Ebatzi: F
\left\{\begin{matrix}F=\frac{4x+7}{25\left(\frac{x^{2}}{10}-\frac{x}{25}+С\right)}\text{, }&С\neq -\frac{x\left(x-0.4\right)}{10}\\F\in \mathrm{R}\text{, }&x=-\frac{7}{4}\text{ and }С=-\frac{301}{800}\end{matrix}\right.
Ebatzi: x
\left\{\begin{matrix}\\x=-1.75\text{, }&\text{unconditionally}\\x=\frac{\sqrt{625СF^{2}+F^{2}+78F+16}+F+4}{5F}\text{; }x=\frac{-\sqrt{625СF^{2}+F^{2}+78F+16}+F+4}{5F}\text{, }&F\neq 0\text{ and }С\geq -\frac{1}{625}-\frac{78}{625F}-\frac{16}{625F^{2}}\end{matrix}\right.
Partekatu
Kopiatu portapapeletan
\left(\frac{x^{2}}{10}-\frac{x}{25}+С\right)F=\frac{4x+7}{25}
Modu arruntean dago ekuazioa.
\frac{\left(\frac{x^{2}}{10}-\frac{x}{25}+С\right)F}{\frac{x^{2}}{10}-\frac{x}{25}+С}=\frac{4x+7}{25\left(\frac{x^{2}}{10}-\frac{x}{25}+С\right)}
Zatitu ekuazioaren bi aldeak \frac{1}{10}x^{2}-0.04x+С balioarekin.
F=\frac{4x+7}{25\left(\frac{x^{2}}{10}-\frac{x}{25}+С\right)}
\frac{1}{10}x^{2}-0.04x+С balioarekin zatituz gero, \frac{1}{10}x^{2}-0.04x+С balioarekiko biderketa desegiten da.
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