Lutasin ang f:y=sqrt{x^2-3x+2}+1/sqrt{3+2x-x^2}

$$f : y = \sqrt { x ^ { 2 } - 3 x + 2 } + \frac { 1 } { \sqrt { 3 + 2 x - x ^ { 2 } } }$$

I-solve ang f (complex solution)

$$f=y\sqrt{x^{2}-3x+2}+\left(3+2x-x^{2}\right)^{-\frac{1}{2}}y$$

$$y\neq 0\text{ and }x\neq -1\text{ and }x\neq 3$$

Mga Hakbang sa Pagso-solve ng Linear Equation

Ang equation ay nasa standard form.

$$\frac{1}{y}f=\sqrt{x^{2}-3x+2}+\frac{1}{\sqrt{3+2x-x^{2}}}$$

I-divide ang magkabilang dulo ng equation gamit ang y^{-1}.

$$\frac{\frac{1}{y}fy}{1}=\frac{\left(\sqrt{x^{2}-3x+2}+\left(3+2x-x^{2}\right)^{-\frac{1}{2}}\right)y}{1}$$

Kapag na-divide gamit ang y^{-1}, ma-a-undo ang multiplication gamit ang y^{-1}.

$$f=\frac{\left(\sqrt{x^{2}-3x+2}+\left(3+2x-x^{2}\right)^{-\frac{1}{2}}\right)y}{1}$$

I-divide ang \sqrt{x^{2}+2-3x}+\left(-x^{2}+3+2x\right)^{-\frac{1}{2}} gamit ang y^{-1}.

$$f=y\sqrt{x^{2}-3x+2}+\left(3+2x-x^{2}\right)^{-\frac{1}{2}}y$$

I-solve ang f

$$f=\frac{y\left(\sqrt{\left(3-x\right)\left(x-2\right)\left(x^{2}-1\right)}+1\right)}{\sqrt{\left(3-x\right)\left(x+1\right)}}$$

$$y\neq 0\text{ and }\left(x\leq 1\text{ or }x\geq 2\right)\text{ and }x<3\text{ and }x>-1$$

Mga Hakbang sa Pagso-solve ng Linear Equation

Ang equation ay nasa standard form.

$$\frac{1}{y}f=\sqrt{x^{2}-3x+2}+\frac{1}{\sqrt{3+2x-x^{2}}}$$

I-divide ang magkabilang dulo ng equation gamit ang y^{-1}.

$$\frac{\frac{1}{y}fy}{1}=\frac{\left(\sqrt{x^{2}-3x+2}+\frac{1}{\sqrt{\left(3-x\right)\left(x+1\right)}}\right)y}{1}$$

Kapag na-divide gamit ang y^{-1}, ma-a-undo ang multiplication gamit ang y^{-1}.

$$f=\frac{\left(\sqrt{x^{2}-3x+2}+\frac{1}{\sqrt{\left(3-x\right)\left(x+1\right)}}\right)y}{1}$$

I-divide ang \sqrt{x^{2}+2-3x}+\frac{1}{\sqrt{\left(3-x\right)\left(1+x\right)}} gamit ang y^{-1}.

$$f=y\sqrt{x^{2}-3x+2}+\frac{y}{\sqrt{\left(3-x\right)\left(x+1\right)}}$$