GPT X+x/(can(x^2-1)) >35/12?

Lớp 9 hả?

ĐK: x^2 -1 >0 <=> x^2 >1 <=> x>1 và x<-1

x + x/căn(x^2-1) >35/12

<=>[ x.căn(x^2-1) +x ]/ căn(x^2-1) > 35/12

=> 12x.căn(x^2-1) +12x > 35.căn(x^2-1)

<=> -23.căn(x^2-1) +12x >0

<=> -23.căn(x^2-1) > -12x

<=> 23.căn(x^2-1) < 12x

<=> 23^2 .(x^2-1) < (12x)^2

<=> 529x^2 -529 < 144x^2

<=> 385x^2 < 529

<=> x^2 < 529/385

<=> -529/385 < x < 529/385

Kết hợp với ĐK

=> 1< x < 529/385

100% rồi đấy bạn.

eo minh lam dai qua xoa di thoi hi xin loi

ban Van Thuan lam nham o cho tru 12xcan(...) voi 35can(...) ko ah coi ky lai di

toi giai tiep theo VT nhe

12x.căn(x^2-1) +12x > 35.căn(x^2-1)

12x(1+can(x^2-1))>35can(x^2-1) (5)

thay rang bpt 5 chi dung voi x>0 vi x<=0 th VT<=0 con VP >0 vay x>0 hay x>1 (theo ())

12x>(can(x^2-1))(35-12x)

2ve >0 nen binh phuong 2 ve

->144x^2>(x^2-1)(35-12x)^2

0>-144x^2+(x^2-1)(35-12x)^2

0>-144x^2+x^2(35-12x)^2-(35-12x)^2

0>-144x^2+x^2(35-12x)^2-(35^2-

-2.35.12x+144x^2)

0>x^2(35-12x)^2-35^2+2.35.12x-2.144x^2

0>x^2(35-12x)^2-35^2+2.12x(35-12x)

0>x^2(35-12x)^2-35^2+2.12x(35-

-12x)+144-144

0>(x(35-12x)-12)^2-35^2-144

0>(x(35-12x)-12)^2-37^2

0>((x(35-12x)-12-37)((x(35-12x)-12+37)

0>((x(35-12x)-49)((x(35-12x)+25)

0>(35x-12x^2-49)(35x-12x^2+25)

tam thuc thu nhat co delta=35^2-4.12.49=1225-2352=-2127<0 nen luon nhan gt am , cho nen can tam thuc thu 2 nhan gt duong la dc

delta=35^2+4.12.25=1225+1200=2425=25.97

x1=(-35-5can97)/-24~3,5>1, x2=(-35+5can97)/-24<~ -0,5<0

nen ngiem la x>x1=(-35-5can97)/-24~3,5>1