Cho a+b=5,ab=-2(a<b).hãy tính a^2+b^2;1/a^3+1/b^3;a-b;a^3-b^3
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Bài 2:
\(a^2+b^2=\left(a+b\right)^2-2ab=5^2-2\cdot\left(-2\right)=9\)
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}=\dfrac{a^3+b^3}{a^3b^3}=\dfrac{\left(a+b\right)^3-3ab\left(a+b\right)}{\left(ab\right)^3}\)
\(=\dfrac{5^3-3\cdot5\cdot\left(-2\right)}{\left(-2\right)^3}=\dfrac{125+30}{8}=\dfrac{155}{8}\)
\(a-b=-\sqrt{\left(a+b\right)^2-4ab}=-\sqrt{5^2-4\cdot\left(-2\right)}=-\sqrt{33}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(A=a^5b-ab^5=a^5b-ab-ab^5+ab\)
\(A=\left(a^5b-ab\right)-\left(ab^5-ab\right)\)
\(A=b\left(a^5-a\right)-a\left(b^5-b\right)\)
Ta có \(m^5-m=m\left(m^4-1\right)=m\left(m^2-1\right)\left(m^2+1\right)\)
\(=m\left(m+1\right)\left(m-1\right)\left(m^2-4+5\right)\)
\(=m\left(m-1\right)\left(m+1\right)\left(m^2-4\right)-5m\left(m-1\right)\left(m+1\right)\)
\(=m\left(m-1\right)\left(m+1\right)\left(m-2\right)\left(m+2\right)-5m\left(m-1\right)\left(m+1\right)\)
\(=\left(m-2\right)\left(m-1\right)m\left(m+1\right)\left(m+2\right)-5\left(m-1\right)m\left(m+1\right)\)
Vì \(m-2;m-1;m;m+1;m+2\) là 5 số nguyên liên tiếp nên chia hết cho 2 ; 3 ; 5
Mà \(\left(2;3;5\right)=1\)
\(\Rightarrow\left(m-2\right)\left(m-1\right)m\left(m+1\right)\left(m+2\right)\) chia hết cho \(2\times3\times5=30\)
\(\Rightarrow m^5-m\) chia hết cho 30
\(\Rightarrow a^5-a\) và \(b^5-b\) Chia hết cho 30
\(\Rightarrow b\left(a^5-a\right)-a\left(b^5-b\right)\) chia hết cho 30
\(\Rightarrow A=a^5b-ab^5\) chia hết cho 30
Vậy A chia hết cho 30
![](https://rs.olm.vn/images/avt/0.png?1311)
Biến đổi vế phải ta có \(\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)\Leftrightarrow a^5+b^5+a^2b^2\left(a+b\right)-\left(a+b\right)\)
\(\Leftrightarrow a^5+b^5+\left(a+b\right)-\left(a+b\right)=a^5+b^5\) (vì ab=1)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét VP: (a3+b3)(a2+b2) - (a+b)
= a5 + b5 + a3b2 + a2b3 - (a+b)
= a5 + b5 + a2b2(a+b) - (a+b)
= a5 + b5 + (a+b) - (a+b)
= a5 + b5 = VP (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
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a: ta có: \(M=\dfrac{a}{\sqrt{ab}+b}+\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\)
\(=\dfrac{a\left(\sqrt{ab}-a\right)+b\left(\sqrt{ab}+b\right)}{\left(\sqrt{ab}+b\right)\left(\sqrt{ab}-a\right)}-\dfrac{a+b}{\sqrt{ab}}\)
\(=\dfrac{-\sqrt{ab}\left(a+b\right)+\left(a-b\right)\left(a+b\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)\cdot\sqrt{a}\cdot\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{a+b}{\sqrt{ab}}\)
\(=\dfrac{-\sqrt{ab}\left(a+b\right)+\left(a-b\right)\left(a+b\right)}{\sqrt{ab}\left(a-b\right)}-\dfrac{a^2-b^2}{\sqrt{ab}\left(a-b\right)}\)
\(=\dfrac{-\sqrt{ab}}{\sqrt{ab}\left(a-b\right)}\)
\(=-\dfrac{1}{a-b}\)
b: Thay \(a=\sqrt{5}+1\) và \(b=\sqrt{5}-1\) vào M, ta được:
\(M=\dfrac{-1}{\sqrt{5}+1-\sqrt{5}+1}=\dfrac{-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=a^5+a^3b^2+b^3a^2+b^5-\left(a+b\right)\)
\(=a^5+b^5+\left(a^3b^2+b^3a^2\right)-\left(a+b\right)\)
\(=a^5+b^5+a^2b^2\left(a+b\right)-\left(a+b\right)\)
\(=a^5+b^5+\left[\left(ab\right)^2-1\right]\left(a+b\right)\)
Mà \(ab=1\Rightarrow\left(ab\right)^2-1=1^2-1=0\)
\(\Rightarrow\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)=a^5+b^5+0=a^5+b^5\)
Vậy ...
\(a^2+b^2=\left(a+b\right)^2-2ab=5^2-2\cdot\left(-2\right)=29\)
\(a-b=\sqrt{\left(a-b\right)^2+4ab}=\sqrt{5^2+4\cdot\left(-2\right)}=\sqrt{17}\)