Sr Examen

Otras calculadoras

absolute(x+2)+absolute(y+1)=0 la ecuación

El profesor se sorprenderá mucho al ver tu solución correcta😉

v

Solución numérica:

Buscar la solución numérica en el intervalo [, ]

Solución

Ha introducido [src] 
|x + 2| + |y + 1| = 0
∣x+2∣+∣y+1∣=0\left|{x + 2}\right| + \left|{y + 1}\right| = 0
Simplificar
Gráfica
Respuesta rápida [src] 
         //-2 - |1 + y|  for |1 + y| <= 0\     //-2 - |1 + y|  for |1 + y| <= 0\
x1 = I*im|<                              | + re|<                              |
         \\    nan          otherwise    /     \\    nan          otherwise    /
x1=re⁡({−∣y+1∣−2for ∣y+1∣≤0NaNotherwise)+iim⁡({−∣y+1∣−2for ∣y+1∣≤0NaNotherwise)x_{1} = \operatorname{re}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} 
x1 = re(Piecewise((-Abs(y + 1 - 2, |y + 1| <= 0), (nan, True))) + i*im(Piecewise((-|y + 1| - 2, |y + 1| <= 0), (nan, True))))
Suma y producto de raíces [src] 
suma
    //-2 - |1 + y|  for |1 + y| <= 0\     //-2 - |1 + y|  for |1 + y| <= 0\
I*im|<                              | + re|<                              |
    \\    nan          otherwise    /     \\    nan          otherwise    /
re⁡({−∣y+1∣−2for ∣y+1∣≤0NaNotherwise)+iim⁡({−∣y+1∣−2for ∣y+1∣≤0NaNotherwise)\operatorname{re}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} 
=
    //-2 - |1 + y|  for |1 + y| <= 0\     //-2 - |1 + y|  for |1 + y| <= 0\
I*im|<                              | + re|<                              |
    \\    nan          otherwise    /     \\    nan          otherwise    /
re⁡({−∣y+1∣−2for ∣y+1∣≤0NaNotherwise)+iim⁡({−∣y+1∣−2for ∣y+1∣≤0NaNotherwise)\operatorname{re}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} 
producto
    //-2 - |1 + y|  for |1 + y| <= 0\     //-2 - |1 + y|  for |1 + y| <= 0\
I*im|<                              | + re|<                              |
    \\    nan          otherwise    /     \\    nan          otherwise    /
re⁡({−∣y+1∣−2for ∣y+1∣≤0NaNotherwise)+iim⁡({−∣y+1∣−2for ∣y+1∣≤0NaNotherwise)\operatorname{re}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} + i \operatorname{im}{\left(\begin{cases} - \left|{y + 1}\right| - 2 & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases}\right)} 
=
/-(2 + |1 + y|)  for |1 + y| <= 0
<                                
\     nan           otherwise
{−(∣y+1∣+2)for ∣y+1∣≤0NaNotherwise\begin{cases} - (\left|{y + 1}\right| + 2) & \text{for}\: \left|{y + 1}\right| \leq 0 \\\text{NaN} & \text{otherwise} \end{cases} 
Piecewise((-(2 + |1 + y|), |1 + y| <= 0), (nan, True))
    © Sr Examen – Калькулятор