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Lời giải:
\(P=|a^3-b^3|=|a-b||a^2+ab+b^2|=|a-b|.|13+ab|\)
Ta có: \(a+b-ab=-1\)
\(\Leftrightarrow a+b+1=ab\).
Do đó:
\(13+2ab=15+2(a+b)\)
\(\Leftrightarrow a^2+b^2+2ab=15+2(a+b)\)
\(\Leftrightarrow (a+b)^2=15+2(a+b)\Leftrightarrow (a+b)^2-2(a+b)-15=0\)
\(\Leftrightarrow (a+b-5)(a+b+3)=0\Rightarrow \left[\begin{matrix} a+b=5\\ a+b=-3\end{matrix}\right.\)
TH1: $a+b=5$\(\Rightarrow ab=a+b+1=6\)
\((a-b)^2=(a+b)^2-4ab=5^2-4.6=1\)
\(\Rightarrow |a-b|=1\)
\(P=|a-b|.|13+ab|=1.|13+6|=19\)
TH2: \(a+b=-3\Rightarrow ab=a+b+1=-2\)
\((a-b)^2=(a+b)^2-4ab=(-3)^2-4(-2)=17\)
\(\Rightarrow |a-b|=\sqrt{17}\)
\(P=|a-b|.|13+ab|=\sqrt{17}|13-2|=11\sqrt{17}\)
Lời giải:
\(P=|a^3-b^3|=|a-b||a^2+ab+b^2|=|a-b|.|13+ab|\)
Ta có: \(a+b-ab=-1\)
\(\Leftrightarrow a+b+1=ab\).
Do đó:
\(13+2ab=15+2(a+b)\)
\(\Leftrightarrow a^2+b^2+2ab=15+2(a+b)\)
\(\Leftrightarrow (a+b)^2=15+2(a+b)\Leftrightarrow (a+b)^2-2(a+b)-15=0\)
\(\Leftrightarrow (a+b-5)(a+b+3)=0\Rightarrow \left[\begin{matrix} a+b=5\\ a+b=-3\end{matrix}\right.\)
TH1: $a+b=5$\(\Rightarrow ab=a+b+1=6\)
\((a-b)^2=(a+b)^2-4ab=5^2-4.6=1\)
\(\Rightarrow |a-b|=1\)
\(P=|a-b|.|13+ab|=1.|13+6|=19\)
TH2: \(a+b=-3\Rightarrow ab=a+b+1=-2\)
\((a-b)^2=(a+b)^2-4ab=(-3)^2-4(-2)=17\)
\(\Rightarrow |a-b|=\sqrt{17}\)
\(P=|a-b|.|13+ab|=\sqrt{17}|13-2|=11\sqrt{17}\)
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:
\(P=\frac{a^4-a-b^4+b}{(b^3-1)(a^3-1)}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a^4-b^4)-(a-b)}{a^3b^3-(a^3+b^3)+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{(a-b)[(a+b)(a^2+b^2)-1]}{a^3b^3-[(a+b)^3-3ab(a+b)]+1}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a-b)[(a^2+b^2)-(a+b)^2]}{a^3b^3-[1-3ab]+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{-2ab(a-b)}{a^3b^3+3ab}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{-2(a-b)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)
\(a^2+b^2=13\Leftrightarrow a^2+b^2+2ab-2ab=13\Leftrightarrow\left(a+b\right)^2-2ab=13\)
Mà \(a+b-ab=-1\Leftrightarrow ab=a+b+1\)Thay vào phương trình trêm ta có:
\(\left(a+b\right)^2-2\left(a+b+1\right)=13\)
<=> \(\left(a+b\right)^2-2\left(a+b\right)+1=16\)
<=> \(\left(a+b+1\right)^2=4^2\)
<=> \(a+b+1=\pm4\)=> \(ab=\pm4\)
Ta lại có: \(a^2+b^2=13\Leftrightarrow\left(a-b\right)^2+2ab=13\)
+) Với ab=4
thay vào ta có: \(\left(a-b\right)^2+8=13\Leftrightarrow\left(a-b\right)^2=5\Leftrightarrow\left|a-b\right|=\sqrt{5}\)
=> \(P=\left|a^3-b^3\right|=\left|\left(a-b\right)\left(a^2+b^2+ab\right)\right|=\left|a-b\right|\left|a^2+b^2+ab\right|\)
\(=\sqrt{5}\left(13+4\right)=17\sqrt{5}\)
+) Với ab=-4 . Em làm tương tự nhé!