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\(a>b>0\Rightarrow a+b>0\)
\(\left(a+b\right)^2=\left(a-b\right)^2+4ab=7^2+4.60=289\Rightarrow a+b=17\)
\(\Rightarrow a^2-b^2=\left(a-b\right)\left(a+b\right)=7.17=119\)
\(a^2+b^2=\left(a-b\right)^2+2ab=7^2+2.60=169\)
\(\Rightarrow a^4+b^4=\left(a^2+b^2\right)^2-2\left(ab\right)^2=169^2-2.60^2=21361\)
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\(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(=7\cdot\sqrt{\left(a-b\right)^2+4ab}\)
\(=7\cdot\sqrt{7^2+4\cdot60}=119\)
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\(a,a^2+b^2=\left(a+b\right)^2-2ab=9^2-2\cdot20=41\\ b,a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2=41^2-2\left(ab\right)^2\\ =1681-2\cdot400=881\\ c,\left(a-b\right)^2=a^2+b^2-2ab=41-2\cdot20=1\\ \Rightarrow a-b=1\\ \Rightarrow C=a^2-b^2=\left(a-b\right)\left(a+b\right)=9\cdot1=9\)
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\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Rightarrow ab+bc+ca=-5\)
\(\Rightarrow\left(ab+bc+ca\right)^2=25\)
\(\Rightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=25\)
\(\Rightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=25\)
\(\Rightarrow a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left[\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2\right]\)
\(=10^2-2.25=50\)
Ta có: a+b+c=0 ⇒(a+b+c)2=0
Hay a2+b2+c2+2ab+2bc+2ca=0
1+2(ac+bc+ca)=0
ab+bc+ca=\(\dfrac{-1}{2}\)
\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=100\left(1\right)\)
\(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+b^2ac+c^2ab+a^bc=a^2b^2+b^2c^2+c^2+a^2+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+c^2a^2=25\)
hay \(2\left(a^2b^2+b^2c^2+c^2a^2\right)=50\left(2\right)\)
Từ (1) và (2) ⇒a4+b4+c4=50
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\(a^2+b^2=\left(a+b\right)^2-2ab=7^2-24=25\)
\(\left(a-b\right)^2=\left(a+b\right)^2-4ab=7^2-4.12=1\)
\(\Rightarrow a-b=-1\)
\(\Rightarrow A=\left(-1\right)^5=?\)
\(B=\left(a^2+b^2\right)^2-2\left(ab\right)^2=25^2-2.12^2=?\)
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Lời giải:
$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)$
$=[(a+b+c)^2-2(ab+bc+ac)]^2-2[(ab+bc+ac)^2-2abc(a+b+c)]$
$=[1^2-2(-1)]^2-2[(-1)^2-2(-1).1]=3$
\(a^4+b^4=a^4+4a^2b^2+b^4-4a^2b^2\)
\(=\left(a^2+b^2\right)-4a^2b^2\)
\(=\left[\left(a-b\right)^2-2ab\right]^2-4\cdot\left(ab\right)^2\)
\(=\left(1^2-2\cdot12\right)^2-4\cdot12^2\)
\(=\left(1-24\right)^2-4\cdot144\)
\(=\left(-23\right)^2-576=-47\)
\(a^2+b^2=\left(a-b\right)^2+2ab=1^2+2.12=25\)
\(a^4+b^4=\left(a^2+b^2\right)-2\left(ab\right)^2=25^2-2.12^2=337\)