\(\left(x+1\right)\left(x-1\right)< 0\)

\(\Leftrightarrow x^2-1< 0\)

\(\Leftrightarrow x^2< 1\)

\(\Leftrightarrow\hept{\begin{cases}x< 1\\x>-1\end{cases}}\)

Vậy giá trị thỏa mãn của x là 0

bạn hỏi về tok đó hả

ừmmm chữ mờ kéo dài

tự làm là hạnh phúc của mỗi công dân.

Áp dụng bđt cô si ta có:

\(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2=2\left(ab+b+1\right)\) 

\(b^2+2c^2+3\ge2\left(bc+c+1\right)\)

\(c^2+2a^2+3\ge2\left(ac+a+1\right)\)

=> \(M\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\)

\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{bcab+abc+ab}+\frac{b}{abc+ab+b}\right)\)

\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{b+1+ab}+\frac{b}{1+ab+b}\right)\)

\(=\frac{1}{2}.\frac{ab+b+1}{ab+b+1}=\frac{1}{2}\)

Bổ sung: 

Dấu "=" xảy ra <=> a = b = c =  1

Vậy GTLN của M = 1/2 tại a = b = c = 1.

\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\ge\frac{\left(a+b+c+d\right)^2}{a+b+b+c+c+d+d+a}=\frac{1}{2}\)

Dấu "=" xảy ra <=> a = b = c = d = 1/4

\(\frac{1}{a+1}+\frac{1}{b+1}=\left(\frac{1}{a+1}+\frac{4}{9}.\frac{a+1}{1}\right)+\left(\frac{1}{b+1}+\frac{4}{9}.\frac{b+1}{1}\right)-\frac{4}{9}\left(a+1\right)-\frac{4}{9}\left(b+1\right)\)

\(\ge2\sqrt{\frac{1}{a+1}.\frac{4}{9}\frac{a+1}{1}}+2\sqrt{\frac{1}{b+1}.\frac{4}{9}\frac{b+1}{1}}-\frac{4}{9}\left(a+b+2\right)\)( BĐT cô si)

\(=\frac{4}{3}+\frac{4}{3}-\frac{4}{3}=\frac{4}{3}\)

Dấu "=" <=> a = b = 1/2

Cách khác: 

\(\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{a+1+b+1}=\frac{4}{3}\)( hệ quả bđt cô - si )

Ta có : \(A=a^2+b^2=\frac{a^2}{1}+\frac{b^2}{1}\ge\frac{\left(a+b\right)^2}{2}=8\)

Vậy \(Min_A=8\)

Khi \(a=b=2\)

Ta có: \(a^5-a=a\left(a^4-1\right)=a\left(a^2-1\right)\left(a^2+1\right)=a\left(a-1\right)\left(a+1\right)\left(a^2-4+5\right)\)

\(=a\left(a-1\right)\left(a+1\right)\left(a^2-4\right)+5a\left(a-1\right)\left(a+1\right)\)

\(=a\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)+5a\left(a-1\right)\left(a+1\right)⋮5\)

vì \(a-2;a-1;a;a+1;a+2\) là 5 số nguyên liên tiếp 

=> \(a\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)⋮5\) 

và \(5a\left(a-1\right)\left(a+1\right)⋮5\)

=> \(a^5-a⋮5\)tương tự ta cũng có: \(b^5-b⋮5\) và \(c^5-c⋮5\)

=> \(\left(a^5-a\right)+\left(b^5-b\right)+\left(c^5-c\right)⋮5\)

=> \(\left(a^5+b^5+c^5\right)-\left(a+b+c\right)⋮5\)

=> \(a^5+b^5+c^5⋮5\)