Solve rx-1/x^r+1*x+1/4x-r= | Microsoft Math Solver

Evaluate

\frac{\left(x+1\right)\left(rx-1\right)}{\left(4x-r\right)\left(x^{r}+1\right)}

Differentiate w.r.t. x

\frac{4rx^{2}-2xr^{2}+r^{3}x^{r+1}+r^{3}x^{r}-r^{2}x^{r-1}-4r^{2}x^{r+2}-6r^{2}x^{r+1}-2r^{2}x^{r}+4rx^{r+1}+4rx^{r+2}+5rx^{r}+4x^{r}-r^{2}+r+4}{\left(\left(4x-r\right)\left(x^{r}+1\right)\right)^{2}}

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\frac { r x - 1 } { x ^ { r } + 1 } \times \frac { x + 1 } { 4 x - r } =

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How do you solve \displaystyle{1.7}\times{10}^{{-{1}}}=\frac{{{x}^{{2}}}}{{{0.60}-{x}}} for \displaystyle{x} ?

https://socratic.org/questions/5999b368b72cff213f6b6084

See below. Explanation: You have: \displaystyle{1.7}\times{10}^{{-{{1}}}}=\frac{{x}^{{2}}}{{{0.60}-{x}}} Multiply both sides by the denominator of the right: \displaystyle{\left({0.60}-{x}\right)}{\left({1.7}\times{10}^{{-{{1}}}}\right)}={\left(\frac{{x}^{{2}}}{\cancel{{{\left({0.60}-{x}\right)}}}}\right)}\cancel{{{\left({0.60}-{x}\right)}}} ...

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https://socratic.org/questions/how-do-you-factor-and-simplify-frac-x-2-2x-8-x-2-9-times-frac-x-3-x-4

\displaystyle\frac{{{x}-{2}}}{{{x}+{3}}} Explanation: \displaystyle\frac{{{x}^{{2}}+{2}{x}-{8}}}{{{x}^{{2}}-{9}}}\cdot\frac{{{x}+{3}}}{{{x}+{4}}} This factors out to \displaystyle\frac{{\cancel{{{\left({x}+{4}\right)}}}{\left({x}-{2}\right)}}}{{{\left({x}-{3}\right)}\cancel{{{\left({x}+{3}\right)}}}}}\cdot\frac{\cancel{{{x}+{3}}}}{\cancel{{{x}+{4}}}} ...

Proof that K[x]/(x-1)^2 is not isomorphic to ((x-1)/(x-1)^2)^2 [closed]

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As rings it is easy to show these are not isomorphic: K[x]/(x - 1)^2 is a ring with unity (1 + (x - 1)^2) and R = (x - 1)/(x - 1)^2\times (x - 1)/(x - 1)^2 has no 1. (Every element of R is ...

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Term (-10)/x^2 is always negative. (1 + 1/x)^9 = [(1 + 1/x)^8] * (1 + 1/x) (above term always positive) So final thing left is (1 + 1/x) which we need to test. ...

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Factor out constants: \frac{\text{d}}{\text{d}x}\left(\frac{\sqrt{x-1}}{3x^2}\right)=\frac{1}{3}\cdot\frac{\text{d}}{\text{d}x}\left(\frac{\sqrt{x-1}}{x^2}\right) Use the procut rule \frac{\text{d}}{\text{d}x}(uv)=v\cdot\frac{\text{d}u}{\text{d}x}+u\cdot\frac{\text{d}v}{\text{d}x} ...

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