a+b=1

=>b=1-a; a=1-b

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

=-a(b^2+b+1)

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

=-b(a^2+a+1)

\(\frac{a}{b^3-1}-\frac{b}{a^3-1}\)

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

\(=\frac{-1}{b^2+b+1}+\frac{1}{a^2+a+1}\)

\(=\frac{(b^2 + b + 1) - (a^2 + a + 1)}{(a^2 + a + 1)(b^2 + b + 1)}\)

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

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

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

\(=\frac{-2\left(a-b\right)}{=a^2b^2+ab\cdot1+ab+1^2-2ab+1+1}=\frac{-2\left(a-b\right)}{a^2b^2+3}\)

=>\(P=\frac{a}{b^3-1}-\frac{b}{a^3-1}+\frac{2\left(a-b\right)}{a^2b^2+3}=\frac{-2\left(a-b\right)+2\left(a-b\right)}{a^2b^2+3}\) =0

Lời giải:

\(P=\frac{a^4-a-b^4+b}{(b^3-1)(a^3-1)}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{(a^4-b^4)-(a-b)}{a^3b^3-(a^3+b^3)+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{(a-b)[(a+b)(a^2+b^2)-1]}{a^3b^3-[(a+b)^3-3ab(a+b)]+1}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{(a-b)[(a^2+b^2)-(a+b)^2]}{a^3b^3-[1-3ab]+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{-2ab(a-b)}{a^3b^3+3ab}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{-2(a-b)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)

Lời giải:

\(P=\frac{a^4-a-b^4+b}{(b^3-1)(a^3-1)}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{(a^4-b^4)-(a-b)}{a^3b^3-(a^3+b^3)+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{(a-b)[(a+b)(a^2+b^2)-1]}{a^3b^3-[(a+b)^3-3ab(a+b)]+1}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{(a-b)[(a^2+b^2)-(a+b)^2]}{a^3b^3-[1-3ab]+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{-2ab(a-b)}{a^3b^3+3ab}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{-2(a-b)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)

a^3+b^3+c^3=3abc

=>(a+b)^3+c^3-3ab(a+b)-3bac=0

=>(a+b+c)(a^2+2ab+b^2-ac-bc+c^2)-3ab(a+b+c)=0

=>(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=0

=>a^2+b^2+c^2-ab-bc-ac=0

=>2a^2+2b^2+2c^2-2ab-2bc-2ac=0

=>(a-c)^2+(a-b)^2+(b-c)^2=0

=>a=b=c

=>A=(1+b/b)(1+b/b)(1+c/c)

=2*2*2=8

a) \(\left(\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\right):\left(\frac{1-x}{x+1}\right)\)

= \(\left(\frac{x+3}{x-2}-\frac{x+2}{x-3}+\frac{x+2}{x^2-2x-3x+6}\right):\left(\frac{1-x}{x+1}\right)\)

= \(\left(\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\right):\left(\frac{1-x}{x+1}\right)\)

= \(\left(\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}\right).\frac{x+1}{1-x}\)

=\(\frac{-3+x}{\left(x-2\right)\left(x-3\right)}.\frac{x+1}{1-x}\)

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

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

b) Để A >1 \(\Leftrightarrow\frac{x+1}{\left(x-2\right)\left(1-x\right)}>1\)

\(\Leftrightarrow\frac{-\left(1-x\right)\left(3-x\right)}{\left(x-2\right)\left(1-x\right)}\)

\(\Leftrightarrow\frac{x-3}{x-2}>0\)

\(\Rightarrow\orbr{\begin{cases}x-3\ge0\\x-2>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\ge3\\x>2\end{cases}\Leftrightarrow}x\ge3}\)

\(\Rightarrow\orbr{\begin{cases}x-3< 0\\x-2< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 3\\x< 2\end{cases}\Leftrightarrow}x< 2}\)

Vậy ...

Với ab = 1 , a + b ¹ 0, ta có:

P = a 3 + b 3 ( a + b ) 3 ( a b ) 3 + 3 ( a 2 + b 2 ) ( a + b ) 4 ( a b ) 2 + 6 ( a + b ) ( a + b ) 5 ( a b ) = a 3 + b 3 ( a + b ) 3 + 3 ( a 2 + b 2 ) ( a + b ) 4 + 6 ( a + b ) ( a + b ) 5 = a 2 + b 2 − 1 ( a + b ) 2 + 3 ( a 2 + b 2 ) ( a + b ) 4 + 6 ( a + b ) 4 = ( a 2 + b 2 − 1 ) ( a + b ) 2 + 3 ( a 2 + b 2 ) + 6 ( a + b ) 4 = ( a 2 + b 2 − 1 ) ( a 2 + b 2 + 2 ) + 3 ( a 2 + b 2 ) + 6 ( a + b ) 4 = ( a 2 + b 2 ) 2 + 4 ( a 2 + b 2 ) + 4 ( a + b ) 4 = ( a 2 + b 2 + 2 ) 2 ( a + b ) 4 = ( a 2 + b 2 + 2 a b ) 2 ( a...