Question

1) Tìm min A= \(\dfrac{3}{2+\sqrt{2x-x^2+7}}\)

2)Tìm max B =\(x+\sqrt{2\left(1-x\right)}\)

Giúp em với ạ, giải chi tiết cho em dễ hiểu được khog ạ 

Answer 1

a.

\(2x-x^2+7=-\left(x^2-2x+1\right)+8=-\left(x-1\right)^2+8\le8\)

\(\Rightarrow2+\sqrt{2x-x^2+7}\le2+\sqrt{8}=2+2\sqrt{2}\)

\(\Rightarrow\dfrac{3}{2+\sqrt{2x-x^2+7}}\ge\dfrac{3}{2+2\sqrt{2}}=\dfrac{3\sqrt{2}-3}{2}\)

\(A_{min}=\dfrac{3\sqrt{2}-3}{2}\) khi \(x=1\)

b. ĐKXĐ: \(x\le1\)

\(B=-\left(1-x-\sqrt{2\left(1-x\right)}+\dfrac{1}{2}-\dfrac{1}{2}-1\right)\)

\(B=-\left(1-x-\sqrt{2\left(1-x\right)}+\dfrac{1}{2}\right)+\dfrac{3}{2}\)

\(B=-\left(\sqrt{1-x}-\dfrac{\sqrt{2}}{2}\right)^2+\dfrac{3}{2}\le\dfrac{3}{2}\)

\(B_{max}=\dfrac{3}{2}\) khi\(x=\dfrac{1}{2}\)