Resolver BS=(0.04-x)^2/(0.05-x)^2

$$B S = \frac { ( 0.04 - x ) ^ { 2 } } { ( 0.05 - x ) ^ { 2 } }$$

Ebatzi: B (complex solution)

$$\left\{\begin{matrix}B=\frac{16\times \left(\frac{25x-1}{20x-1}\right)^{2}}{25S}\text{, }&x\neq \frac{1}{20}\text{ and }S\neq 0\\B\in \mathrm{C}\text{, }&S=0\text{ and }x=\frac{1}{25}\end{matrix}\right.$$

Ekuazio lineala ebazteko urratsak

\left(0.04-x\right)^{2} zabaltzeko, erabili Newton-en binomioa \left(a-b\right)^{2}=a^{2}-2ab+b^{2}.

$$BS=\frac{0.0016-0.08x+x^{2}}{\left(0.05-x\right)^{2}}$$

\left(0.05-x\right)^{2} zabaltzeko, erabili Newton-en binomioa \left(a-b\right)^{2}=a^{2}-2ab+b^{2}.

$$BS=\frac{0.0016-0.08x+x^{2}}{0.0025-0.1x+x^{2}}$$

Modu arruntean dago ekuazioa.

$$SB=\frac{x^{2}-\frac{2x}{25}+0.0016}{x^{2}-\frac{x}{10}+0.0025}$$

Zatitu ekuazioaren bi aldeak S balioarekin.

$$\frac{SB}{S}=\frac{16\left(25x-1\right)^{2}}{25\left(20x-1\right)^{2}S}$$

S balioarekin zatituz gero, S balioarekiko biderketa desegiten da.

$$B=\frac{16\left(25x-1\right)^{2}}{25\left(20x-1\right)^{2}S}$$

Zatitu \frac{16\left(25x-1\right)^{2}}{25\left(20x-1\right)^{2}} balioa S balioarekin.

$$B=\frac{16\left(25x-1\right)^{2}}{25S\left(20x-1\right)^{2}}$$

Ebatzi: S (complex solution)

$$\left\{\begin{matrix}S=\frac{16\times \left(\frac{25x-1}{20x-1}\right)^{2}}{25B}\text{, }&x\neq \frac{1}{20}\text{ and }B\neq 0\\S\in \mathrm{C}\text{, }&B=0\text{ and }x=\frac{1}{25}\end{matrix}\right.$$

Ekuazio lineala ebazteko urratsak

\left(0.04-x\right)^{2} zabaltzeko, erabili Newton-en binomioa \left(a-b\right)^{2}=a^{2}-2ab+b^{2}.

$$BS=\frac{0.0016-0.08x+x^{2}}{\left(0.05-x\right)^{2}}$$

\left(0.05-x\right)^{2} zabaltzeko, erabili Newton-en binomioa \left(a-b\right)^{2}=a^{2}-2ab+b^{2}.

$$BS=\frac{0.0016-0.08x+x^{2}}{0.0025-0.1x+x^{2}}$$

Modu arruntean dago ekuazioa.

$$BS=\frac{x^{2}-\frac{2x}{25}+0.0016}{x^{2}-\frac{x}{10}+0.0025}$$

Zatitu ekuazioaren bi aldeak B balioarekin.

$$\frac{BS}{B}=\frac{16\left(25x-1\right)^{2}}{25\left(20x-1\right)^{2}B}$$

B balioarekin zatituz gero, B balioarekiko biderketa desegiten da.

$$S=\frac{16\left(25x-1\right)^{2}}{25\left(20x-1\right)^{2}B}$$

Zatitu \frac{16\left(25x-1\right)^{2}}{25\left(20x-1\right)^{2}} balioa B balioarekin.

$$S=\frac{16\left(25x-1\right)^{2}}{25B\left(20x-1\right)^{2}}$$

Ebatzi: B

$$\left\{\begin{matrix}B=\frac{16\times \left(\frac{25x-1}{20x-1}\right)^{2}}{25S}\text{, }&x\neq \frac{1}{20}\text{ and }S\neq 0\\B\in \mathrm{R}\text{, }&S=0\text{ and }x=\frac{1}{25}\end{matrix}\right.$$

Ekuazio lineala ebazteko urratsak

\left(0.04-x\right)^{2} zabaltzeko, erabili Newton-en binomioa \left(a-b\right)^{2}=a^{2}-2ab+b^{2}.

$$BS=\frac{0.0016-0.08x+x^{2}}{\left(0.05-x\right)^{2}}$$

\left(0.05-x\right)^{2} zabaltzeko, erabili Newton-en binomioa \left(a-b\right)^{2}=a^{2}-2ab+b^{2}.

$$BS=\frac{0.0016-0.08x+x^{2}}{0.0025-0.1x+x^{2}}$$

Modu arruntean dago ekuazioa.

$$SB=\frac{x^{2}-\frac{2x}{25}+0.0016}{x^{2}-\frac{x}{10}+0.0025}$$

Zatitu ekuazioaren bi aldeak S balioarekin.

$$\frac{SB}{S}=\frac{16\left(25x-1\right)^{2}}{25\left(20x-1\right)^{2}S}$$

S balioarekin zatituz gero, S balioarekiko biderketa desegiten da.

$$B=\frac{16\left(25x-1\right)^{2}}{25\left(20x-1\right)^{2}S}$$

Zatitu \frac{16\left(25x-1\right)^{2}}{25\left(20x-1\right)^{2}} balioa S balioarekin.

$$B=\frac{16\left(25x-1\right)^{2}}{25S\left(20x-1\right)^{2}}$$

Ebatzi: S

$$\left\{\begin{matrix}S=\frac{16\times \left(\frac{25x-1}{20x-1}\right)^{2}}{25B}\text{, }&x\neq \frac{1}{20}\text{ and }B\neq 0\\S\in \mathrm{R}\text{, }&B=0\text{ and }x=\frac{1}{25}\end{matrix}\right.$$

Ekuazio lineala ebazteko urratsak

\left(0.04-x\right)^{2} zabaltzeko, erabili Newton-en binomioa \left(a-b\right)^{2}=a^{2}-2ab+b^{2}.

$$BS=\frac{0.0016-0.08x+x^{2}}{\left(0.05-x\right)^{2}}$$

\left(0.05-x\right)^{2} zabaltzeko, erabili Newton-en binomioa \left(a-b\right)^{2}=a^{2}-2ab+b^{2}.

$$BS=\frac{0.0016-0.08x+x^{2}}{0.0025-0.1x+x^{2}}$$

Modu arruntean dago ekuazioa.

$$BS=\frac{x^{2}-\frac{2x}{25}+0.0016}{x^{2}-\frac{x}{10}+0.0025}$$

Zatitu ekuazioaren bi aldeak B balioarekin.

$$\frac{BS}{B}=\frac{16\left(25x-1\right)^{2}}{25\left(20x-1\right)^{2}B}$$

B balioarekin zatituz gero, B balioarekiko biderketa desegiten da.

$$S=\frac{16\left(25x-1\right)^{2}}{25\left(20x-1\right)^{2}B}$$

Zatitu \frac{16\left(25x-1\right)^{2}}{25\left(20x-1\right)^{2}} balioa B balioarekin.

$$S=\frac{16\left(25x-1\right)^{2}}{25B\left(20x-1\right)^{2}}$$