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\(P=\left(4x^2\right)-3x+\left(\frac{1}{4x}\right)+2015\)
\(=\left(4x^2-4x+1\right)+x+\frac{1}{4x}+2014\)
\(=\left(2x-1\right)^2+\left(x+\frac{1}{4x}\right)+2014\)
Áp dụng bđt Cauchy cho 2 số không âm ;
\(x+\frac{1}{4x}\ge2\sqrt[2]{\frac{1}{4}}=1\)
\(< =>\left(2x-1\right)^2+\left(x+\frac{1}{4x}\right)+2014\ge0+1+2014=2015\)
Vậy \(Min_p=2015\)xảy ra khi \(x=\frac{1}{2}\)
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\(M=\)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=\(\frac{1}{2}\)
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a.
\(A=x^2+\dfrac{2021}{x}=x^2+\dfrac{2021}{2x}+\dfrac{2021}{2x}\ge3\sqrt[3]{\dfrac{2021^2}{4x^2}}=3\sqrt[3]{\dfrac{2021^2}{4}}\)
Dấu "=" xảy ra khi \(x=\sqrt[3]{\dfrac{2021}{3}}\)
b.
\(B=4\left(x-1\right)+\dfrac{25}{x-1}+4\ge2\sqrt{\dfrac{100\left(x-1\right)}{x-1}}+4=24\)
Dấu "=" xảy ra khi \(x=\dfrac{7}{2}\)
c.
\(C=3x+\dfrac{16}{x^3}=x+x+x+\dfrac{16}{x^3}\ge4\sqrt[4]{\dfrac{16x^3}{x^3}}=8\)
\(A_{min}=8\) khi \(x=2\)
d.
\(D=x+\dfrac{1}{x}=\left(\dfrac{x}{4}+\dfrac{1}{x}\right)+\dfrac{3}{4}.x\ge2\sqrt{\dfrac{x}{4x}}+\dfrac{3}{4}.2=\dfrac{5}{2}\)
Dấu "=" xảy ra khi \(x=2\)
e.
\(E=\dfrac{9\left(x-2\right)+18}{2-x}+\dfrac{2}{x}=2\left(\dfrac{1}{x}+\dfrac{9}{2-x}\right)-9\ge\dfrac{2.\left(1+3\right)^2}{x+2-x}-9=7\)
\(E_{min}=7\) khi \(x=\dfrac{1}{5}\)
f.
\(F=\dfrac{3}{1-x}+\dfrac{4}{x}\ge\dfrac{\left(\sqrt{3}+2\right)^2}{1-x+x}=7+4\sqrt{3}\)
Dấu "=" xảy ra khi \(x=4-2\sqrt{3}\)
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a/ \(P=3x+\frac{1}{2x}=\frac{x}{2}+\frac{5x}{2}+\frac{1}{2x}\) \(\ge2\sqrt{\frac{x}{2}.\frac{1}{2x}}+\frac{5.1}{2}=\frac{5}{2}\)
"="\(\Leftrightarrow x=1\)
b/ \(B=\frac{3x}{2}+\frac{1}{x+1}=\frac{3\left(x+1\right)}{2}-\frac{3}{2}+\frac{1}{x+1}\)
\(\ge2\sqrt{\frac{3\left(x+1\right)}{2}.\frac{1}{x+1}}-\frac{3}{2}=\frac{-3+2\sqrt{6}}{2}\)
"="\(\Leftrightarrow3\left(x+1\right)^2=2\Leftrightarrow x=\frac{-3+\sqrt{6}}{3}\)
c/ \(C=\frac{x}{3}+\frac{5}{2x-1}=\frac{2x-1}{6}+\frac{1}{6}+\frac{5}{2x-1}\)
\(\ge2\sqrt{\frac{2x-1}{6}.\frac{5}{2x-1}}+\frac{1}{6}=\frac{1+4\sqrt{15}}{6}\)
"="\(\Leftrightarrow x=\frac{6+\sqrt{30}}{12}\)
d/ \(D=\frac{x^2+4x+4}{x}=x+4+\frac{4}{x}\)\(\ge2\sqrt{x.\frac{4}{x}}+4=8\)
"="\(\Leftrightarrow x=2\)
a/ \(\frac{x}{2}+\frac{1}{2x}+\frac{5}{2}x\ge2\sqrt{\frac{x}{2}.\frac{1}{2x}}+\frac{5}{2}.1=\frac{7}{2}\)
\("="\Leftrightarrow x=1\)
b/ \(B=\frac{3\left(x+1\right)}{2}+\frac{1}{x+1}-\frac{3}{2}\ge2\sqrt{\frac{3\left(x+1\right)}{2\left(x+1\right)}}-\frac{3}{2}=\frac{-3+2\sqrt{6}}{2}\)
\("="\Leftrightarrow\left(x+1\right)^2=\frac{2}{3}\Rightarrow x=\frac{-3+\sqrt{6}}{3}\)
c/ \(C=\frac{2x-1}{6}+\frac{5}{2x-1}+\frac{1}{6}\ge2\sqrt{\frac{\left(2x-1\right).5}{6\left(2x-1\right)}}+\frac{1}{6}=\frac{1+2\sqrt{30}}{6}\)
\("="\Leftrightarrow\left(2x-1\right)^2=30\Rightarrow x=...\)
d/ \(D=x+\frac{4}{x}+4\ge2\sqrt{x.\frac{4}{x}}+4=8\)
\("="\Leftrightarrow x^2=4\Rightarrow x=...\)
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\(M=4x^2-10x+\frac{9}{2x}+2018\)
\(=4x^2-12x+2x+\frac{9}{2x}+2018\)
\(=\left(4x^2-12x+9\right)+\left(2x+\frac{9}{2x}\right)+2009\)
\(=\left[\left(2x\right)^2-2.2x.3+3^2\right]+\left(2x+\frac{9}{2x}\right)+2009\)
\(=\left(2x-3\right)^2+\left(2x+\frac{9}{2x}\right)+2009\)
Ta có : \(2x+\frac{9}{2x}\ge2\sqrt{2x\cdot\frac{9}{2x}}=2.\sqrt{9}=6\)
\(\Rightarrow M\ge\left(2x-3\right)^2+6+2009\ge2015\)
Dấu "=" xảy ra <=> \(x=\frac{3}{2}\)
Vậy GTNN của M là \(2015\) tại \(x=\frac{3}{2}\)
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Lâu rồi không show cách này:)
Sửa đề: \(M=4x^2-3x+\frac{1}{4x}+2017\)
Ta có: \(M=\frac{\left(4x+1\right)\left(2x-1\right)^2}{4x}+2017\ge2017\)
Đẳng thức xảy ra khi \(x=\frac{1}{2}\)
Em kiểm tra lại đề nhé! Hàm số của biểu thức : \(M=4^2-3x+\frac{1}{4x}+2017\) có đồ thị đi xuống nên sẽ không tồn tại GTNN em nhé!
(4x2 -4x+1) + (x+ \(\frac{1}{4x}\)-2)+ 2016=(2x-1)2 +(√x -√ \(\frac{1}{4x}\))2 >=2016 đạt giá trị nhỏ nhất khi x=0,5