Vyřešit 25^x-4/2x^x-4 | Microsoft Math Solver

Vyhodnotit

\frac{25^{x}-4}{2\left(x^{x}-2\right)}

Derivovat vzhledem k x

\frac{2\ln(5)\times 5^{2x}x^{x}-\ln(x)\times \left(25x\right)^{x}+4\ln(x)x^{x}+4x^{x}-\left(25x\right)^{x}-4\ln(5)\times 5^{2x}}{2\left(x^{x}-2\right)^{2}}

Graf

Kvíz

Polynomial

5 úloh podobných jako:

\frac { 25 ^ { x } - 4 } { 2 x ^ { x } - 4 }

Podobné úlohy z vyhledávání na webu

(2x^(2)-x-6)/(x-2)=a

https://www.tiger-algebra.com/drill/(2x~(2)-x-6)/(x-2)=a/

(2x(2)-x-6)/(x-2)=a Geometric figure: Straight Line Slope = 2 x-intercept = -3/2 = -1.50000 a-intercept = 3/1 = 3.00000 Rearrange: Rearrange the equation by subtracting what is to the ...

x^2-4=(2-x)/(x+3)

https://www.tiger-algebra.com/drill/x~2-4=(2-x)/(x_3)/

x2-4=(2-x)/(x+3) Three solutions were found : x = 2 x =(-5-√-3)/2=(-5-i√ 3 )/2= -2.5000-0.8660i x =(-5+√-3)/2=(-5+i√ 3 )/2= -2.5000+0.8660i Rearrange: Rearrange the equation by subtracting ...

How do you find the vertical, horizontal or slant asymptotes for \displaystyle{f{{\left({x}\right)}}}=\frac{{{2}{x}-{4}}}{{{x}^{{2}}-{4}}} ?

https://socratic.org/questions/how-do-you-find-the-vertical-horizontal-or-slant-asymptotes-for-f-x-2x-4-x-2-4

vertical asymptote x = -2 horizontal asymptote y = 0 Explanation: The first step here is to factorise and simplify f(x). \displaystyle\Rightarrow{f{{\left({x}\right)}}}=\frac{{{2}{x}-{4}}}{{{x}^{{2}}-{4}}}=\frac{{{2}\cancel{{{\left({x}-{2}\right)}}}}}{{\cancel{{{\left({x}-{2}\right)}}}{\left({x}+{2}\right)}}}=\frac{{2}}{{{x}+{2}}} ...

How do you express \displaystyle\frac{{{x}^{{5}}+{1}}}{{{x}^{{6}}-{x}^{{4}}}} in partial fractions?

https://socratic.org/questions/how-do-you-express-x-5-1-x-6-x-4-in-partial-fractions

\displaystyle\frac{{{x}^{{5}}+{1}}}{{{x}^{{4}}{\left({x}+{1}\right)}{\left({x}-{1}\right)}}}=-\frac{{1}}{{x}^{{2}}}-\frac{{1}}{{x}^{{4}}}+\frac{{1}}{{{x}-{1}}} Explanation: \displaystyle\frac{{{x}^{{5}}+{1}}}{{{x}^{{4}}{\left({x}^{{2}}-{1}\right)}}}=\frac{{{x}^{{5}}+{1}}}{{{x}^{{4}}{\left({x}+{1}\right)}{\left({x}-{1}\right)}}} ...

How do you differentiate \displaystyle\frac{{-{x}^{{2}}-{5}{x}-{4}}}{{{2}{x}^{{2}}-{1}}} using the quotient rule?

https://socratic.org/questions/how-do-you-differentiate-x-2-5x-4-2x-2-1-using-the-quotient-rule

\displaystyle\frac{{-{8}{x}^{{3}}-{30}{x}^{{2}}-{14}{x}+{5}}}{{\left({2}{x}^{{2}}-{1}\right)}^{{2}}} Explanation: Applying the quotient rule gives : \displaystyle{f}'{\left({x}\right)}=\frac{{{\left({\left({2}{x}^{{2}}-{1}\right)}.\frac{{d}}{{\left.{d}{x}\right.}}{\left(-{x}^{{2}}-{5}{x}-{4}\right)}-{\left(-{x}^{{2}}-{5}{x}-{4}\right)}.\frac{{d}}{{\left.{d}{x}\right.}}{\left({2}{x}^{{2}}-{1}\right)}\right)}}}{{\left({2}{x}^{{2}}-{1}\right)}^{{2}}} ...

25x-(1/2)x^2=300

https://www.tiger-algebra.com/drill/25x-(1/2)x~2=300/

25x-(1/2)x2=300 Two solutions were found : x = 30 x = 20 Rearrange: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : ...

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