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Question Number 138984 by bramlexs22 last updated on 20/Apr/21

((36)/( (√(x−2)))) +(4/( (√(y−1)))) +4(√(x−2)) +(√(y−1)) = 28  find x & y

$$\frac{\mathrm{36}}{\:\sqrt{\mathrm{x}−\mathrm{2}}}\:+\frac{\mathrm{4}}{\:\sqrt{\mathrm{y}−\mathrm{1}}}\:+\mathrm{4}\sqrt{\mathrm{x}−\mathrm{2}}\:+\sqrt{\mathrm{y}−\mathrm{1}}\:=\:\mathrm{28} \\ $$ $$\mathrm{find}\:\mathrm{x}\:\&\:\mathrm{y} \\ $$

Answered by mitica last updated on 21/Apr/21

((36)/( (√(x−2))))+4(√(x−2))+(4/( (√(y−1))))+(√(y−1))≥  2(√(((36)/( (√(x−2))))∙4(√(x−2))))+2(√((4/( (√(y−1))))∙(√(y−1))))=24+4=28⇒  ((36)/( (√(x−2))))=4(√(x−2))⇒x=11  (4/( (√(y−1))))=(√(y−1))⇒y=5

$$\frac{\mathrm{36}}{\:\sqrt{{x}−\mathrm{2}}}+\mathrm{4}\sqrt{{x}−\mathrm{2}}+\frac{\mathrm{4}}{\:\sqrt{{y}−\mathrm{1}}}+\sqrt{{y}−\mathrm{1}}\geqslant \\ $$ $$\mathrm{2}\sqrt{\frac{\mathrm{36}}{\:\sqrt{{x}−\mathrm{2}}}\centerdot\mathrm{4}\sqrt{{x}−\mathrm{2}}}+\mathrm{2}\sqrt{\frac{\mathrm{4}}{\:\sqrt{{y}−\mathrm{1}}}\centerdot\sqrt{{y}−\mathrm{1}}}=\mathrm{24}+\mathrm{4}=\mathrm{28}\Rightarrow \\ $$ $$\frac{\mathrm{36}}{\:\sqrt{{x}−\mathrm{2}}}=\mathrm{4}\sqrt{{x}−\mathrm{2}}\Rightarrow{x}=\mathrm{11} \\ $$ $$\frac{\mathrm{4}}{\:\sqrt{{y}−\mathrm{1}}}=\sqrt{{y}−\mathrm{1}}\Rightarrow{y}=\mathrm{5} \\ $$

Commented bybramlexs22 last updated on 21/Apr/21

AM−GM

$$\mathrm{AM}−\mathrm{GM} \\ $$

Answered by mr W last updated on 21/Apr/21

4((9/( (√(x−2))))−6+(√(x−2)))+((4/( (√(y−1))))−4+(√(y−1)))=0  4((3/( ((x−2))^(1/4) ))−((x−2))^(1/4) )^2 +((2/( ((y−1))^(1/4) ))−((y−1))^(1/4) )^2 =0  (3/( ((x−2))^(1/4) ))−((x−2))^(1/4) =0 ⇒(√(x−2))=3 ⇒x=11  (2/( ((y−1))^(1/4) ))−((y−1))^(1/4) =0 ⇒(√(y−1))=2 ⇒y=5

$$\mathrm{4}\left(\frac{\mathrm{9}}{\:\sqrt{{x}−\mathrm{2}}}−\mathrm{6}+\sqrt{{x}−\mathrm{2}}\right)+\left(\frac{\mathrm{4}}{\:\sqrt{{y}−\mathrm{1}}}−\mathrm{4}+\sqrt{{y}−\mathrm{1}}\right)=\mathrm{0} \\ $$ $$\mathrm{4}\left(\frac{\mathrm{3}}{\:\sqrt[{\mathrm{4}}]{{x}−\mathrm{2}}}−\sqrt[{\mathrm{4}}]{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\:\sqrt[{\mathrm{4}}]{{y}−\mathrm{1}}}−\sqrt[{\mathrm{4}}]{{y}−\mathrm{1}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$ $$\frac{\mathrm{3}}{\:\sqrt[{\mathrm{4}}]{{x}−\mathrm{2}}}−\sqrt[{\mathrm{4}}]{{x}−\mathrm{2}}=\mathrm{0}\:\Rightarrow\sqrt{{x}−\mathrm{2}}=\mathrm{3}\:\Rightarrow{x}=\mathrm{11} \\ $$ $$\frac{\mathrm{2}}{\:\sqrt[{\mathrm{4}}]{{y}−\mathrm{1}}}−\sqrt[{\mathrm{4}}]{{y}−\mathrm{1}}=\mathrm{0}\:\Rightarrow\sqrt{{y}−\mathrm{1}}=\mathrm{2}\:\Rightarrow{y}=\mathrm{5} \\ $$

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