\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+7x+6x+42}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+7\right)+6\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{\left(x+6\right)+\left(x+4\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)\(\Leftrightarrow\frac{2x+10}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{2\left(x+5\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{2\left(x+7\right)+\left(x+4\right)}{\left(x+4\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3x+18}{\left(x+4\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3\left(x+6\right)}{\left(x+4\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=54\)

\(\Leftrightarrow x^2+11x+28=54\)

\(\Leftrightarrow x^2+11x+30.25=56.25\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+5.5\right)^2=\left(7.5\right)^2\\\left(x+5.5\right)^2=\left(-7.5\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+5.5=7.5\\x+5.5=-7.5\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=7.5-5.5=2\\x=-7.5-5.5=-13\end{cases}}\)

\(\text{Vậy x }\in\left\{2;-13\right\}\)

\(\text{nhớ tích cho mk nha Thanh Do}\)

\(x-x^2-2\)

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

Ta có : \(x^2\ge0\forall x\)

\(\Rightarrow x^2+2\ge0\)

mà \(x\le x^2\)

\(\Rightarrow x< x^2+2\)

\(\Rightarrow x-\left(x^2+2\right)< 0\)

\(\Rightarrow x-x^2-2< 0\)

Ta có: \(x-x^2-2=x-x^2-\frac{1}{4}-\frac{7}{4}\)

\(=\left(x-x^2-\frac{1}{4}\right)-\frac{7}{4}=-\left(x^2-x+\frac{1}{4}\right)-\frac{7}{4}\)

\(=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)-\frac{7}{4}=-\left(x-\frac{1}{2}\right)^2-\frac{7}{4}\)

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow-\left(x-\frac{1}{2}\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-\frac{1}{2}\right)^2-\frac{7}{4}\le-\frac{7}{4}\)\(\forall x\)

\(\Rightarrow x-x^2-2\le-\frac{7}{4}\)\(\forall x\)

hay \(x-x^2-2< 0\)( đpcm ) 

M N P Q E F G H

Bài làm 

Xét tam giác MNQ ta có : 

E là trung điểm MN 

H là trung điểm MQ

=)) EH là đường TB tam giác MNQ

=)) EH // QN và EH = 1/2 QN (1)

Xét tam giác PNQ ta có : 

F là trung điểm MP 

G là trung điểm QP 

=)) FG là đường TB tam giác PNQ

=)) FG // NQ và FE = 1/2 NQ (2)

Từ 1 ; 2 =)) tứ giác EFGH là hình bình hành 

Ta chứng minh bất đẳng thức phụ sau \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)(*)

\(BĐT\)(*) \(< =>\frac{x+y}{xy}\ge\frac{4}{x+y}< =>\left(x+y\right)^2\ge4xy< =>\left(x-y\right)^2\ge0\)*đúng*

Sử dụng bất đẳng thức (*) ta có \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\); \(\frac{1}{b}+\frac{1}{c}\ge\frac{4}{b+c}\); \(\frac{1}{c}+\frac{1}{a}\ge\frac{4}{c+a}\)

Cộng theo vế 3 bất đẳng thức trên ta được \(\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\ge\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\)

\(< =>2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge2\left(\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\right)\)

\(< =>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\)(Dấu "=" xảy ra khi và chỉ khi \(a=b=c\))

Done!

Áp dụng bất đẳng thức \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ta có : ( bạn tự chứng minh )

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)*1*

\(\frac{1}{b}+\frac{1}{c}\ge\frac{4}{b+c}\)*2*

\(\frac{1}{c}+\frac{1}{a}\ge\frac{4}{c+a}\)*3*

Cộng *1* , *2* , *3* theo vế ta có :

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\ge\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\)

<=> \(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

<=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

Vậy ta có điều phải chứng minh

Đẳng thức xảy ra <=> a = b = c > 0

\(P=\left(x^2+4x+1\right)^2-12\left(x+2\right)^2+2093\)

\(=\left(x^2+4x+4-3\right)^2-12\left(x+2\right)^2+2093\)

\(=\left[\left(x+2\right)^2-3\right]^2-12\left(x+2\right)^2+2093\)

\(=\left(x+2\right)^4-6\left(x+2\right)^2+9-12\left(x+2\right)^2+2093\)

\(=\left(x+2\right)^4-18\left(x+2\right)^2+2102\)

\(=\left(x+2\right)^4-18\left(x+2\right)^2+81+2021\)

\(=\left[\left(x+2\right)^4-18\left(x+2\right)^2+81\right]+2021\)

\(=\left[\left(x+2\right)^2-9\right]^2+2021\)

\(=\left[\left(x+2-3\right)\left(x+2+3\right)\right]^2+2021\)

\(=\left[\left(x-1\right)\left(x+5\right)\right]^2+2021\)

Vì \(\left[\left(x-1\right)\left(x+5\right)\right]^2\ge0\forall x\)

\(\Rightarrow\left[\left(x-1\right)\left(x+5\right)\right]^2+2021\ge2021\)\(\forall x\)

hay \(P\ge2021\)

Dấu " = " xảy ra \(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

Vậy \(minP=2021\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

Sai đề nha.

\(0^n+0^n=0^n\)

n={1;2;3}