Câu 2:

a)

Gọi điểm `M(x;y)`

Dễ thấy `a=5,b=3.c=4`

`M in (E)=>MF_1=a-c/ax=5+4/5x,MF_2=a-c/ax=5-4/5x,F_1F_2=2c=8`

`M` nhìn `F_1,F_2` dưới 1 góc `60^o=>\hat{F_1MF_2}=60^o`

Xét `ΔMF_1F_2:``MF_1^2+MF_2^2-2*MF_1*MF_2*cos\hat{F_1MF_2}=F_1F_2^2`

`=>(5+4/5x)^2+(5-4/5x)^2-2(5+4/5x)(5-4/5x)*cos60^o=8^2`

`<=> x=+-(5\sqrt{13})/4`

`=>y=+-(3\sqrt{3})/4`

Vậy `M((5\sqrt{13})/4;(3\sqrt{3})/4),M((5\sqrt{13})/4;-(3\sqrt{3})/4),M(-(5\sqrt{13})/4;(3\sqrt{3})/4),M(-(5\sqrt{13})/4;-(3\sqrt{3})/4)`

b)

`S_(gạch)=400-4*(x(20-x))/2=2x^2-40x+400(cm^2)`

Để `S_(gạch)<=218(cm^2)`

`=>2x^2-40x+400(cm^2)<=218`

`<=>f(x)=x^2-20x+91<=0`

Ta có `Δ'=9>0⇒x_1=10+\sqrt{9}=13,x_2=10-\sqrt{9}=7`

Ta thấy `a=1>0⇒`Để `f(x)<=0` thì `x in [7;13]`

Vậy `x in [7;13]`