Menu Close

11-If-5-2-236-and-10-3-162-then-the-value-of-15-10-20-40-5-80-is-a-5-398-b-4-398-c-3-398-d-6-398-12-If-x-3-1-3-then-x

Tinku Tara 0 Comments
Pin




Question Number 222943 by Rajakumarselvi last updated on 11/Jul/25
11. If (√5) = 2.236 and (√(10)) = 3.162 then the value of ((15)/( (√(10))+(√(20))+(√(40))−(√5)−(√(80)))) is (a) 5.398 (b) 4.398 (c) 3.398 (d) 6.398 12. If x=(((√3)+1)/3) then x^3 +(1/(x^3 ))=? (a) ((28(√3) +15)/8) (b) ((28(√3)−15)/8) (c) ((27(√3)−35)/4) (d) ((27(√3)+35)/4) 13. Simplify ((x^4 ((x^3_ ((x^2 (√x)))^(1/3) ))^(1/4) ))^(1/5) (a) x^((23)/(24)) (b) x^((23)/6) (c) x^(5/6) (d) x^((119)/(120)) 14.
$$\mathrm{11}.\:\mathrm{If}\:\sqrt{\mathrm{5}}\:=\:\mathrm{2}.\mathrm{236}\:\mathrm{and}\:\sqrt{\mathrm{10}}\:=\:\mathrm{3}.\mathrm{162}\:\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{15}}{\:\sqrt{\mathrm{10}}+\sqrt{\mathrm{20}}+\sqrt{\mathrm{40}}−\sqrt{\mathrm{5}}−\sqrt{\mathrm{80}}}\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{5}.\mathrm{398}\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{4}.\mathrm{398}\:\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{3}.\mathrm{398}\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{6}.\mathrm{398} \\ $$$$\mathrm{12}.\:\mathrm{If}\:\mathrm{x}=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{3}}\:\:\mathrm{then}\:\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} \:}=? \\ $$$$\left(\mathrm{a}\right)\:\frac{\mathrm{28}\sqrt{\mathrm{3}}\:+\mathrm{15}}{\mathrm{8}}\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\frac{\mathrm{28}\sqrt{\mathrm{3}}−\mathrm{15}}{\mathrm{8}}\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\frac{\mathrm{27}\sqrt{\mathrm{3}}−\mathrm{35}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\frac{\mathrm{27}\sqrt{\mathrm{3}}+\mathrm{35}}{\mathrm{4}} \\ $$$$\mathrm{13}.\:\mathrm{Simplify}\:\:\sqrt[{\mathrm{5}}]{\mathrm{x}^{\mathrm{4}} \sqrt[{\mathrm{4}}]{\mathrm{x}^{\mathrm{3}_{\:} } \sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}}}}} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{x}^{\frac{\mathrm{23}}{\mathrm{24}}} \:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{x}^{\frac{\mathrm{23}}{\mathrm{6}}} \:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{x}^{\frac{\mathrm{5}}{\mathrm{6}}} \:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{x}^{\frac{\mathrm{119}}{\mathrm{120}}} \\ $$$$\mathrm{14}.\:\: \\ $$
Answered by Rasheed.Sindhi last updated on 11/Jul/25
11. (√5) =2.236 , (√(10)) =3.162 ((15)/( (√(10)) +(√(20)) +(√(40)) −(√5) −(√(80)) )) =((15)/( (√(10)) +2(√5) +2(√(10 −(√5) −4(√5))))) =((15)/(3(√(10)) −3(√5))) =((15)/(3((√(10)) −(√5) ))) =(5/( (√(10)) −(√5) ))∙(((√(10)) +(√5) )/( (√(10)) +(√5) )) =((5((√(10)) +(√5) ))/(10−5)) =(√(10)) +(√5) =3.162+2.236 =5.398 (a)✓
$$\mathrm{11}. \\ $$$$\sqrt{\mathrm{5}}\:=\mathrm{2}.\mathrm{236}\:,\:\sqrt{\mathrm{10}}\:=\mathrm{3}.\mathrm{162} \\ $$$$\frac{\mathrm{15}}{\:\sqrt{\mathrm{10}}\:+\sqrt{\mathrm{20}}\:+\sqrt{\mathrm{40}}\:−\sqrt{\mathrm{5}}\:−\sqrt{\mathrm{80}}\:} \\ $$$$=\frac{\mathrm{15}}{\:\sqrt{\mathrm{10}}\:+\mathrm{2}\sqrt{\mathrm{5}}\:+\mathrm{2}\sqrt{\mathrm{10}\:−\sqrt{\mathrm{5}}\:−\mathrm{4}\sqrt{\mathrm{5}}}} \\ $$$$=\frac{\mathrm{15}}{\mathrm{3}\sqrt{\mathrm{10}}\:−\mathrm{3}\sqrt{\mathrm{5}}} \\ $$$$=\frac{\mathrm{15}}{\mathrm{3}\left(\sqrt{\mathrm{10}}\:−\sqrt{\mathrm{5}}\:\right)} \\ $$$$=\frac{\mathrm{5}}{\:\sqrt{\mathrm{10}}\:−\sqrt{\mathrm{5}}\:}\centerdot\frac{\sqrt{\mathrm{10}}\:+\sqrt{\mathrm{5}}\:}{\:\sqrt{\mathrm{10}}\:+\sqrt{\mathrm{5}}\:} \\ $$$$=\frac{\mathrm{5}\left(\sqrt{\mathrm{10}}\:+\sqrt{\mathrm{5}}\:\right)}{\mathrm{10}−\mathrm{5}} \\ $$$$=\sqrt{\mathrm{10}}\:+\sqrt{\mathrm{5}}\: \\ $$$$=\mathrm{3}.\mathrm{162}+\mathrm{2}.\mathrm{236} \\ $$$$=\mathrm{5}.\mathrm{398}\: \\ $$$$\left(\mathrm{a}\right)\checkmark \\ $$
Answered by Rasheed.Sindhi last updated on 11/Jul/25
12. x=(((√3) +1)/2) , x^3 +(1/x^3 )=? x+(1/x)=(((√3) +1)/2)+(2/( (√3) +1))∙(((√3) −1)/( (√3) −1)) =(((√3) +1)/2)+((2((√3) −1))/(3−1)) =(((√3) +1+2(√3) −2)/2) =((3(√3) −1)/2) x^3 +(1/x^3 )=(x+(1/x))^3 −3(x+(1/x)) =(((3(√3) −1)/2))^3 −3(((3(√3) −1)/2)) =(((3(√3) −1)/2))((((3(√3) −1)/2))^2 −3) =(((3(√3) −1)/2))(((27−6(√3) +1)/4)−3) =(((3(√3) −1)/2))(((28−6(√3) −12)/4)) =(((3(√3) −1)/2))(((16−6(√3) )/4)) =((48(√3) −54−16+6(√3))/8) =((54(√3) −70)/( 8)) =((27(√3) −35)/4)⇒(c)✓
$$\mathrm{12}. \\ $$$${x}=\frac{\sqrt{\mathrm{3}}\:+\mathrm{1}}{\mathrm{2}}\:,\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=? \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\frac{\sqrt{\mathrm{3}}\:+\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}\:+\mathrm{1}}\centerdot\frac{\sqrt{\mathrm{3}}\:−\mathrm{1}}{\:\sqrt{\mathrm{3}}\:−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}}\:+\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right)}{\mathrm{3}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}}\:+\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{2}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{3}\sqrt{\mathrm{3}}\:−\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}\:−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} −\mathrm{3}\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}\:−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}\:−\mathrm{1}}{\mathrm{2}}\right)\left(\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}\:−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}\:−\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{27}−\mathrm{6}\sqrt{\mathrm{3}}\:\:+\mathrm{1}}{\mathrm{4}}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}\:−\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{28}−\mathrm{6}\sqrt{\mathrm{3}}\:\:−\mathrm{12}}{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}\:−\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{16}−\mathrm{6}\sqrt{\mathrm{3}}\:\:}{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{48}\sqrt{\mathrm{3}}\:−\mathrm{54}−\mathrm{16}+\mathrm{6}\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{54}\sqrt{\mathrm{3}}\:−\mathrm{70}}{\:\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{27}\sqrt{\mathrm{3}}\:−\mathrm{35}}{\mathrm{4}}\Rightarrow\left({c}\right)\checkmark \\ $$
Answered by Rasheed.Sindhi last updated on 11/Jul/25
15. ((3+2(√5))/(4−2(√5)))=p+q(√5) ((3+2(√5))/(4−2(√5)))∙((4+2(√5) )/(4+2(√5) ))=p+q(√5) ((32+14(√5) )/(16−20))=p+q(√5) ((32)/(−4))+((14(√5) )/(−4))=p+q(√5) p=((32)/(−4))=−8 q=((14)/(−4))=−(7/2) ⇒(a)✓
$$\mathrm{15}. \\ $$$$\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{5}}}={p}+{q}\sqrt{\mathrm{5}}\: \\ $$$$\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{5}}}\centerdot\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{5}}\:}{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{5}}\:}={p}+{q}\sqrt{\mathrm{5}}\: \\ $$$$\frac{\mathrm{32}+\mathrm{14}\sqrt{\mathrm{5}}\:}{\mathrm{16}−\mathrm{20}}={p}+{q}\sqrt{\mathrm{5}}\: \\ $$$$\frac{\mathrm{32}}{−\mathrm{4}}+\frac{\mathrm{14}\sqrt{\mathrm{5}}\:}{−\mathrm{4}}={p}+{q}\sqrt{\mathrm{5}} \\ $$$${p}=\frac{\mathrm{32}}{−\mathrm{4}}=−\mathrm{8} \\ $$$${q}=\frac{\mathrm{14}}{−\mathrm{4}}=−\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{a}\right)\checkmark \\ $$
Answered by Rasheed.Sindhi last updated on 12/Jul/25
17. { ((x=3(√5) +2(√2) )),((y=3(√5) −2(√2) )) :} (x^2 −y^2 )^2 =? (x^2 −y^2 )^2 =((x−y)(x+y))^2 x+y=6(√5) x−y=4(√2) ((6(√5) )(4(√2) ))^2 =(24(√(10)) )^2 =6760
$$\mathrm{17}. \\ $$$$\begin{cases}{{x}=\mathrm{3}\sqrt{\mathrm{5}}\:+\mathrm{2}\sqrt{\mathrm{2}}\:}\\{{y}=\mathrm{3}\sqrt{\mathrm{5}}\:−\mathrm{2}\sqrt{\mathrm{2}}\:}\end{cases}\:\:\:\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} =? \\ $$$$\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} =\left(\left({x}−{y}\right)\left({x}+{y}\right)\right)^{\mathrm{2}} \\ $$$${x}+{y}=\mathrm{6}\sqrt{\mathrm{5}}\:\: \\ $$$${x}−{y}=\mathrm{4}\sqrt{\mathrm{2}} \\ $$$$\left(\left(\mathrm{6}\sqrt{\mathrm{5}}\:\right)\left(\mathrm{4}\sqrt{\mathrm{2}}\:\right)\right)^{\mathrm{2}} =\left(\mathrm{24}\sqrt{\mathrm{10}}\:\right)^{\mathrm{2}} =\mathrm{6760} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *