Selesaikan {l}{a^2+4a+1=0}{b^2+4b+1=0}

\left\{ \begin{array} { l } { a ^ { 2 } + 4 a + 1 = 0 } \\ { b ^ { 2 } + 4 b + 1 = 0 } \end{array} \right.

Selesaikan untuk a, b (complex solution)

a=-\left(\sqrt{3}+2\right)\approx -3.732050808\text{, }b=-\left(\sqrt{3}+2\right)\approx -3.732050808

a=\sqrt{3}-2\approx -0.267949192\text{, }b=-\left(\sqrt{3}+2\right)\approx -3.732050808

a=-\left(\sqrt{3}+2\right)\approx -3.732050808\text{, }b=\sqrt{3}-2\approx -0.267949192

a=\sqrt{3}-2\approx -0.267949192\text{, }b=\sqrt{3}-2\approx -0.267949192

Selesaikan untuk a, b

a=-\sqrt{3}-2\approx -3.732050808\text{, }b=-\left(\sqrt{3}+2\right)\approx -3.732050808

a=\sqrt{3}-2\approx -0.267949192\text{, }b=-\left(\sqrt{3}+2\right)\approx -3.732050808

a=-\sqrt{3}-2\approx -3.732050808\text{, }b=\sqrt{3}-2\approx -0.267949192

a=\sqrt{3}-2\approx -0.267949192\text{, }b=\sqrt{3}-2\approx -0.267949192

Kuiz

5 masalah yang serupa dengan:

\left\{ \begin{array} { l } { a ^ { 2 } + 4 a + 1 = 0 } \\ { b ^ { 2 } + 4 b + 1 = 0 } \end{array} \right.

Masalah Sama dari Carian Web

Let f(t) be defined by f(t) = \left\{ \begin{array}{c} t, &0 \le t < 1 \\ b - t^2, &1 \le t \le 2 \end{array} \right.

https://math.stackexchange.com/questions/777108/let-ft-be-defined-by-ft-left-beginarrayc-t-0-le-t-1-b

This seems correct to me. Now you must check for which b the right and left derivatives at x = 1 are equal. The left derivative is 1 and the right is -1+b so 1 = -1+b iff b =2.

Is \sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n, (p,q,n)\in\mathbb{N} ^3 solvable?

https://math.stackexchange.com/q/374619

The solutions are of the form \displaystyle(p, q)= \left(\frac{3t^2nr+n^3}{8},\,\frac{3n^2t+t^3r}{8}\right), for any rational parameter t. To prove it, we start with \left(p+q\sqrt{r}\right)^{1/3}+\left(p-q\sqrt{r}\right)^{1/3}=n\tag{ ...

Transformation related to a particular random variable

https://math.stackexchange.com/q/1620488

Well, aside from the fact that your given density function is negative for y<0 (and its integral over the support is therefore zero; is there a typo somewhere?), your next step would be to take the ...

Matrix to the power t.

https://math.stackexchange.com/q/2903784

For a diagonal n\times n matrix, it is not very hard to check that \bigl(D(a_1,\dots,a_n)\bigr)^t=D(a_1^t,\dots,a_n^t). For A_3 and A_4, just compute the first values to find that A_3^2=0,\quad A_4^2=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix},\quad A_4^3=0. ...

Find the real roots for \displaystyle \sqrt[4]{386-x}+\sqrt[4]{x}=6.

https://math.stackexchange.com/q/2529147

Very nice! The only thing that I can add to this is that I would have dealt with u^4+v^4 in a way which is, I think, simpler than yours. Observe ...

Elementary use of Hölder inequality

https://math.stackexchange.com/q/2609187

You can compute straight away \|a^2b\|_{L^2}^2 = \int a^3 ab^2 \leq \frac{1}{2} \int a^6 + a^2b^4 = C( \|a\|_{L^6}^6 + \|ab^2\|_{L^2}^2) \leq C( \|a\|_{L^6}^6 + \|a\|_{L^\infty}^2\|b^2\|_{L^2}^2). ...

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