I=mgsqrt{1+(1/n)^2} eching

$$I = m g \sqrt { 1 + ( \frac { 1 } { n } ) ^ { 2 } }$$

g uchun yechish

$$\left\{\begin{matrix}g=\frac{I|n|}{m\sqrt{n^{2}+1}}\text{, }&m\neq 0\text{ and }n\neq 0\\g\in \mathrm{R}\text{, }&I=0\text{ and }m=0\text{ and }n\neq 0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\frac{1}{n}ni darajaga oshirish uchun, surat va maxrajni darajaga oshirib, keyin bo‘ling.

$$I=mg\sqrt{1+\frac{1^{2}}{n^{2}}}$$

Ifodalarni qo‘shish yoki ayirish uchun ularni yoyib, maxrajlarini bir xil qiling. 1 ni \frac{n^{2}}{n^{2}} marotabaga ko'paytirish.

$$I=mg\sqrt{\frac{n^{2}}{n^{2}}+\frac{1^{2}}{n^{2}}}$$

\frac{n^{2}}{n^{2}} va \frac{1^{2}}{n^{2}} da bir xil maxraji bor, ularning suratini qo‘shish orqali qo‘shing.

$$I=mg\sqrt{\frac{n^{2}+1^{2}}{n^{2}}}$$

n^{2}+1^{2} kabi iboralarga o‘xshab birlashtiring.

$$I=mg\sqrt{\frac{n^{2}+1}{n^{2}}}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$mg\sqrt{\frac{n^{2}+1}{n^{2}}}=I$$

Tenglama standart shaklda.

$$\sqrt{\frac{n^{2}+1}{n^{2}}}mg=I$$

Ikki tarafini m\sqrt{\left(n^{2}+1\right)n^{-2}} ga bo‘ling.

$$\frac{\sqrt{\frac{n^{2}+1}{n^{2}}}mg}{\sqrt{\frac{n^{2}+1}{n^{2}}}m}=\frac{I}{\sqrt{\frac{n^{2}+1}{n^{2}}}m}$$

m\sqrt{\left(n^{2}+1\right)n^{-2}} ga bo'lish m\sqrt{\left(n^{2}+1\right)n^{-2}} ga ko'paytirishni bekor qiladi.

$$g=\frac{I}{\sqrt{\frac{n^{2}+1}{n^{2}}}m}$$

I ni m\sqrt{\left(n^{2}+1\right)n^{-2}} ga bo'lish.

$$g=\frac{I|n|}{m\sqrt{n^{2}+1}}$$

I uchun yechish

$$I=\frac{gm\sqrt{n^{2}+1}}{|n|}$$

$$n\neq 0$$