cách giải là gì vậy mấy bạn

\(x^2+3x+1\)

=\(\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{5}{4}\)

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

Ta có:\(\left(x+\dfrac{3}{2}\right)^2\ge0\) Với mọi x

 =>\(\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\)

Dấu "=" xảy ra <=>\(\left(x+\dfrac{3}{2}\right)^2=0\)

                        <=>\(x+\dfrac{3}{2}=0\)

                        <=>\(x=\dfrac{-3}{2}\)

min =1 

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

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

Dấu '=' xảy ra khi \(\left(x+1\right)^2=4\)

=>x+1=2 hoặc x+1=-2

=>x=1 hoặc x=-3

\(x>0\)

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

-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 \(C_{min}=\dfrac{9}{8}\)

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

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

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

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

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

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=\(-\frac{1}{5}\)

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

Có: \(\left(2x+1\right)^2+\left(x-1\right)^2\ge0\)

Dấu = xảy ra khi: \(\hept{\begin{cases}2x+1=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}}\) ( k hợp lý => loại )

Ta xét: \(2x+1=0\Rightarrow A=\frac{1}{4}\)

\(x-1=0\Rightarrow A=16\)

Vì: \(\frac{1}{4}< 16\Rightarrow x=-\frac{1}{2}\)

Vậy: \(Min_A=\frac{1}{4}\) tại \(x=-\frac{1}{2}\)

\(y-x=1\Rightarrow x=y-1\)

\(\Rightarrow x^2+y^2=\left(y-1\right)^2+y^2\)

\(=y^2-2y+1+y^2\)

\(=2y^2-2y+1\)

\(=2\left(y^2-y+\frac{1}{2}\right)\)

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

\(=2\left(y-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\forall y\)

Dấu"=" xảy ra khi \(2\left(y-\frac{1}{2}\right)^2=0\Rightarrow y=\frac{1}{2}\)

Vì \(y-x=1\)nên

\(\Rightarrow\frac{1}{2}-x=1\Rightarrow x=-\frac{1}{2}\)

Vậy \(Min_A=\frac{1}{2}\Leftrightarrow x=-\frac{1}{2};y=\frac{1}{2}\)