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Question Number 51592 by Gulay last updated on 28/Dec/18

Commented by Gulay last updated on 28/Dec/18

a=3.7     b=2  sir plz help me

$$\mathrm{a}=\mathrm{3}.\mathrm{7}\:\:\:\:\:\mathrm{b}=\mathrm{2} \\ $$$$\mathrm{sir}\:\mathrm{plz}\:\mathrm{help}\:\mathrm{me} \\ $$

Commented by Abdo msup. last updated on 29/Dec/18

2)  ((a/b) −(b/a)).((ab)/(a+b)) =((a^2 −b^2 )/(ab)) .((ab)/(a+b)) =(((a−b)(a+b))/(a+b))  =a−b =2,7 −2 =0,7  3)((a^3  +b^3  +3ab(a+b))/(a^2  +2ab +b^2 )) =(((a+b)^3 )/((a+b)^2 )) =a+b =2,7+2=4,7  1)((a^3 b^2  −ab^4 )/((a+b)(ab−b^2 ))) =((b(a^3 b−ab^3 ))/(b(a+b)(a−b)))  =((ab(a^2 −b^2 ))/(a^2 −b^2 )) =ab =2,7.2 =5,4

$$\left.\mathrm{2}\right)\:\:\left(\frac{{a}}{{b}}\:−\frac{{b}}{{a}}\right).\frac{{ab}}{{a}+{b}}\:=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{ab}}\:.\frac{{ab}}{{a}+{b}}\:=\frac{\left({a}−{b}\right)\left({a}+{b}\right)}{{a}+\mathrm{b}} \\ $$$$=\mathrm{a}−\mathrm{b}\:=\mathrm{2},\mathrm{7}\:−\mathrm{2}\:=\mathrm{0},\mathrm{7} \\ $$$$\left.\mathrm{3}\right)\frac{{a}^{\mathrm{3}} \:+{b}^{\mathrm{3}} \:+\mathrm{3}{ab}\left({a}+{b}\right)}{{a}^{\mathrm{2}} \:+\mathrm{2}{ab}\:+{b}^{\mathrm{2}} }\:=\frac{\left({a}+{b}\right)^{\mathrm{3}} }{\left({a}+{b}\right)^{\mathrm{2}} }\:={a}+{b}\:=\mathrm{2},\mathrm{7}+\mathrm{2}=\mathrm{4},\mathrm{7} \\ $$$$\left.\mathrm{1}\right)\frac{{a}^{\mathrm{3}} {b}^{\mathrm{2}} \:−{ab}^{\mathrm{4}} }{\left({a}+{b}\right)\left({ab}−{b}^{\mathrm{2}} \right)}\:=\frac{{b}\left({a}^{\mathrm{3}} {b}−{ab}^{\mathrm{3}} \right)}{{b}\left({a}+{b}\right)\left({a}−{b}\right)} \\ $$$$=\frac{{ab}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:={ab}\:=\mathrm{2},\mathrm{7}.\mathrm{2}\:=\mathrm{5},\mathrm{4} \\ $$$$ \\ $$

Commented by Abdo msup. last updated on 29/Dec/18

forgive change in my answer 2,7 by 3,7...

$${forgive}\:{change}\:{in}\:{my}\:{answer}\:\mathrm{2},\mathrm{7}\:{by}\:\mathrm{3},\mathrm{7}... \\ $$

Commented by Gulay last updated on 30/Dec/18

thank you so much sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$

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