Question

Answer 1

Câu 5:ĐKXĐ: \(4+5x-x^2\ge0\)

=>\(x^2-5x-4\le0\)

=>\(x^2-5x+\frac{25}{4}-\frac{41}{4}\le0\)

=>\(\left(x-\frac52\right)^2\le\frac{41}{4}\)

=>\(-\frac{\sqrt{41}}{2}\le x-\frac52\le\frac{\sqrt{41}}{2}\)

=>\(\frac{-\sqrt{41}+5}{2}\le x\le\frac{\sqrt{41}+5}{2}\)

\(\sqrt{4+5x-x^2}-5x=-x^2+2\)

=>\(\sqrt{4+5x-x^2}=-x^2+5x+2\) (1)

Đặt \(\sqrt{4+5x-x^2}=a\) (a>=0)

=>\(a^2=4+5x-x^2\)

=>\(a^2-2=4+5x-x^2-2=2+5x-x^2\)

(1) sẽ trở thành: \(a^2-2=a\)

=>\(a^2-a-2=0\)

=>(a-2)(a+1)=0

=>a=2(nhận) hoặc a=-1(loại)

a=2

=>\(4+5x-x^2=2^2=4\)

=>\(5x-x^2=0\)

=>x(5-x)=0

=>x=0(nhận) hoặc x=5(nhận)