Question-164417 Tinku Tara June 4, 2023 Algebra 0 Comments FacebookTweetPin Question Number 164417 by HongKing last updated on 16/Jan/22 Answered by Rasheed.Sindhi last updated on 17/Jan/22 y=4x−3x2+y2+6y+9+x2+y2−2x−2y+2=x2+(y+3)2+x2−2x+1+y2−2y+1=x2+(y+3)2+(x−1)2+(y−1)2=x2+(4x−3+3)2+(x−1)2+(4x−3−1)2=x2+(4x)2+(x−1)2+4(x−1)2=∣x∣17+∣x−1∣5=x17+(1−x)5[∵x∈(0,1)]=(17−5)x+5=(17−5)(0,1)+5=(0,17−5)+5=(5,17) Commented by Rasheed.Sindhi last updated on 17/Jan/22 ThankstoguidememrWsir!I′vecorrectednow! Commented by HongKing last updated on 18/Jan/22 thankyoudearSircool Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: let-f-n-x-nx-2-e-x-n-x-0-1-study-the-simple-convergence-of-f-n-x-2-study-the-uniform-convergence-of-f-n-x-Next Next post: find-the-range-f-x-log-4-log-2-log-1-2-x- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![y=4x−3 (√(x^2 +y^2 +6y+9)) +(√(x^2 +y^2 −2x−2y+2)) =(√(x^2 +(y+3)^2 ))+(√(x^2 −2x+1+y^2 −2y+1)) =(√(x^2 +(y+3)^2 ))+(√((x−1)^2 +(y−1)^2 )) =(√(x^2 +(4x−3+3)^2 ))+(√((x−1)^2 +(4x−3−1)^2 )) =(√(x^2 +(4x)^2 ))+(√((x−1)^2 +4(x−1)^2 )) =∣x∣(√(17)) +∣x−1∣(√5) =x(√(17)) +(1−x)(√5) [∵ x∈(0,1)] =((√(17)) −(√5) )x+(√5) =((√(17)) −(√5) )(0,1)+(√5) =(0,(√(17)) −(√5) )+(√5) =((√5) ,(√(17)) )](https://www.tinkutara.com/question/Q164435.png)
