Câu hỏi của αβγ δεζ ηθι

`ĐK: -7 = 0)`` 1-x+x+7+2\sqrt{(1-x)(x+7)}=t^2`` 8+2\sqrt{-x^2-6x+7}=t^2`` 2\sqrt{-x^2-6x+7}=t^2-8`Khi đó ptr có dạng: `t+t^2-8=8`` t^2+t-16=0`Ptr có:`\Delta=1^2-4.(-16)=65 > 0``=>` Ptr có `2` nghiệm pb`t_1=[-b+\sqrt{\Delta}]/[2a]=[-1+\sqrt{65}]/2` (t/m)`t_2=[-b-\sqrt{\Delta}]/[2a]=[-1-\sqrt{65}]/2` (ko t/m) `@t=[-1+\sqrt{65}]/2=>\sqrt{1-x}+\sqrt{x+7}=[-1+\sqrt{65}]/2` ` 1-x+x+7+2\sqrt{-x^2-6x+7}=[33-\sqrt{65}]/2` ` \sqrt{-x^2-6x+7}=[17-\sqrt{65}]/4` ` -x^2-6x+7=[354-34\sqrt{65}]/16` ` x^2+6x+[242-34\sqrt{65}]/16=0`Ptr có:`\Delta'=3^2-[242-34\sqrt{65}]/16=[34\sqrt{65}-98]/16 > 0``=>`Ptr có `2` nghiệm pb`x_1=[-b'+\sqrt{\Delta'}]/a=-3+[\sqrt{34\sqrt{65}-98]/4=[-12+\sqrt{34\sqrt{65}-98}]/4``x_2=[-b'-\sqrt{\Delta'}]/a=-3-[\sqrt{34\sqrt{65}-98]/4=[-12-\sqrt{34\sqrt{65}-98}]/4`Vậy `S={[-12+-\sqrt{34\sqrt{65}-98}]/4}` Số ra khủng khiếp qué :)Câu trả lời: `ĐK: -7 = 0)`` 1-x+x+7+2\sqrt{(1-x)(x+7)}=t^2`` 8+2\sqrt{-x^2-6x+7}=t^2`` 2\sqrt{-x^2-6x+7}=t^2-8`Khi đó ptr có dạng: `t+t^2-8=8`` t^2+t-16=0`Ptr có:`\Delta=1^2-4.(-16)=65 > 0``=>` Ptr có `2` nghiệm pb`t_1=[-b+\sqrt{\Delta}]/[2a]=[-1+\sqrt{65}]/2` (t/m)`t_2=[-b-\sqrt{\Delta}]/[2a]=[-1-\sqrt{65}]/2` (ko t/m) `@t=[-1+\sqrt{65}]/2=>\sqrt{1-x}+\sqrt{x+7}=[-1+\sqrt{65}]/2` ` 1-x+x+7+2\sqrt{-x^2-6x+7}=[33-\sqrt{65}]/2` ` \sqrt{-x^2-6x+7}=[17-\sqrt{65}]/4` ` -x^2-6x+7=[354-34\sqrt{65}]/16` ` x^2+6x+[242-34\sqrt{65}]/16=0`Ptr có:`\Delta'=3^2-[242-34\sqrt{65}]/16=[34\sqrt{65}-98]/16 > 0``=>`Ptr có `2` nghiệm pb`x_1=[-b'+\sqrt{\Delta'}]/a=-3+[\sqrt{34\sqrt{65}-98]/4=[-12+\sqrt{34\sqrt{65}-98}]/4``x_2=[-b'-\sqrt{\Delta'}]/a=-3-[\sqrt{34\sqrt{65}-98]/4=[-12-\sqrt{34\sqrt{65}-98}]/4`Vậy `S={[-12+-\sqrt{34\sqrt{65}-98}]/4}` Số ra khủng khiếp qué :)