Mình cần gấp lắm, giúp mình với !!!!

1) B\(\ge\left|x-2013+2015-x\right|+\left|x-2014\right|\ge2\)

dấu bg xảy ra khi (x-2013)(2015-x)\(\ge\)0 và x-2014=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 <=> \(x=y=\frac{1}{2}\)

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

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

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

Lại có \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}=\frac{4}{1}=4\)(2)

Từ (1) và (2) => \(A=\left(1+\frac{1}{x}\right)^2+\left(1+\frac{1}{y}\right)^2\ge\frac{\left(2+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\ge\frac{\left(2+4\right)^2}{2}=18\)

Đẳng thức xảy ra <=> x = y = 1/2

Vậy MinA = 18 

\(\text{Ta có:}A+2=x^2+3x+\frac{1}{4}+2=x^2+3x+\frac{9}{4}=x^2+2.\frac{3}{2}x+\left(\frac{3}{2}\right)^2=\left(x+\frac{3}{2}\right)^2\ge0\)

\(\Rightarrow A+2\ge0\Rightarrow A\ge-2\)

Dấu "=" xảy ra khi: \(x+\frac{3}{2}=0\Leftrightarrow x=\frac{-3}{2}\)