertical, Horizontal, and Oblique Asymptote given \displaystyle\frac{{{x}^{{2}}-{4}}}{{{x}^{{3}}+{4}{x}^{{2}}}} ?

https://socratic.org/questions/how-do-you-find-the-vertical-horizontal-and-oblique-asymptote-given-x-2-4-x-3-4x

The vertical asymptotes are \displaystyle{x}={0} and \displaystyle{x}=-{4} The horizontal asymptote is \displaystyle{y}={0} Explanation: Let rewrite the function \displaystyle\frac{{{x}^{{2}}-{4}}}{{{x}^{{3}}+{4}{x}^{{2}}}}=\frac{{{x}^{{2}}-{4}}}{{{x}^{{2}}{\left({x}+{4}\right)}}} ...

How do you long divide \displaystyle\frac{{{x}^{{2}}-{x}-{12}}}{{{x}-{4}}} ?

https://socratic.org/questions/how-do-you-long-divide-x-2-x-12-x-4

I found: \displaystyle\frac{{{x}^{{2}}-{x}-{12}}}{{{x}-{4}}}={x}+{3} Explanation: Have a look:

Which of the following functions on R are uniformly continous?

https://math.stackexchange.com/questions/830993/which-of-the-following-functions-on-r-are-uniformly-continous

Hint If a function f is lipschitzian then it's uniformly continuous and notice that if f' is bounded then f is lipschitzian. We can apply the above remark to a,b,c). Now for d) let x_n=2n\pi ...

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