Riješi x^2-z(z+1)x+z^3/sqrt{2+x-x^2}=0 | Microsoftov alat za rješavanje matematike

Izračunaj x (complex solution)

\left\{\begin{matrix}x=z\text{, }&z\neq -1\text{ and }z\neq 2\\x=z^{2}\text{, }&z\neq -i\text{ and }z\neq i\text{ and }z\neq -\sqrt{2}\text{ and }z\neq \sqrt{2}\end{matrix}\right,

Izračunaj z (complex solution)

z=\sqrt{x}

z=x

z=-\sqrt{x}\text{, }x\neq -1\text{ and }x\neq 2

Izračunaj x

\left\{\begin{matrix}x=z\text{, }&z>-1\text{ and }z<2\\x=z^{2}\text{, }&|z|<\sqrt{2}\end{matrix}\right,

Izračunaj z

\left\{\begin{matrix}z=x\text{, }&x>-1\text{ and }x<2\\z=\sqrt{x}\text{; }z=-\sqrt{x}\text{, }&x\geq 0\text{ and }x<2\end{matrix}\right,

Kviz

Algebra

\frac{ { x }^{ 2 } -z(z+1)x+ { z }^{ 3 } }{ \sqrt{ 2+x- { x }^{ 2 } } } =0

Slični problemi iz pretraživanja weba

What's the mistake on my answer for this inequality \frac{\left(x+1\right)}{\sqrt{x^2+1}}>\frac{\left(x+2\right)}{\sqrt{x^2+4}}

https://math.stackexchange.com/questions/1842720/whats-the-mistake-on-my-answer-for-this-inequality-frac-leftx1-right-s

The first part is correct. The inequality holds iff (x+1)\sqrt{x^2+4}>(x+2)\sqrt{x^2+1}\ \ (*) Hence (squaring), provided x+1>0 (and hence x+2>0) we have (x+1)^2(x^2+4)>(x+2)^2(x^2+1) which ...

Intervals of a derivative

https://math.stackexchange.com/questions/3116846/intervals-of-a-derivative

First, note that the denominator is ALWAYS positive when it exists, because it's inside a square root. So now if you look at the numerator, for -(2x+3) , for (-2,-1) , we see that it's positive ...

Simplifying Differentiation

https://math.stackexchange.com/q/401671

I will assume that we are looking at \frac{(x^2+1)^3}{2\sqrt{x-1}} +\sqrt{x-1}(6x)(x^2+1)^2. Take out the common factor (x^2+1)^2, and multiply and divide by 2\sqrt{x-1}. We get \frac{(x^2+1)^2}{2\sqrt{x-1}}\left((x^2+1)+(2)(x+1)(6x) \right). ...

ideas for finding the roots of an annoying polynomial

https://math.stackexchange.com/q/440203

Here's something that's slightly better than the estimate you gave above. To simplify notation let y=x^{1/4} and B=\sqrt{b}. Then \begin{align} f(y) &=a-2By^2-\sqrt{3}y^3-4y^4 \\ &=2B\left(\frac{a}{2B}-y^2\left(1+\frac{\sqrt{3}}{2B}y+\frac{2}{B}y^2\right)\right)\\ &=2B\left(\frac{a}{2B}-y^2\left(1+\frac{\sqrt{3}}{4B}y\right)^2-\left(\frac{2}{B}-\frac{3}{16B^2}\right)y^4\right) \end{align} ...

Prove this relation between the length of the graph and the area below it

https://math.stackexchange.com/questions/118893/prove-this-relation-between-the-length-of-the-graph-and-the-area-below-it

Let h,\ k:[a,b]\to \mathbb{R} be two C^1(a,b) functions s.t.: k\geq h in [a,b], k(a)=h(a) and k(b)=h(b); the graph of h is a concave arc of a circle, that is there esist R>0, x_0\in [a,b] ...

Find (a,b) if x^2-bx+a = 0, x^2-ax+b = 0 both have distinct positive integers roots

https://math.stackexchange.com/q/494679

Let the roots of x^2 - ax +b be r and s, and the roots of x^2 - bx + a be u and v. Then \begin{align} r+s &= a\\ rs &= b\\ u+v &= b\\ uv &= a. \end{align} Let, without loss of ...

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