Ta có : 

\(a+b=1\Rightarrow\left(a+b\right)^2=1\)

Mà \(\left(a+b\right)^2=a^2+2ab+b^2\)

\(\Rightarrow a^2+2ab+b^2=1\left(1\right)\)

Ta lại có :

\(\left(a-b\right)^2\ge0\Leftrightarrow a^2-2ab+b^2\ge0\left(2\right)\)

Ta đi cộng vế ( 1 ) và vế ( 2 ) , ta được :

\(2\left(a^2+b^2\right)\ge1\Leftrightarrow a^2+b^2\ge\frac{1}{2}\)

Ta có :

\((a^2+b^2)^2=a^4+2a^2b^2+b^4=\frac{1}{4}\left(3\right)\)

 \(\left(a^2-b^2\right)^2\ge0\Leftrightarrow a^4-2a^2b^2+b^4\ge0\left(4\right)\)

Cộng tiếp đẳng thức ( 3 ); ( 4 ) , ta lại được :

\(2\left(a^4+b^4\right)\ge\frac{1}{4}\Rightarrow a^4+b^4\ge\frac{1}{8}\)

Vậy ..................

A=10.11+11.12+...+19.20

3A=10.11.3+11.12.3+...+19.20.3=10.11(12-9)+11.12(13-10)+12.13(14-11)+...+19.20(21-18)

3A=10.11.12 - 9.10.11 + 11.12.13 - 10.11.12 + 12.13.14 - 11.12.13 +...+ 19.20.21 - 18.19.20

3A=19.20.21-9.10.11 => A=(19.20.21-9.10.11)/3

Nhân tung tóe + rút gọn ta được: \(\Sigma_{cyc}a^3b^2+\Sigma_{cyc}ab^3\ge abc\left(ab+bc+ca+a+b+c\right)\)

\(\Leftrightarrow\)\(\Sigma\frac{a^2b}{c}+\Sigma\frac{a^2}{b}\ge ab+bc+ca+a+b+c\) (*) 

(*) đúng do \(\hept{\begin{cases}\frac{a^2b}{c}+bc\ge2ab\\\frac{a^2}{b}+b\ge2a\end{cases}}\Rightarrow\hept{\begin{cases}\Sigma\frac{a^2b}{c}\ge ab+bc+ca\\\Sigma\frac{a^2}{b}\ge a+b+c\end{cases}}\)

"=" \(\Leftrightarrow\)\(a=b=c\)

miniworld ID:35762642

lychiqu

CẢNH BÁO!!

Nguy cơ bị NGHIỆP QUẬT sml =))

a) \(A=\left(3x+2\right)^2-9x\left(x+1\right)\)

\(A=9x^2+12x+4-9x^2-9x\)

\(A=3x+4\)

\(B=\left(2x-1\right)^2-2\left(2x-1\right)\left(5x-1\right)+\left(5x-1\right)^2\)

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

\(B=\left(2x-1-5x+1\right)^2\)

\(B=\left(-3x\right)^2\)

\(B=9x^2\)

Từ đẳng thức : \(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\)

\(\Rightarrow1:\frac{40}{x-30}=1:\frac{20}{y-15}=1:\frac{28}{z-21}\)

\(\Rightarrow\frac{x-30}{40}=\frac{y-15}{20}=\frac{z-21}{28}\)

\(\Rightarrow\frac{x}{40}-\frac{3}{4}=\frac{y}{20}-\frac{3}{4}=\frac{z}{28}-\frac{3}{4}\)

\(\Rightarrow\frac{x}{40}=\frac{y}{20}=\frac{z}{28}\)

Đặt \(\frac{x}{40}=\frac{y}{20}=\frac{z}{28}=k\Rightarrow\hept{\begin{cases}x=40k\\y=20k\\z=28k\end{cases}}\)

Khi đó : xyz = 22400

<=> 40k.20k.28k = 22400

=> 22400.k³ = 22400

=> k³ = 1

=> k³ = 1³

=> k = 1

Khi đó : x = 40.1 = 40 ; 

              y = 20.1 = 20;

              z = 28.1 = 28 

Vậy x = 40 ; y = 20 ; z = 28

Ta có:\(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\)

hay\(\frac{x-30}{40}=\frac{y-15}{20}=\frac{z-21}{28}\)

\(=\frac{x}{40}-\frac{3}{4}=\frac{y}{20}-\frac{3}{4}=\frac{z}{28}-\frac{3}{4}\)

\(\Rightarrow\)\(\frac{x}{40}=\frac{y}{20}=\frac{z}{28}\)

\(\Rightarrow\)\(\frac{x}{40}=\frac{y}{20}=\frac{z}{28}=\frac{x.y.z}{40.20.28}=\frac{22400}{22400}=1\)

\(\Rightarrow\)\(\hept{\begin{cases}\frac{x}{40}=1\\\frac{y}{20}=1\\\frac{z}{28}=1\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}x=40\\y=20\\z=28\end{cases}}\)

Vậy x=40; y=20; z=28