-a/bx+a^2+b^2/b=3/4x ਨੂੰ ਹੱਲ ਕਰੋ

$$- \frac{ a }{ b } x+ \frac{ { a }^{ 2 } + { b }^{ 2 } }{ b } = \frac{ 3 }{ 4 } x$$

a ਲਈ ਹਲ ਕਰੋ (ਜਟਿਲ ਹੱਲ)

$$a=\frac{\sqrt{\left(x-b\right)\left(x+4b\right)}+x}{2}$$

$$a=\frac{-\sqrt{\left(x-b\right)\left(x+4b\right)}+x}{2}\text{, }b\neq 0$$

b ਲਈ ਹਲ ਕਰੋ (ਜਟਿਲ ਹੱਲ)

$$\left\{\begin{matrix}b=\frac{-\sqrt{\left(9x-8a\right)\left(x+8a\right)}+3x}{8}\text{, }&\left(a\neq 0\text{ and }x\neq a\right)\text{ or }\left(x\neq 0\text{ and }arg(x)\geq \pi \right)\text{ or }\left(a\neq 0\text{ and }arg(x)\geq \pi \right)\text{ or }\left(x\neq a\text{ and }arg(x)\geq \pi \right)\\b=\frac{\sqrt{\left(9x-8a\right)\left(x+8a\right)}+3x}{8}\text{, }&\left(a\neq 0\text{ and }x\neq a\right)\text{ or }\left(x\neq 0\text{ and }arg(x)<\pi \right)\text{ or }\left(a\neq 0\text{ and }arg(x)<\pi \right)\text{ or }\left(x\neq a\text{ and }arg(x)<\pi \right)\end{matrix}\right.$$

a ਲਈ ਹਲ ਕਰੋ

$$a=\frac{\sqrt{\left(x-b\right)\left(x+4b\right)}+x}{2}$$

$$a=\frac{-\sqrt{\left(x-b\right)\left(x+4b\right)}+x}{2}\text{, }b\geq \frac{-5|x|+3x}{8}\text{ and }b\leq \frac{5|x|+3x}{8}\text{ and }b\neq 0$$

b ਲਈ ਹਲ ਕਰੋ

$$\left\{\begin{matrix}b=\frac{-\sqrt{\left(9x-8a\right)\left(x+8a\right)}+3x}{8}\text{, }&\left(a\neq x\text{ and }a\neq 0\text{ and }a\geq -\frac{5|x|}{8}+\frac{x}{2}\text{ and }a\leq \frac{5|x|}{8}+\frac{x}{2}\right)\text{ or }\left(x<0\text{ and }a\geq \frac{9x}{8}\text{ and }a\leq -\frac{x}{8}\right)\\b=\frac{\sqrt{\left(9x-8a\right)\left(x+8a\right)}+3x}{8}\text{, }&\left(a\neq x\text{ and }a\neq 0\text{ and }a\geq -\frac{5|x|}{8}+\frac{x}{2}\text{ and }a\leq \frac{5|x|}{8}+\frac{x}{2}\right)\text{ or }\left(x>0\text{ and }a\geq -\frac{x}{8}\text{ and }a\leq \frac{9x}{8}\right)\end{matrix}\right.$$