\(a>0;b>0\Rightarrow a^2>0;b^2>0\Rightarrow a^2-b^2< =a^2+b^2=1\)

\(\Rightarrow\left(a-b\right)\left(a+b\right)< 1\Rightarrow a-b< \frac{1}{a+b}\Rightarrow\frac{1}{a+b}>a-b\)

với a>0;b>0 ta có: \(a^3b+ab^3>=2\sqrt{a^3bab^3}=2\sqrt{a^4b^4}=2a^2b^2\)(bđt cosi)

\(\Rightarrow a^4+a^3b+ab^3+b^4>=a^4+2a^2b^2+b^4=\left(a^2+b^2\right)^2=1^2=1\)

\(\Rightarrow a^3\left(a+b\right)+b^3\left(a+b\right)=\left(a^3+b^3\right)\left(a+b\right)>=1\Rightarrow a^3+b^3>=\frac{1}{a+b}\)mà \(\frac{1}{a+b}>a-b\)

\(\Rightarrow a^3+b^3>a-b\left(đpcm\right)\)

a/ \(\left(x+2\right)^2-9=0\)

<=> \(\left(x+2-3\right)\left(x+2+3\right)=0\)

<=> \(\left(x-1\right)\left(x+5\right)=0\)

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

<=> \(\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

b/ \(x^2-2x+1=25\)

<=> \(\left(x-1\right)^2=25\)

<=> \(\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)

<=> \(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)

a) (x+2)²=0

 ==> x+2=0

 ==> x=0-2

==> x=-2

tất cả đống này là hằng đẳng thức : \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right).\)

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

\(x^3+4^3=\left(x+4\right)\left(x^2-4x+16\right)\)

\(x^6+2^3=\left(x^2+2\right)\left(x^4-2x^2+4\right)\)

\(\left(3x\right)^3+2^3=\left(3x+2\right)\left(9x^2-6x+4\right)\)

\(x^4+2x^2-24=x^4-4x^2+2x^2-24+4x^2=\left(x^4-4x^2\right)+\left(6x^2-24\right)\)

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

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

\(=\left(a-2b-b\sqrt{5}\right)\left(a-2b+b\sqrt{5}\right)\)