Zgjidh {l}{x^{2}-ab-a^{2}=bx}{x^2=2a^2b^2+abx}{ax^2=a+b-bx}

Gjej x, a, b (complex solution)

x=0\text{, }a=0\text{, }b=0

x=1\text{, }a=\frac{\sqrt{5}+1}{2}\approx 1.618033989\text{, }b=\frac{1-\sqrt{5}}{2}\approx -0.618033989

x=1\text{, }a=\frac{1-\sqrt{5}}{2}\approx -0.618033989\text{, }b=\frac{\sqrt{5}+1}{2}\approx 1.618033989

x=1\text{, }a=\frac{1}{2}+\frac{1}{2}i=0.5+0.5i\text{, }b=\frac{1}{2}-\frac{1}{2}i=0.5-0.5i

x=1\text{, }a=\frac{1}{2}-\frac{1}{2}i=0.5-0.5i\text{, }b=\frac{1}{2}+\frac{1}{2}i=0.5+0.5i

x=-\frac{2}{3}\approx -0.666666667\text{, }a=-1\text{, }b=\frac{1}{3}\approx 0.333333333

x=1\text{, }a=-1\text{, }b=1

x=\frac{-\sqrt{5}-1}{2}\approx -1.618033989\text{, }a=\frac{\sqrt{5}+1}{2}\approx 1.618033989\text{, }b=1

x=\frac{\sqrt{5}-1}{2}\approx 0.618033989\text{, }a=\frac{1-\sqrt{5}}{2}\approx -0.618033989\text{, }b=1

x=1\text{, }a=-1\text{, }b=-\frac{1}{2}=-0.5

x=-\frac{1}{2}-\frac{1}{2}i=-0.5-0.5i\text{, }a=\frac{1}{2}+\frac{1}{2}i=0.5+0.5i\text{, }b=-\frac{1}{2}=-0.5

x=-\frac{1}{2}+\frac{1}{2}i=-0.5+0.5i\text{, }a=\frac{1}{2}-\frac{1}{2}i=0.5-0.5i\text{, }b=-\frac{1}{2}=-0.5

Gjej x, a, b

x=0\text{, }a=0\text{, }b=0

x=1\text{, }a=\frac{\sqrt{5}+1}{2}\approx 1.618033989\text{, }b=\frac{1-\sqrt{5}}{2}\approx -0.618033989

x=1\text{, }a=\frac{1-\sqrt{5}}{2}\approx -0.618033989\text{, }b=\frac{\sqrt{5}+1}{2}\approx 1.618033989

x=-\frac{2}{3}\approx -0.666666667\text{, }a=-1\text{, }b=\frac{1}{3}\approx 0.333333333

x=1\text{, }a=-1\text{, }b=1

x=\frac{-\sqrt{5}-1}{2}\approx -1.618033989\text{, }a=\frac{\sqrt{5}+1}{2}\approx 1.618033989\text{, }b=1

x=\frac{\sqrt{5}-1}{2}\approx 0.618033989\text{, }a=\frac{1-\sqrt{5}}{2}\approx -0.618033989\text{, }b=1

x=1\text{, }a=-1\text{, }b=-\frac{1}{2}=-0.5

Grafiku

Kuiz

Algebra

5 probleme të ngjashme me:

\left. \begin{array} { l } { x ^ { 2 } - a b - a ^ { 2 } = b x } \\ { x ^ { 2 } = 2 a ^ { 2 } b ^ { 2 } + a b x } \\ { a x ^ { 2 } = a + b - b x } \end{array} \right.

Probleme të ngjashme nga kërkimi në ueb

a^{100}-1 is divisible by 1000.

https://math.stackexchange.com/questions/1773562/a100-1-is-divisible-by-1000

By the Chinese remainder theorem, the solutions of x^{100} \equiv 1 \bmod 1000 are exactly the solutions of the system x^{100} \equiv 1 \bmod 8, \qquad x^{100} \equiv 1 \bmod 125 Now, 100 ...

Finding, writing down and visualizing images of functions

https://math.stackexchange.com/questions/40285/finding-writing-down-and-visualizing-images-of-functions

If you want an algebraic solution, follow Qiaochu Yuan and parametrize the region in polar coordinates by (x,y)=(r \cos t, r \sin t), for r\in [0,1) and t \in (0,\pi/2). Then f(x,y)=r^2(\cos^2 t - \sin^2 t, 2 \sin t \cos t) = r^2(\cos 2t, \sin 2t) ...

Determining the Jordan form of a matrix given the invariant factors

https://math.stackexchange.com/q/56820

Edit : I'm sorry, but my first answer was definitely incorrect and I really hope I didn't cause any confusion. Won't speedread problems in the future :) Since we have V = F[x] / (x) \oplus F[x]/ (x(x-1)(x-2)^2) \oplus F[x]/(x^2(x-1)^2(x-3)^3) ...

Continuity of the locus of the maximum of a two variables real function

https://math.stackexchange.com/questions/1331657/continuity-of-the-locus-of-the-maximum-of-a-two-variables-real-function

g is indeed continuous (and the condition that for each x the maximum of f(x,y) is attained at a unique point y is essential). Let g(x_0) = y_0, i.e. M := f(x_0, y_0) = \max_{y \in [0,1]} f(x_0,y ...

Prove that L = \{ a^nb^m : m = n^3 \} is not Context Free using Pumping Lemma

https://math.stackexchange.com/questions/2464188/prove-that-l-anbm-m-n3-is-not-context-free-using-pumping-lemma

Note that in the given language a word length completely determines the actual word. We can enumerate the words, making s_n = a^nb^{n^3}, \ L = \{s_n: n \in \Bbb N \}. Now let's apply the pumping ...

Find the line between two planes (not using vector product).

https://math.stackexchange.com/questions/811493/find-the-line-between-two-planes-not-using-vector-product

You must just get a generalized solution for \begin{equation} \begin{bmatrix} 1&2&-3&| &3 \\ 1&-1&-3&|&6 \end{bmatrix}. \end{equation}After row reduction I get \begin{equation} \begin{bmatrix} ...

Kopjuar në clipboard

Shembuj

Ekuacioni quadratik

{ x } ^ { 2 } - 4 x - 5 = 0

Trigonometria

4 \sin \theta \cos \theta = 2 \sin \theta

\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]

Ekuacioni i njëkohshëm

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

Diferencimi

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

Integrimi

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

Limitet

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

Temat

Temat

Probleme të ngjashme nga kërkimi në ueb

Share

Shembuj