Question

Cho \(A=\dfrac{x-\sqrt{x}+2}{\sqrt{x}+3}\) và \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

Cho M = B : A ( với \(x\ge0;x\ne4\)). Tính giá trị x để M có giá trị lớn nhất.

Answer 1

\(M=\dfrac{B}{A}=\dfrac{\dfrac{\sqrt{x}+1}{\sqrt{x}+3}}{\dfrac{x-\sqrt{x}+2}{\sqrt{x}+3}}=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+2}\)\(=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}}\)

Dễ thấy: \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\forall x\)

Và \(\sqrt{x}+1\ge1\forall x\)\(\Rightarrow M=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}}\ge1\)

Xảy ra khi \(x=1\)

Answer 2

Lời giải:

Ta có:

\(B:A=\frac{\sqrt{x}+1}{\sqrt{x}+3}:\frac{x-\sqrt{x}+2}{\sqrt{x}+3}=\frac{\sqrt{x}+1}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{x-\sqrt{x}+2}=\frac{\sqrt{x}+1}{x-\sqrt{x}+2}\)

Đặt \(\sqrt{x}+1=t\Rightarrow \sqrt{x}=t-1\)

Khi đó:

\(M=B:A=\frac{t}{(t-1)^2-(t-1)+2}=\frac{t}{t^2-3t+4}\) \((t\ge 1)\)

\(\Rightarrow M(t^2-3t+4)-t=0\)

\(\Leftrightarrow Mt^2-t(3M+1)+4M=0\)

Nếu \(M=0\rightarrow t=0\) (vô lý vì \(t\geq 1\) ) \(\rightarrow M\neq 0\)

Khi đó: \(\Delta=(3M+1)^2-16M^2\geq 0\)

\(\Leftrightarrow -7M^2+6M+1\geq 0\)

\(\Leftrightarrow -\frac{1}{7}\leq M\le 1\), tức là M đạt max bằng $1$

Khi đó \(t^2-4t+4=0\Leftrightarrow t=2\) \(\Leftrightarrow x=1\) (thỏa mãn)

Vậy \(x=1\)

Answer 3

\(M=\dfrac{B}{A}=\dfrac{\dfrac{\sqrt{x}+1}{\sqrt{x}+3}}{\dfrac{x-\sqrt{x}+2}{\sqrt{x}+3}}=\dfrac{\sqrt{x}+1}{x- \sqrt{x}+2}\)\(=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}}\)

Dễ thấy: \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\forall x\)

Và \(\sqrt{x}+1\ge1\forall x\)\(\Rightarrow M=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7} {4}}\ge1\)

Xảy ra khi \(x=1\)