Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 185599 by mathlove last updated on 24/Jan/23

x!=x^3 −x          faind x=?

$${x}!={x}^{\mathrm{3}} −{x}\:\:\:\:\:\:\:\:\:\:{faind}\:{x}=? \\ $$

Commented by liuxinnan last updated on 24/Jan/23

5!=120  5^3 −5=120  x<5    x!<x^3 −x  x>5     x!>x^3 −x

$$\mathrm{5}!=\mathrm{120} \\ $$$$\mathrm{5}^{\mathrm{3}} −\mathrm{5}=\mathrm{120} \\ $$$${x}<\mathrm{5}\:\:\:\:{x}!<{x}^{\mathrm{3}} −{x} \\ $$$${x}>\mathrm{5}\:\:\:\:\:{x}!>{x}^{\mathrm{3}} −{x} \\ $$

Commented by Rasheed.Sindhi last updated on 24/Jan/23

please write your solution as answer  not as comment.

$${please}\:{write}\:{your}\:{solution}\:{as}\:{answer} \\ $$$${not}\:{as}\:{comment}. \\ $$

Commented by liuxinnan last updated on 24/Jan/23

next must

$${next}\:{must} \\ $$

Answered by Frix last updated on 24/Jan/23

There′s no formula, you must try  1!=1     1^3 −1=0  2!=2     2^3 −2=6=3!  3!=6     3^3 −3=24=4!  ...

$$\mathrm{There}'\mathrm{s}\:\mathrm{no}\:\mathrm{formula},\:\mathrm{you}\:\mathrm{must}\:\mathrm{try} \\ $$$$\mathrm{1}!=\mathrm{1}\:\:\:\:\:\mathrm{1}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2}!=\mathrm{2}\:\:\:\:\:\mathrm{2}^{\mathrm{3}} −\mathrm{2}=\mathrm{6}=\mathrm{3}! \\ $$$$\mathrm{3}!=\mathrm{6}\:\:\:\:\:\mathrm{3}^{\mathrm{3}} −\mathrm{3}=\mathrm{24}=\mathrm{4}! \\ $$$$... \\ $$

Answered by ajfour last updated on 24/Jan/23

(x−1)!+1=x^2   (x−1)!+1=(x−1)^2 +2(x−1)+1  (x−2)!=x−2+3  (x−3)!=1+(3/(x−2))  x>3  and also  (x−3)!<4  ⇒ x−3<3  x<6  so let us try   x=4  and x=5  I guess x=5 satisfies  5!=120  5^3 −5=120  hence x=5

$$\left({x}−\mathrm{1}\right)!+\mathrm{1}={x}^{\mathrm{2}} \\ $$$$\left({x}−\mathrm{1}\right)!+\mathrm{1}=\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\left({x}−\mathrm{1}\right)+\mathrm{1} \\ $$$$\left({x}−\mathrm{2}\right)!={x}−\mathrm{2}+\mathrm{3} \\ $$$$\left({x}−\mathrm{3}\right)!=\mathrm{1}+\frac{\mathrm{3}}{{x}−\mathrm{2}} \\ $$$${x}>\mathrm{3}\:\:{and}\:{also} \\ $$$$\left({x}−\mathrm{3}\right)!<\mathrm{4} \\ $$$$\Rightarrow\:{x}−\mathrm{3}<\mathrm{3} \\ $$$${x}<\mathrm{6} \\ $$$${so}\:{let}\:{us}\:{try}\:\:\:{x}=\mathrm{4}\:\:{and}\:{x}=\mathrm{5} \\ $$$${I}\:{guess}\:{x}=\mathrm{5}\:{satisfies} \\ $$$$\mathrm{5}!=\mathrm{120} \\ $$$$\mathrm{5}^{\mathrm{3}} −\mathrm{5}=\mathrm{120} \\ $$$${hence}\:{x}=\mathrm{5} \\ $$

Commented by mathlove last updated on 24/Jan/23

thanks

$${thanks} \\ $$

Answered by Rasheed.Sindhi last updated on 24/Jan/23

(x−2)!(x−1)x=x(x−1)(x+1)   (x−2)!=x+1  (x−3)!=((x+1)/(x−2))−1+1 ; x>2           =(3/(x−2))+1  x−2 ∣ 3  x−2=1,3  x=3,5  3 fails,5 is successful  x=5^✓ : (5−3)!=(3/(5−2))+1                      2=2

$$\left({x}−\mathrm{2}\right)!\left({x}−\mathrm{1}\right){x}={x}\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right) \\ $$$$\:\left({x}−\mathrm{2}\right)!={x}+\mathrm{1} \\ $$$$\left({x}−\mathrm{3}\right)!=\frac{{x}+\mathrm{1}}{{x}−\mathrm{2}}−\mathrm{1}+\mathrm{1}\:;\:{x}>\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{{x}−\mathrm{2}}+\mathrm{1} \\ $$$${x}−\mathrm{2}\:\mid\:\mathrm{3} \\ $$$${x}−\mathrm{2}=\mathrm{1},\mathrm{3} \\ $$$${x}=\mathrm{3},\mathrm{5} \\ $$$$\mathrm{3}\:{fails},\mathrm{5}\:{is}\:{successful} \\ $$$${x}=\mathrm{5}^{\checkmark} :\:\left(\mathrm{5}−\mathrm{3}\right)!=\frac{\mathrm{3}}{\mathrm{5}−\mathrm{2}}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}=\mathrm{2} \\ $$

Answered by SEKRET last updated on 24/Jan/23

 x∙(x−1)∙(x−2)!=x∙(x−1)(x+1)     (x−2)!=x−2+3     a!=a+3    a>0     1=4      2=5      6=6     a=3   24=7  >>    x−2=3    x=5      ✅

$$\:\boldsymbol{\mathrm{x}}\centerdot\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)\centerdot\left(\boldsymbol{\mathrm{x}}−\mathrm{2}\right)!=\boldsymbol{\mathrm{x}}\centerdot\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right) \\ $$$$\:\:\:\left(\boldsymbol{\mathrm{x}}−\mathrm{2}\right)!=\boldsymbol{\mathrm{x}}−\mathrm{2}+\mathrm{3} \\ $$$$\:\:\:\boldsymbol{\mathrm{a}}!=\boldsymbol{\mathrm{a}}+\mathrm{3}\:\:\:\:\boldsymbol{\mathrm{a}}>\mathrm{0} \\ $$$$\:\:\:\mathrm{1}=\mathrm{4} \\ $$$$\:\:\:\:\mathrm{2}=\mathrm{5} \\ $$$$\:\:\:\:\mathrm{6}=\mathrm{6}\:\:\:\:\:\boldsymbol{\mathrm{a}}=\mathrm{3} \\ $$$$\:\mathrm{24}=\mathrm{7}\:\:>> \\ $$$$\:\:\boldsymbol{\mathrm{x}}−\mathrm{2}=\mathrm{3} \\ $$$$\:\:\boldsymbol{\mathrm{x}}=\mathrm{5} \\ $$$$ \\ $$$$ \\ $$✅

Terms of Service

Privacy Policy

Contact: info@tinkutara.com