Question

সমাধান কর: $\frac{a+x-\sqrt{a^2-x^2}}{a+x+\sqrt{a^2-x^2}}=\frac{b}{x}$,$2a>b>0$ এবং $x\neq0$

Answer 1

$\frac{a+x-\sqrt{a^2-x^2}}{a+x+\sqrt{a^2-x^2}}=$$\frac{b}{x}$

বা, $\frac{\left(a+x-\sqrt{a^2-x^2}\right)+\left(a+x+\sqrt{a^2-x^2}\right)}{\left(a+x-\sqrt{a^2-x^2}\right)-\left(a+x+\sqrt{a^2-x^2}\right)}=$$\frac{b+x}{b-x}$

বা, $\frac{a+x-\sqrt{a^2-x^2}+a+x+\sqrt{a^2-x^2}}{a+x-\sqrt{a^2-x^2}-a-x-\sqrt{a^2-x^2}}=$$\frac{b+x}{b-x}$

বা, $\frac{2a+2x}{-2\sqrt{a^2-x^2}}=$$\frac{b+x}{b-x}$

বা, $\frac{2\left(a+x\right)}{-2\sqrt{a^2-x^2}}=$$\frac{b+x}{b-x}$

বা, $\frac{\left(a+x\right)}{-\sqrt{a^2-x^2}}=$$\frac{b+x}{b-x}$

বা, $\left(\frac{a+x}{-\sqrt{a^2-x^2}}\right)^2=$$\left(\frac{b+x}{b-x}\right)^2$

বা, $\frac{\left(a+x\right)^2}{a^2-x^2}=$$\frac{\left(b+x\right)^2}{\left(b-x\right)^2}$

বা, $\frac{a^2+2ax+x^2}{a^2-x^2}=$$\frac{b^2+2bx+x^2}{b^2-2bx+x^2}$

বা, $\frac{\left(a^2+2ax+x^2\right)+\left(a^2-x^2\right)}{\left(a^2+2ax+x^2\right)-\left(a^2-x^2\right)}=$$\frac{\left(b^2+2bx+x^2\right)+\left(b^2-2bx+x^2\right)}{\left(b^2+2bx+x^2\right)-\left(b^2-2bx+x^2\right)}$

বা, $\frac{a^2+2ax+x^2+a^2-x^2}{a^2+2ax+x^2-a^2+x^2}=$$\frac{b^2+2bx+x^2+b^2-2bx+x^2}{b^2+2bx+x^2-b^2+2bx-x^2}$

বা, $\frac{2a^2+2ax}{2ax+2x^2}=$$\frac{2b^2+2x^2}{4bx}$

বা, $\frac{2a\left(a+x\right)}{2x\left(a+x\right)}=$$\frac{2\left(b^2+x^2\right)}{4bx}$

বা, $\frac{a}{x}=$$\frac{b^2+x^2}{2bx}$

বা, $\frac{a}{x}\cdot x=$$\frac{b^2+x^2}{2bx}\cdot x$

বা, $a=\frac{b^2+x^2}{2b}$

বা, $b^2+x^2=2ab$

বা, $x^2=2ab-b^2$

$\therefore x=\pm\sqrt{2ab-b^2}$

সুতরাং নির্নেয় সমাধান $\pm\sqrt{2ab-b^2}$ [Answer]