Question

giải phương trình

1. y² - 2y + 3 = \(\dfrac{6}{x^2+2x+4}\)

2. \(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\) biết rằng a + b + c = 3

Answer 1

a)\(VT=y^2-2y+3=\left(y-1\right)^2+2\ge2\)

\(VP=\dfrac{6}{x^2+2x+4}=\dfrac{6}{\left(x+1\right)^2+3}\le\dfrac{6}{3}=2\)

Dấu "=" xảy ra khi: \(y=1;x=-1\)

b) Áp dụng bất đẳng thức AM-GM:

\(\sqrt{x-a}\le\dfrac{x-a+1}{2}\)

\(\sqrt{y-b}\le\dfrac{y-b+1}{2}\)

\(\sqrt{z-c}\le\dfrac{z-c+1}{2}\)

Cộng theo vế:

\(VT\le\dfrac{x-a+1+y-b+1+z-c+1}{2}=\dfrac{x+y+z}{2}=VP\)

Dấu "=" xảy ra khi: \(x=y=z=2\)