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a: \(a^2+4a=b^2+4b+1\)
=>\(a^2+4a-b^2-4b=0\)
=>(a-b)(a+b)+4(a-b)=0
=>(a-b)(a+b+4)=0
mà a-b<>0
nên a+b+4=0
=>a+b=-4
b: Đặt \(X=a^3+b^3\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=\left(-4\right)^3-3ab\cdot\left(-4\right)=-64+12ab\)
\(a^2+4a=1\)
=>\(a^2+4a-1=0\)
=>\(a^2+4a+4-5=0\)
=>\(\left(a+2\right)^2=5\)
=>\(\left[\begin{array}{l}a+2=\sqrt5\\ a+2=-\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}a=\sqrt5-2\\ a=-\sqrt5-2\end{array}\right.\)
\(b^2+4b=1\)
=>\(b^2+4b-1=0\)
=>\(b^2+4b+4-5=0\)
=>\(\left(b+2\right)^2=5\)
=>\(\left[\begin{array}{l}b+2=\sqrt5\\ b+2=-\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}b=\sqrt5-2\\ b=-\sqrt5-2\end{array}\right.\)
Vì a<>b nên sẽ có hai trường hợp sau:
TH1: \(a=\sqrt5-2;b=-\sqrt5-2\)
=>\(ab=\left(\sqrt5-2\right)\left(-\sqrt5-2\right)=-\left(\sqrt5-2\right)\left(\sqrt5+2\right)=-1\)
X=-64+12ab
=-64-12
=-76
TH2: \(a=-\sqrt5-2;b=\sqrt5-2\)
=>\(ab=\left(\sqrt5-2\right)\left(-\sqrt5-2\right)=-\left(\sqrt5-2\right)\left(\sqrt5+2\right)=-1\)
X=-64+12ab
=-64-12
=-76
Vậy: X=-76
c: Đặt \(Y=a^4+b^4\)
\(=\left(a^2+b^2\right)^2-2a^2b^2\)
\(=\left\lbrack\left(a+b\right)^2-2ab\right\rbrack^2-2\cdot\left(ab\right)^2\)
\(=\left\lbrack\left(-4\right)^2-2\cdot\left(-1\right)\right\rbrack^2-2\cdot\left(-1\right)^2=\left\lbrack16+2\right\rbrack^2-2\)
\(=18^2-2\)
=324-2
=322
Thực hiện phép nhân đa thức với đa thức ở vế trái.
=> VT = VP (đpcm)
\(a^2+b^2=\left(a+b\right)^2-2ab=\left(-3\right)^2-2\cdot\left(-2\right)=9+4=13\)
\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=\left(-3\right)^3-3\cdot\left(-2\right)\cdot\left(-3\right)\)
\(=-27-18=-45\)
Ta có: a + b + c = 0
\(\Rightarrow\) (a + b + c)2 = 0
\(\Leftrightarrow\) a2 + b2 + c2 + 2ab + 2bc + 2ac = 0
\(\Leftrightarrow\) 2009 + 2(ab + bc + ac) = 0
\(\Leftrightarrow\) ab + bc + ac = \(\dfrac{-2009}{2}\)
\(\Leftrightarrow\) (ab + bc + ac)2 = \(\left(\dfrac{-2009}{2}\right)^2\)
\(\Leftrightarrow\) a2b2 + b2c2 + a2c2 + 2abc(a + b + c) = \(\left(\dfrac{-2009}{2}\right)^2\)
\(\Leftrightarrow\) a2b2 + b2c2 + c2a2 = \(\left(\dfrac{-2009}{2}\right)^2\) (Vì a + b + c = 0)
Lại có: a2 + b2 + c2 = 2009
\(\Rightarrow\) (a2 + b2 + c2)2 = 20092
\(\Leftrightarrow\) a4 + b4 + c4 + 2(a2b2 + b2c2 + c2a2) = 20092
\(\Leftrightarrow\) a4 + b4 + c4 + 2.\(\dfrac{2009^2}{4}\) = 20092
\(\Leftrightarrow\) a4 + b4 + c4 = 20092 - \(\dfrac{2009^2}{2}\) = 2018040,5
Chúc bn học tốt!
1: Ta có: \(a^2+b^2+c^2\)
\(=\left(a+b+c\right)^2-2\cdot\left(ab+bc+ca\right)\)
\(=5^2-2\cdot174=-323\)
\(a^2+b^2=a^3+b^3=a^4+b^4\)
\(\Rightarrow\left(a^3+b^3\right)^2=\left(a^2+b^2\right)\left(a^4+b^4\right)\)
\(\Rightarrow a^6+b^6+2a^3b^3=a^6+b^6+a^2b^4+a^4b^2\)
\(\Rightarrow2a^3b^3=a^2b^2\left(a^2+b^2\right)\)
\(\Rightarrow2ab=a^2+b^2\)
\(\Rightarrow\left(a-b\right)^2=0\)
\(\Rightarrow a=b\)
Thế vào \(a^2+b^2=a^3+b^3\)
\(\Rightarrow a^2+a^2=a^3+a^3\Rightarrow2a^3=2a^2\Rightarrow a=b=1\)
\(\Rightarrow a+b=2\)