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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: \(a+b=-3\Rightarrow ab=a+b+1=-2\)
\((a-b)^2=(a+b)^2-4ab=(-3)^2-4(-2)=17\)
\(\Rightarrow |a-b|=\sqrt{17}\)
\(P=|a-b|.|13+ab|=\sqrt{17}|13-2|=11\sqrt{17}\)
Ta có: \(\left(a-1\right)^3=9a\)
=>\(a^3-3a^2+3a-1=9a\)
=>\(a^3-1=3a^2-3a+9a=3a^2+6a\)
=>\(\left(a-1\right)\left(a^2+a+1\right)=3a+3\left(a^2+a\right)\)
=>(a-1)(b+1)=3a+3b=3(a+b)
=>\(\left(a-1\right)^3\cdot\left(b+1\right)^3=\left\lbrack3\left(a+b\right)\right\rbrack^3=27\cdot\left(a+b\right)^3\)
Ta có: \(a^3-3a^2+3a-1=9a\)
=>1+3(a+1)+\(3\left(a+1\right)^2+\left(a+1\right)^3=9\left(a+1\right)^2\)
=>\(a^3+3a^3\left(a+1\right)+3a^3\left(a+1\right)^2+a^3\cdot\left(a+1\right)^3=a^3\cdot9\left(a+1\right)^2\)
=>\(a^3+3a^2\cdot\left\lbrack a\left(a+1\right)\right\rbrack+3\cdot a\cdot\left\lbrack a\left(a+1\right)\right\rbrack^2+\left\lbrack a\left(a+1\right)\right\rbrack^3=a^2\cdot\left(a+1\right)^2\cdot9a\)
=>\(a^3+3a^2b+3ab^2+b^3=9ab^2\)
=>\(\left(a+b\right)^3=9ab^2=\left(a-1\right)^3\cdot b^2\)
=>\(27\left(a+b\right)^3=27\left(a-1\right)^3\cdot b^2\)
=>\(\left(a-1\right)^3\cdot\left(b+1\right)^3=27\left(a-1\right)^3\cdot b^2\)
=>\(\left(b+1\right)^3=27b^2\)
\(a^2+b^2=13\Leftrightarrow a^2+b^2+2ab-2ab=13\Leftrightarrow\left(a+b\right)^2-2ab=13\)
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!