Question

Daca a supra b este egal cu: 1) 2 supra 5; 2) 1,2; 3) 2,(2) calculati 2a+b supra 4b-a
Va rog ajutati. Ma si pe mine la acest ex

Answer 1

 

$\displaystyle\\\text{Pentru rezolvarea problemei vom simplifica {\underline{\bf fortat cu b}} fractia: } \frac{2a-b}{4b-a}\\\\\\\frac{2a-b^{\b(\b b}}{4b-a~~}=\frac{\dfrac{2a-b}{b} }{\dfrac{4b-a}{b}}=\frac{\dfrac{2a}{b}-\dfrac{-b}{b} }{\dfrac{4b}{b}-\dfrac{-a}{b}}=\boxed{\frac{2\times\dfrac{a}{b}+1}{4+\dfrac{a}{b}}}$

$\displaystyle\\1)~~\frac{a}{b}=\frac{2}{5}\\\\\\\implies\frac{2\times\dfrac{a}{b}+1}{4+\dfrac{a}{b}}=\frac{2\times\dfrac{2}{5}+1}{4+\dfrac{2}{5}}=\frac{\dfrac{4}{5}+\dfrac{5}{5} }{\dfrac{20}{5}+\dfrac{2}{5}} =\frac{\dfrac{9}{5}}{\dfrac{22}{5}}=\boxed{\frac{9}{22}}$

$\displaystyle\\2)~~\frac{a}{b}=1,2=\frac{12}{10}=\frac{6}{5}\\\\\\\implies\frac{2\times\dfrac{a}{b}+1}{4+\dfrac{a}{b}}=\frac{2\times\dfrac{6}{5}+1}{4+\dfrac{6}{5}}=\frac{2\times\dfrac{6}{5}+1}{4+\dfrac{6}{5}}=\frac{\dfrac{12}{5}+\dfrac{5}{5}}{\dfrac{20}{5}+\dfrac{6}{5}} =\frac{\dfrac{17}{5}}{\dfrac{26}{5}}=\boxed{\frac{17}{5}}$

$\displaystyle\\3)~~\frac{a}{b}=2,(2)=\frac{22-2}{9}=\frac{20}{9}\\\\\\\implies\frac{2\times\dfrac{a}{b}+1}{4+\dfrac{a}{b}}=\frac{2\times\dfrac{20}{9}+1}{4+\dfrac{20}{9}}=\frac{\dfrac{40}{9}+\dfrac{9}{9}}{\dfrac{36}{9}+\dfrac{20}{9}} =\frac{\dfrac{49}{9}}{\dfrac{56}{9}}=\frac{49}{56}=\boxed{\frac{7}{8}}$