Ta có :

\(y=\frac{2}{1-x}+\frac{1}{x}\)

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

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

\(\Rightarrow y=\frac{2x}{1-x}+\frac{1-x}{x}+3\)

Vì \(0< x< 1\Rightarrow\hept{\begin{cases}\frac{2x}{1-x}>0\\\frac{1}{x}>0\end{cases}}\)

Áp dụng BĐT Cô si cho 2 số dương , ta có :

\(\Rightarrow y=\frac{2x}{1-x}+\frac{1-x}{x}+3\ge2\sqrt{\frac{2x}{1-x}.\frac{1-x}{x}}+3=2\sqrt{2}+3\)

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

                                                                                                                                              (       vì\(0< x< 1\) )

               Vậy \(Min_y=2\sqrt{2}+3\) khi \(x=\sqrt{2}-1\)

\(y=\frac{2}{1-x}+\frac{1}{x}\ge\frac{\left(\sqrt{2}+1\right)^2}{1-x+x}=3+2\sqrt{2}\)

Dấu = xảy ra khi 

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

\(\Leftrightarrow x=\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\)

1.  x≥1 <=> \(\frac{1}{x}\le1\Leftrightarrow\frac{1}{x}+1\le2\Leftrightarrow A\le2\Rightarrow MaxA=2\Leftrightarrow x=1\)

2. Áp dụng bđt cosi cho x>0. ta có: \(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\Leftrightarrow P\ge2\Rightarrow MinP=2\Leftrightarrow x=\frac{1}{x}\Leftrightarrow x=1\)

3: \(A=\frac{x^2+x+4}{x+1}=\frac{\left(x^2+2x+1\right)-\left(x+1\right)+4}{x+1}=x+1-1+\frac{4}{x+1}\)

áp dụng cosi cho 2 số dương ta có: \(x+1+\frac{4}{x+1}\ge2\sqrt{x+1.\frac{4}{x+1}}=2\Leftrightarrow A+1\ge2\Rightarrow A\ge3\Rightarrow MinA=3\Leftrightarrow x+1=\frac{4}{x+1}\Leftrightarrow x=1\)

x=1 nhe nhap minh di ma ket ban voi minh nhe

1.

\(f\left(x\right)=\dfrac{4}{x}+\dfrac{x-1+1}{1-x}=\dfrac{2^2}{x}+\dfrac{1}{1-x}-1\ge\dfrac{\left(2+1\right)^2}{x+1-x}-1=8\)

\(f\left(x\right)_{min}=8\) khi \(x=\dfrac{2}{3}\)

2.

\(f\left(x\right)=\dfrac{1}{x}+\dfrac{1}{1-x}\ge\dfrac{4}{x+1-x}=4\)

\(f\left(x\right)_{min}=4\) khi \(x=\dfrac{1}{2}\)

x=23x=23

2.

x=12

Bài này dùng Cô si ngược dấu:

Áp dụng BĐT Cô si:\(\frac{1}{x^2+1}=1-\frac{x^2}{x^2+1}\ge1-\frac{x^2}{2x}=1-\frac{x}{2}\)

Tương tự với ba BĐT còn lại và cộng theo vế ta được:\(VT\ge4-\frac{x+y+z+t}{2}=2\)

Dấu "=' xảy ra tại a = b = c = 1

Vậy min A = 2 khi và chỉ khi a = b = c = 1

tth ngược dấu nhé 

\(A=\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}+\frac{1}{t^2+1}\)

\(\Leftrightarrow\)\(-A+4=\left(1-\frac{1}{x^2+1}\right)+\left(1-\frac{1}{y^2+1}\right)+\left(1-\frac{1}{z^2+1}\right)+\left(1-\frac{1}{t^2+1}\right)\)

\(\Leftrightarrow\)\(-A+4\ge1-\frac{x}{2}+1-\frac{y}{2}+1-\frac{z}{2}+1-\frac{t}{2}=4-\frac{x+y+z+t}{2}=2\)

\(\Leftrightarrow\)\(-A+4\ge2\)

\(\Leftrightarrow\)\(A\le2\)

Phùng Minh Quân ông ms ngược dấu á!bài người ta tìm gtnn mừ

1. \(1=x^2+y^2\ge2xy\Rightarrow xy\le\frac{1}{2}\)

 \(A=-2+\frac{2}{1+xy}\ge-2+\frac{2}{1+\frac{1}{2}}=-\frac{2}{3}\)

max A = -2/3 khi x=y=\(\frac{\sqrt{2}}{2}\)

\(\frac{1}{xy}+\frac{1}{xz}=\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{x}.\frac{4}{y+z}=\frac{4}{\left(4-t\right)t}=\frac{4}{4-\left(t-2\right)^2}\ge1\) với t = y+z => x =4 -t

\(A=x^2+y^2=\frac{\left(1^2+1^2\right)\left(x^2+y^2\right)}{2}\ge\frac{\left(1.x+1.y\right)^2}{2}=\frac{1}{2}\)A min = 1 khi x =y = 1/2

\(\sqrt{A}=\sqrt{x^2+y^2}\le\sqrt{x^2}+\sqrt{y^2}=x+y=1\)( \(\sqrt{a+b}\le\sqrt{a}+\sqrt{b}\))

=> A\(\le1\) => Max A = 1 khi x =0;y =1 hoặc x =1 ; y =0

By Titu's Lemma we easy have:

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

\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)

\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{17}{4}\)

Mk xin b2 nha!

\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)

\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)

\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

1) có \(2y\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)

\(\Rightarrow\left(\sqrt{xy}+\frac{1}{4\sqrt{xy}}\right)^2+\frac{15}{16xy}+\frac{1}{2}\ge\frac{15}{16}\cdot4+\frac{1}{2}=\frac{17}{4}\)

Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)