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\(a^2+b^2=13\Leftrightarrow a^2+b^2+2ab-2ab=13\Leftrightarrow\left(a+b\right)^2-2ab=13\)
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!
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)+\(3\left(a+1\right)^2+\left(a+1\right)^3=9\left(a+1\right)^2\)
=>\(a^3+3a^3\left(a+1\right)+3a^3\left(a+1\right)^2+a^3\cdot\left(a+1\right)^3=a^3\cdot9\left(a+1\right)^2\)
=>\(a^3+3a^2\cdot\left\lbrack a\left(a+1\right)\right\rbrack+3\cdot a\cdot\left\lbrack a\left(a+1\right)\right\rbrack^2+\left\lbrack a\left(a+1\right)\right\rbrack^3=a^2\cdot\left(a+1\right)^2\cdot9a\)
=>\(a^3+3a^2b+3ab^2+b^3=9ab^2\)
=>\(\left(a+b\right)^3=9ab^2=\left(a-1\right)^3\cdot b^2\)
=>\(27\left(a+b\right)^3=27\left(a-1\right)^3\cdot b^2\)
=>\(\left(a-1\right)^3\cdot\left(b+1\right)^3=27\left(a-1\right)^3\cdot b^2\)
=>\(\left(b+1\right)^3=27b^2\)
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:
\(P=\frac{a^4-a-b^4+b}{(b^3-1)(a^3-1)}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a^4-b^4)-(a-b)}{a^3b^3-(a^3+b^3)+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{(a-b)[(a+b)(a^2+b^2)-1]}{a^3b^3-[(a+b)^3-3ab(a+b)]+1}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a-b)[(a^2+b^2)-(a+b)^2]}{a^3b^3-[1-3ab]+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{-2ab(a-b)}{a^3b^3+3ab}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{-2(a-b)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)
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=|a3−b3|=|a−b||a2+ab+b2|=|a−b|.|13+ab|P=|a3−b3|=|a−b||a2+ab+b2|=|a−b|.|13+ab|
Ta có: a+b−ab=−1a+b−ab=−1
⇔a+b+1=ab⇔a+b+1=ab.
Do đó:
13+2ab=15+2(a+b)13+2ab=15+2(a+b)
⇔a2+b2+2ab=15+2(a+b)⇔a2+b2+2ab=15+2(a+b)
⇔(a+b)2=15+2(a+b)⇔(a+b)2−2(a+b)−15=0⇔(a+b)2=15+2(a+b)⇔(a+b)2−2(a+b)−15=0
⇔(a+b−5)(a+b+3)=0⇒[a+b=5a+b=−3⇔(a+b−5)(a+b+3)=0⇒[a+b=5a+b=−3
TH1: a+b=5a+b=5⇒ab=a+b+1=6⇒ab=a+b+1=6
(a−b)2=(a+b)2−4ab=52−4.6=1(a−b)2=(a+b)2−4ab=52−4.6=1
⇒|a−b|=1⇒|a−b|=1
P=|a−b|.|13+ab|=1.|13+6|=19P=|a−b|.|13+ab|=1.|13+6|=19
TH2: a+b=−3⇒ab=a+b+1=−2a+b=−3⇒ab=a+b+1=−2
(a−b)2=(a+b)2−4ab=(−3)2−4(−2)=17(a−b)2=(a+b)2−4ab=(−3)2−4(−2)=17
⇒|a−b|=√17⇒|a−b|=17
P=|a−b|.|13+ab|=√17|13−2|=11√17
chúc bạn học tốt hem