\(\left(2x+1\right)^2+\left(x-1\right)^2\)

\(=4x^2+4x+1+x^2-2x+1\)

\(=5x^2+2x+2\)

\(=\left(\sqrt{5}.x\right)^2+2.\sqrt{5}.x.\frac{\sqrt{5}}{5}+\left(\frac{\sqrt{5}}{5}\right)^2+\frac{9}{5}\)

\(=\left(\sqrt{5}x+\frac{\sqrt{5}}{5}\right)^2+\frac{9}{5}\)

Ta có

\(\left(\sqrt{5}.x+\frac{\sqrt{5}}{5}\right)^2\ge0\)

\(\Rightarrow\left(\sqrt{5}.x+\frac{\sqrt{5}}{5}\right)^2+\frac{9}{5}\ge\frac{9}{5}\)

Dấu " = " xảy ra khi \(\sqrt{5}.x+\frac{\sqrt{5}}{5}=0\Leftrightarrow x=-\frac{1}{5}\)

Vậy biểu thức đạt giá trị nhỏ nhất là \(\frac{9}{5}\) khi x=\(-\frac{1}{5}\)

\(x>0\)

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

-Ta đặt \(A=T=4x^2+1;B=4x\) thì ta có: 

\(A\ge B\Rightarrow A+T\ge B+T\) (do \(T>0\))\(\Rightarrow\dfrac{A+T}{B+T}\ge1\)

-Do đó: \(C=\dfrac{4x^2+1}{4x}+\dfrac{x}{\left(2x+1\right)^2}\ge\text{​​​​}\dfrac{4x^2+1+4x^2+1}{4x+4x^2+1}+\dfrac{x}{\left(2x+1\right)^2}=\dfrac{2\left(4x^2+1\right)}{\left(2x+1\right)^2}+\dfrac{8x}{\left(2x+1\right)^2}-\dfrac{7x}{\left(2x+1\right)^2}=\dfrac{2\left(2x+1\right)^2}{\left(2x+1\right)^2}-\dfrac{7x}{\left(2x+1\right)^2}=2-\dfrac{7x}{\left(2x+1\right)^2}\)

-Áp dụng BĐT AM-GM ta có:

\(C\ge2-\dfrac{7x}{\left(2x+1\right)^2}\ge2-\dfrac{7x}{4.2x}=2-\dfrac{7}{8}=\dfrac{9}{8}\)

\(C=\dfrac{9}{8}\Leftrightarrow x=\dfrac{1}{2}\)

-Vậy \(C_{min}=\dfrac{9}{8}\)

a) Vì \(\left(2x+\frac{1}{4}\right)^4\ge0\forall x\)

\(\Rightarrow A\ge1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow2x+\frac{1}{4}=0\Leftrightarrow x=\frac{-1}{8}\)

b) \(B=-\left(\frac{4}{9}x-\frac{2}{15}\right)^6+3\)

\(B=3-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\)

Vì \(\left(\frac{4}{9}x-\frac{2}{15}\right)^6\ge0\forall x\)

\(\Rightarrow B\le3\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{4}{9}x-\frac{2}{15}=0\Leftrightarrow x=\frac{3}{10}\)

với mọi x thì (2x+1/4)⁴>=0 (lớn  hơn hoặc bằng )

A=(2x+1/4)⁴-1>=-1

để A đạt GTNN thì (2x+1/4)⁴=0

2x+1/4=0 =>x=-1/8

\(M=x^2+2y^2+2xy-2x-3y+1\)

=> \(M=x^2+2x\left(y-1\right)+\left(y-1\right)^2-\left(y-1\right)^2+2y^2-3y+1\)

=> \(M=\left(x+y-1\right)^2-y^2+2y-1+2y^2-3y+1\)

=> \(M=\left(x+y-1\right)^2+y^2-y\)

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

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

Có \(\left(x+y-1\right)^2\ge0\)với mọi x, y

\(\left(y-\frac{1}{2}\right)^2\ge0\)với mọi y

=> \(M=\left(x+y-1\right)^2+\left(y-\frac{1}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\)với mọi x, y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y-1=0\\y-\frac{1}{2}=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{2}\end{cases}}\)

KL: M_min = \(\frac{-1}{4}\)<=> \(x=y=\frac{1}{2}\)

cảm ơn Giang