\(M=4x^2-3x+\dfrac{1}{4x}+2011\)

\(M=4x^2-4x+1+x+\dfrac{1}{4x}+2011\)

\(M=\left(2x-1\right)^2+\left(x+\dfrac{1}{4x}\right)+2010\)

Vì \(\left(2x-1\right)^2\ge0\) và \(x>0\)

\(\Rightarrow\dfrac{1}{4x}>0\)

Lợi dụng BĐT Cauchy cho 2 số nguyên dương ta có:

\(x+\dfrac{1}{4x}\ge2\sqrt{x\dfrac{1}{4x}}=2.\dfrac{1}{2}=1\)

\(\Rightarrow M=\left(2x-1\right)^2+\left(x+\dfrac{1}{4x}\right)+2010\ge0+1+2010=2011\)

\(\Rightarrow M\ge2011\)

Dấu " = " xảy ra khi:

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\x=\dfrac{1}{4x}\\x>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x^2=\dfrac{1}{4}\\x>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\\x>0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(M_{min}=2011\) đạt được khi \(x=\dfrac{1}{2}\)

x>o nhá

\(M=\)như trên

\(=>M=4x^2-4x+1+x+\frac{1}{4x}+2010\)

\(=>M=\left(4x^2-4x+1\right)+\left(x+\frac{1}{4x}\right)+2010\)

\(=>M=\left(2x-1\right)^2+\left(x+\frac{1}{4x}\right)+2010\)

Áp dụng BĐT Cô- si cho 2 số không âm, ta có: 

\(x+\frac{1}{4x}\ge2\sqrt{x.\frac{1}{4x}}=2\sqrt{\frac{1}{4}}=1\)

\(=>M=\left(2x-1\right)^2+\left(x+\frac{1}{4x}\right)+2010\ge0+1+2010=2011\\ \)

=>minM=2011 khi x=\(\frac{1}{2}\)

1.

\(G=\dfrac{2}{x^2+8}\le\dfrac{2}{8}=\dfrac{1}{4}\)

\(G_{max}=\dfrac{1}{4}\) khi \(x=0\)

\(H=\dfrac{-3}{x^2-5x+1}\) biểu thức này ko có min max

2.

\(D=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{3}{2}\)

\(D_{min}=\dfrac{3}{2}\) khi \(x=4\)

\(E=\dfrac{4x^4-x^2-1}{\left(x^2+1\right)^2}=\dfrac{-\left(x^4+2x^2+1\right)+5x^4+x^2}{\left(x^2+1\right)^2}=-1+\dfrac{5x^4+x^2}{\left(x^2+1\right)^2}\ge-1\)

\(E_{min}=-1\) khi \(x=0\)

\(G=\dfrac{3\left(x^2-4x+5\right)-5}{x^2-4x+5}=3-\dfrac{5}{\left(x-2\right)^2+1}\ge3-\dfrac{5}{1}=-2\)

\(G_{min}=-2\) khi \(x=2\)

\(N=\dfrac{57x^2+38x+95}{19\left(4x^2+4x+1\right)}=\dfrac{14\left(4x^2+4x+1\right)+\left(x^2-18x+81\right)}{19\left(4x^2+4x+1\right)}=\dfrac{14}{19}+\left(\dfrac{x-9}{2x+1}\right)^2\ge\dfrac{14}{19}\)

\(N_{min}=\dfrac{14}{19}\) khi \(x=9\)

Nếu đặt ẩn: \(N=\dfrac{3x^2+2x+5}{\left(2x+1\right)^2}\)

Đặt \(2x+1=t\Leftrightarrow x=\dfrac{t-1}{2}\)

\(\Rightarrow N=\dfrac{3\left(\dfrac{t-1}{2}\right)^2+2\left(\dfrac{t-1}{2}\right)+5}{t^2}=\dfrac{3t^2-2t+19}{4t^2}=\dfrac{19}{4t^2}-\dfrac{1}{2t}+\dfrac{3}{4}\)

\(N=\dfrac{19}{4}\left(\dfrac{1}{t}-\dfrac{1}{19}\right)^2+\dfrac{14}{19}\ge\dfrac{14}{19}\)

Giusp e ạ !

Bài 1:

Ta có: \(M=4x^2-3x+\dfrac{1}{4x}+2011=4x^2-4x+1+x+\dfrac{1}{4x}+2010\)

\(=\left(4x^2-4x+1\right)+\left(x+\dfrac{1}{4x}\right)+2010\)

\(=\left(2x-1\right)^2+\left(x+\dfrac{1}{4x}\right)+2010\)

Áp dụng BĐT Cô- si cho 2 số không âm, ta có:

\(x+\dfrac{1}{4x}\ge2\sqrt{x.\dfrac{1}{4x}}=2\sqrt{\dfrac{1}{4}}=1\)

Suy ra: \(M=\left(2x-1\right)^2+\left(x+\dfrac{1}{4x}\right)+2010\ge0+1+2010=2011\)

Vậy: \(Min_M=2011\Leftrightarrow x=\dfrac{1}{2}\)

Bài 2: Tham khảo: với hai số thực không âm a, b thỏa a2 + b2 = 4, tìm giá trị lớn nhất của biểu thức M= ab /(a+b+2) | Câu hỏi ôn tập thi vào lớp 10

\(M=\left(x^2+\frac{1}{8x}+\frac{1}{8x}\right)+3\left(x^2-x\right)\ge3\sqrt[3]{x^2.\frac{1}{8x}.\frac{1}{8x}}+3\left(x-\frac{1}{2}\right)^2-\frac{3}{4}\)

\(\ge\frac{3}{4}+0-\frac{3}{4}=0\)

Dấu bằng xảy ra khi \(x=\frac{1}{2}.\)

\(KL:Min\text{ }M=0\)