Question Number 96460 by bobhans last updated on 01/Jun/20
Commented by MJS last updated on 01/Jun/20
$$−\mathrm{4} \\ $$$$\mathrm{only}\:\mathrm{step}\:\mathrm{needed}\:\mathrm{several}\:\mathrm{times}\:\mathrm{is} \\ $$$$\frac{{a}}{{b}+\sqrt{{c}}}=\frac{{a}\left({b}−\sqrt{{c}}\right)}{{b}^{\mathrm{2}} −{c}} \\ $$
Commented by bobhans last updated on 01/Jun/20
$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{short}\:\mathrm{cut}? \\ $$
Commented by MJS last updated on 01/Jun/20
$$\frac{−\mathrm{4}+\sqrt{\mathrm{11}}}{\mathrm{1}+\sqrt{−\mathrm{3}+\sqrt{\mathrm{11}}}}=\frac{\left(−\mathrm{4}+\sqrt{\mathrm{11}}\right)\left(\mathrm{1}−\sqrt{−\mathrm{3}+\sqrt{\mathrm{11}}}\right)}{\mathrm{1}−\left(−\mathrm{3}+\sqrt{\mathrm{11}}\right)}= \\ $$$$=\frac{\left(−\mathrm{4}+\sqrt{\mathrm{11}}\right)\left(\mathrm{1}−\sqrt{−\mathrm{3}+\sqrt{\mathrm{11}}}\right)}{\mathrm{4}−\sqrt{\mathrm{11}}}= \\ $$$$=\frac{\left(−\mathrm{4}+\sqrt{\mathrm{11}}\right)\left(\mathrm{1}−\sqrt{−\mathrm{3}+\sqrt{\mathrm{11}}}\right)\left(\mathrm{4}+\sqrt{\mathrm{11}}\right)}{\mathrm{16}−\mathrm{11}}= \\ $$$$=\frac{\left(−\mathrm{16}+\mathrm{11}\right)\left(\mathrm{1}−\sqrt{−\mathrm{3}+\sqrt{\mathrm{11}}}\right)}{−\mathrm{5}}= \\ $$$$=\mathrm{1}−\sqrt{−\mathrm{3}+\sqrt{\mathrm{11}}} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{with}\:\mathrm{the}\:\mathrm{2nd}\:\mathrm{fraction} \\ $$
Commented by john santu last updated on 01/Jun/20
$$\mathrm{waw}…..\mathrm{great}…. \\ $$
Commented by john santu last updated on 01/Jun/20
$$\mathrm{second}\:\mathrm{fraction}\: \\ $$$$\frac{−\mathrm{12}+\sqrt{\mathrm{11}}}{−\mathrm{3}+\sqrt{\sqrt{\mathrm{11}}−\mathrm{3}}}\:×\frac{−\mathrm{3}−\sqrt{\sqrt{\mathrm{11}}−\mathrm{3}}}{−\mathrm{3}−\sqrt{\sqrt{\mathrm{11}}−\mathrm{3}}}\:= \\ $$$$\frac{\left(−\mathrm{12}+\sqrt{\mathrm{11}}\right)\left(−\mathrm{3}−\sqrt{\sqrt{\mathrm{11}}−\mathrm{3}}\right)}{\mathrm{9}−\left(\sqrt{\mathrm{11}}−\mathrm{3}\right)}= \\ $$$$\frac{−\left(\mathrm{12}−\sqrt{\mathrm{11}}\right)\left(−\mathrm{3}−\sqrt{\sqrt{\mathrm{11}}−\mathrm{3}}\right)}{\mathrm{12}−\sqrt{\mathrm{11}}}= \\ $$$$\mathrm{3}+\sqrt{\sqrt{\mathrm{11}}−\mathrm{3}}\: \\ $$
Commented by john santu last updated on 01/Jun/20
$$\mathrm{but}\:\mathrm{wrong}\:\mathrm{sir}\:\mathrm{in}\:\mathrm{part}\: \\ $$$$\frac{\left(−\mathrm{4}+\sqrt{\mathrm{11}}\right)\left(\mathrm{1}−\sqrt{\left.−\mathrm{3}+\sqrt{\mathrm{11}}\right)}\:\right.}{\mathrm{4}−\sqrt{\mathrm{11}}}\:= \\ $$$$\frac{−\left(\mathrm{4}−\sqrt{\mathrm{11}}\right)\left(\mathrm{1}−\sqrt{−\mathrm{3}+\sqrt{\mathrm{11}}}\right)}{\mathrm{4}−\sqrt{\mathrm{11}}}\:= \\ $$$$−\mathrm{1}+\sqrt{−\mathrm{3}+\sqrt{\mathrm{11}}}\: \\ $$
Commented by MJS last updated on 01/Jun/20
$$\mathrm{you}'\mathrm{re}\:\mathrm{right} \\ $$
Commented by john santu last updated on 01/Jun/20
$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by john santu last updated on 01/Jun/20