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Question Number 139288 by 7770 last updated on 25/Apr/21

IF ((x^3 +1)/(x^2 −1))=x+(√(6/x)) 6x^2 −5x+4=??

$$\boldsymbol{{IF}}\:\frac{\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{1}}=\boldsymbol{{x}}+\sqrt{\frac{\mathrm{6}}{\boldsymbol{{x}}}} \\ $$$$\mathrm{6}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{{x}}+\mathrm{4}=?? \\ $$

Answered by MJS_new last updated on 25/Apr/21

(((x+1)(x^2 −x+1))/((x−1)(x+1)))=x+(√(6/x)) ((x^2 −x+1)/(x−1))=x+(√(6/x)) (1/(x−1))=(√(6/x)) (√(6/x))=t>0 ⇔ x=(6/t^2 )∧t>0 (t^2 /(6−t^2 ))=t ⇒ (t=−3∨t=0∨t=2)∧t>0 ⇒ t=2 ⇒ x=(3/2) ⇒ 6x^2 −5x+4=10

$$\frac{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)}={x}+\sqrt{\frac{\mathrm{6}}{{x}}} \\ $$$$\frac{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{{x}−\mathrm{1}}={x}+\sqrt{\frac{\mathrm{6}}{{x}}} \\ $$$$\frac{\mathrm{1}}{{x}−\mathrm{1}}=\sqrt{\frac{\mathrm{6}}{{x}}} \\ $$$$\sqrt{\frac{\mathrm{6}}{{x}}}={t}>\mathrm{0}\:\Leftrightarrow\:{x}=\frac{\mathrm{6}}{{t}^{\mathrm{2}} }\wedge{t}>\mathrm{0} \\ $$$$\frac{{t}^{\mathrm{2}} }{\mathrm{6}−{t}^{\mathrm{2}} }={t} \\ $$$$\Rightarrow\:\left({t}=−\mathrm{3}\vee{t}=\mathrm{0}\vee{t}=\mathrm{2}\right)\wedge{t}>\mathrm{0}\:\Rightarrow\:{t}=\mathrm{2} \\ $$$$\Rightarrow\:{x}=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\:\mathrm{6}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}=\mathrm{10} \\ $$

Answered by lyubita last updated on 25/Apr/21

10

$$\mathrm{10} \\ $$

Answered by liberty last updated on 25/Apr/21

((x^2 −x+1)/(x−1)) = x+(√(6/x)) ((x^2 −x+1−x^2 +x)/(x−1)) = (√(6/x)) (1/(x−1)) = (√(6/x)) (1/(x^2 −2x+1)) = (6/x) ⇒ 6x^2 −12x+6 = x ⇒6x^2 −13x+6 = 0 (x−(9/6))(x−(4/6))=0 → { ((x=(3/2))),((x=(2/3))) :} ⇔ 6x^2 −5x+4=6((9/4))−((15)/2)+((16)/4) = ((54−30+16)/4) = ((40)/4)=10 or 6x^2 −5x+4=6((4/9))−((10)/3)+4 =(8/3)−((10)/3)+((12)/3)=((10)/3)

$$\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{1}}\:=\:\mathrm{x}+\sqrt{\frac{\mathrm{6}}{\mathrm{x}}} \\ $$$$\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}−\mathrm{x}^{\mathrm{2}} +\mathrm{x}}{\mathrm{x}−\mathrm{1}}\:=\:\sqrt{\frac{\mathrm{6}}{\mathrm{x}}} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}\:=\:\sqrt{\frac{\mathrm{6}}{\mathrm{x}}} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{1}}\:=\:\frac{\mathrm{6}}{\mathrm{x}} \\ $$$$\Rightarrow\:\mathrm{6x}^{\mathrm{2}} −\mathrm{12x}+\mathrm{6}\:=\:\mathrm{x} \\ $$$$\Rightarrow\mathrm{6x}^{\mathrm{2}} −\mathrm{13x}+\mathrm{6}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{x}−\frac{\mathrm{9}}{\mathrm{6}}\right)\left(\mathrm{x}−\frac{\mathrm{4}}{\mathrm{6}}\right)=\mathrm{0}\:\rightarrow\begin{cases}{\mathrm{x}=\frac{\mathrm{3}}{\mathrm{2}}}\\{\mathrm{x}=\frac{\mathrm{2}}{\mathrm{3}}}\end{cases} \\ $$$$\Leftrightarrow\:\mathrm{6x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{4}=\mathrm{6}\left(\frac{\mathrm{9}}{\mathrm{4}}\right)−\frac{\mathrm{15}}{\mathrm{2}}+\frac{\mathrm{16}}{\mathrm{4}} \\ $$$$=\:\frac{\mathrm{54}−\mathrm{30}+\mathrm{16}}{\mathrm{4}}\:=\:\frac{\mathrm{40}}{\mathrm{4}}=\mathrm{10} \\ $$$$\mathrm{or}\:\mathrm{6x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{4}=\mathrm{6}\left(\frac{\mathrm{4}}{\mathrm{9}}\right)−\frac{\mathrm{10}}{\mathrm{3}}+\mathrm{4} \\ $$$$=\frac{\mathrm{8}}{\mathrm{3}}−\frac{\mathrm{10}}{\mathrm{3}}+\frac{\mathrm{12}}{\mathrm{3}}=\frac{\mathrm{10}}{\mathrm{3}} \\ $$

Commented by lyubita last updated on 25/Apr/21

x = (2/3) tidak memenuhi

$${x}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:{tidak}\:{memenuhi} \\ $$

Answered by qaz last updated on 25/Apr/21

((x^3 +1)/(x^2 −1))=((x^2 −x+1)/(x−1))=x+(1/(x−1))=x+(√(6/x)) ≪1≫:(1/(x−1))=(√(6/x)) ⇒6x^2 −13x+6=0 x=(3/2) or (2/3) 6x^2 −5x+4=(13x−6)−5x+4 =8x−2 =8×(3/2)−2 or 8×(2/3)−2 =10 or ((10)/3) ≪2≫:x=(√(6/x)) ∧ x=(1/(x−1)) ⇒x^3 =6 ∧ x^2 −x=1 ⇎x=(6)^(1/3) ∧ x=((1±(√5))/2) ............... Impossible ≪1≫&≪2≫ 6x^2 −5x+4=10 or ((10)/3)

$$\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{{x}−\mathrm{1}}={x}+\frac{\mathrm{1}}{{x}−\mathrm{1}}={x}+\sqrt{\frac{\mathrm{6}}{{x}}} \\ $$$$\ll\mathrm{1}\gg:\frac{\mathrm{1}}{{x}−\mathrm{1}}=\sqrt{\frac{\mathrm{6}}{{x}}} \\ $$$$\Rightarrow\mathrm{6}{x}^{\mathrm{2}} −\mathrm{13}{x}+\mathrm{6}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}}\:{or}\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{6}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}=\left(\mathrm{13}{x}−\mathrm{6}\right)−\mathrm{5}{x}+\mathrm{4} \\ $$$$=\mathrm{8}{x}−\mathrm{2} \\ $$$$=\mathrm{8}×\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{2}\:{or}\:\mathrm{8}×\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{2} \\ $$$$=\mathrm{10}\:{or}\:\frac{\mathrm{10}}{\mathrm{3}} \\ $$$$\ll\mathrm{2}\gg:{x}=\sqrt{\frac{\mathrm{6}}{{x}}}\:\:\wedge\:{x}=\frac{\mathrm{1}}{{x}−\mathrm{1}} \\ $$$$\Rightarrow{x}^{\mathrm{3}} =\mathrm{6}\:\wedge\:{x}^{\mathrm{2}} −{x}=\mathrm{1} \\ $$$$\nLeftrightarrow{x}=\sqrt[{\mathrm{3}}]{\mathrm{6}}\:\:\:\:\wedge\:\:{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\:...............\:{Impossible} \\ $$$$\ll\mathrm{1}\gg\&\ll\mathrm{2}\gg \\ $$$$\mathrm{6}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}=\mathrm{10}\:{or}\:\frac{\mathrm{10}}{\mathrm{3}} \\ $$

Commented by lyubita last updated on 25/Apr/21

x = (2/3) is not solution Substitute x = (2/3) to the equation. LHS ≠ LRS

$${x}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:{is}\:{not}\:{solution} \\ $$$${Substitute}\:{x}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:{to}\:{the}\:{equation}.\: \\ $$$${LHS}\:\neq\:{LRS} \\ $$

Commented by qaz last updated on 26/Apr/21

yes im wrong,x≠(2/3),because (1/(x−1))<0 if x=(2/3).

$${yes}\:{im}\:{wrong},{x}\neq\frac{\mathrm{2}}{\mathrm{3}},{because}\:\frac{\mathrm{1}}{{x}−\mathrm{1}}<\mathrm{0}\:{if}\:{x}=\frac{\mathrm{2}}{\mathrm{3}}. \\ $$

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