Question-159527

Question Number 159527 by HongKing last updated on 18/Nov/21

Commented by mr W last updated on 18/Nov/21

$${only}\:{x}=\frac{\mathrm{23}}{\mathrm{24}} \\ $$

Commented by HongKing last updated on 18/Nov/21

yes my dear Ser, but solution if possible please

$$\mathrm{yes}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{but}\:\mathrm{solution}\:\mathrm{if}\:\mathrm{possible}\:\mathrm{please} \\ $$

Answered by mr W last updated on 18/Nov/21

first of all: −1≤x≤1 ((√(1−x))+(√(2−x))+(√(3−x)))^2 =6−3x+2(...LHS...) ((√(1−x))+(√(2−x))+(√(3−x)))^2 =6−3x+2(1+x) ((√(1−x))+(√(2−x))+(√(3−x)))^2 =8−x (√(1−x))+(√(2−x))+(√(3−x))=(√(8−x)) (√(2−x))+(√(3−x))=(√(8−x))−(√(1−x)) 2−x+3−x+2(√((2−x)(3−x)))=8−x+1−x−2(√((1−x)(8−x))) (√((2−x)(3−x)))=2−(√((1−x)(8−x))) (2−x)(3−x)=4+(1−x)(8−x)−4(√((1−x)(8−x))) 3−2x=2(√((1−x)(8−x))) (3−2x)^2 =4(1−x)(8−x) 24x=23 ⇒x=((23)/(24)) ✓

$${first}\:{of}\:{all}:\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\ $$$$\left(\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{3}−{x}}\right)^{\mathrm{2}} =\mathrm{6}−\mathrm{3}{x}+\mathrm{2}\left(…{LHS}…\right) \\ $$$$\left(\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{3}−{x}}\right)^{\mathrm{2}} =\mathrm{6}−\mathrm{3}{x}+\mathrm{2}\left(\mathrm{1}+{x}\right) \\ $$$$\left(\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{3}−{x}}\right)^{\mathrm{2}} =\mathrm{8}−{x} \\ $$$$\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{3}−{x}}=\sqrt{\mathrm{8}−{x}} \\ $$$$\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{3}−{x}}=\sqrt{\mathrm{8}−{x}}−\sqrt{\mathrm{1}−{x}} \\ $$$$\mathrm{2}−{x}+\mathrm{3}−{x}+\mathrm{2}\sqrt{\left(\mathrm{2}−{x}\right)\left(\mathrm{3}−{x}\right)}=\mathrm{8}−{x}+\mathrm{1}−{x}−\mathrm{2}\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{8}−{x}\right)} \\ $$$$\sqrt{\left(\mathrm{2}−{x}\right)\left(\mathrm{3}−{x}\right)}=\mathrm{2}−\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{8}−{x}\right)} \\ $$$$\left(\mathrm{2}−{x}\right)\left(\mathrm{3}−{x}\right)=\mathrm{4}+\left(\mathrm{1}−{x}\right)\left(\mathrm{8}−{x}\right)−\mathrm{4}\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{8}−{x}\right)} \\ $$$$\mathrm{3}−\mathrm{2}{x}=\mathrm{2}\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{8}−{x}\right)} \\ $$$$\left(\mathrm{3}−\mathrm{2}{x}\right)^{\mathrm{2}} =\mathrm{4}\left(\mathrm{1}−{x}\right)\left(\mathrm{8}−{x}\right) \\ $$$$\mathrm{24}{x}=\mathrm{23} \\ $$$$\Rightarrow{x}=\frac{\mathrm{23}}{\mathrm{24}}\:\checkmark \\ $$

Commented by HongKing last updated on 18/Nov/21

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{cool} \\ $$

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