[tex]d)\\\\d=\dfrac{b}{k-1} ,~~zal.~k-1\neq 0~~\Rightarrow ~~k\neq 1\\\\\\d=\dfrac{b}{k-1} ~~\mid \cdot (k-1)\\\\d(k-1)=b\\\\dk-d=b\\\\dk=b+d~~\mid \div d\\\\k=\dfrac{b+d}{d} =\dfrac{b}{d} +\dfrac{d}{d} =\dfrac{b}{d} +1,~~zal.~d\neq 0[/tex]
[tex]e)\\\\e=\dfrac{a-1}{k+1} ,~~zal.~k+1\neq 0~~\Rightarrow ~~k\neq -1\\\\e=\dfrac{a-1}{k+1} ~~\mid \cdot (k+1)\\\\e(k+1)=a-1\\\\ek+e=a-1\\\\ek=a-e-1~~\mid \div e\\\\k=\dfrac{a-e-1}{e} =\dfrac{a}{e} -\dfrac{e}{e} -\dfrac{1}{e} =\dfrac{a}{e} -1-\dfrac{1}{e} ,~~zal.~~e\neq 0[/tex]
[tex]f)\\\\5f=\dfrac{2c-3}{1-2k} ,~~zal.~1-2k\neq 0~~\Rightarrow ~~k\neq \dfrac{1}{2} \\\\5f=\dfrac{2c-3}{1-2k}~~\mid \cdot ( 1-2k)\\\\5f(1-2k)=2c-3\\\\5f-10fk=2c-3\\\\-10fk=2c-5f-3~~\mid ~~\div (-10f)\\\\k=\dfrac{2c-5f-3}{-10f} =-\dfrac{2c}{10f} +\dfrac{5f}{10f} +\dfrac{3}{10f} =-\dfrac{c}{5f} +\dfrac{1}{2} +\dfrac{3}{10f} ,~~zal.~f\neq 0[/tex]
[tex]g)\\\\k(a+b)-1=c\\\\k(a+b)=c+1~~\mid \div (a+b)\\\\k=\dfrac{c+1}{a+b} ,~~zal.~~a+b\neq 0~~\Rightarrow ~~a\neq -b[/tex]
[tex]h)\\\\3+2k(M-2)=D\\\\2k(M-2)=D-3\\\\k(2M-4)=D-3~~\mid\div (2M-4)\\\\k=\dfrac{D-3}{2M-4} ,~~zal.~2M-4\neq 0~~\Rightarrow ~~M\neq 2[/tex]
[tex]i)\\\\F+2=H-10k(E+2)\\\\10k(E+2)=H-F-2\\\\k(10E+20)=H-F-2 \mid \div (10E+20)\\\\k=\dfrac{H-F-2}{10E+20} , ~~zal.~10E+20\neq 0~~\Rightarrow ~~E\neq -2[/tex]
Aby rozwiązać równanie , przekształcamy je w taki sposób, żeby po jednej jego stronie była tylko niewiadoma a po drugiej pozostałe wartości.
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