Cho 3 số a,b,c thỏa mãn a + b + c = 0
$a^{2}$ + $b^{2}$ + $c^{2}$ =2009
Tính A = $a^{4}$ + $b^{4}$ + $c^{4}$
Cho 3 số a,b,c thỏa mãn a + b + c = 0
$a^{2}$ + $b^{2}$ + $c^{2}$ =2009
Tính A = $a^{4}$ + $b^{4}$ + $c^{4}$
Đáp án:
$a^4+b^4+c^4=\dfrac{2009^2}{2}$
Giải thích các bước giải:
$a+b+c=0$
$\rightarrow (a+b+c)^2=0$
$\rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0$
$\rightarrow 2(ab+bc+ca)=-(a^2+b^2+c^2)$
$\rightarrow ab+bc+ca=-\dfrac{2009}{2}$
$\rightarrow (ab+bc+ca)^2=\dfrac{2009^2}{4}$
$\rightarrow a^2b^2+b^2c^2+c^2a^2+2ab.bc+2bc.ca+2ca.ab=\dfrac{2009^2}{4}$
$\rightarrow a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)=\dfrac{2009^2}{4}$
$\rightarrow a^2b^2+b^2c^2+c^2a^2=\dfrac{2009^2}{4}$
Ta có:
$a^2+b^2+c^2=2009$
$\rightarrow (a^2+b^2+c^2)^2=2009^2$
$\rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=2009^2$
$\rightarrow a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)=2009^2$
$\rightarrow a^4+b^4+c^4+2.\dfrac{2009^2}{4}=2009^2$
$\rightarrow a^4+b^4+c^4+\dfrac{2009^2}{2}=2009^2$
$\rightarrow a^4+b^4+c^4=\dfrac{2009^2}{2}$
Ta có : a+b+c = 0
=> (a+b+c)^2= 0
a^2 + b^2 + c^2 + 2(ab+bc+ca) = 0
=> 2009 + 2(ab+bc+ca) = 0
=> ab+bc+ca = -2009/2
=> (ab + bc + ca)^2 = 2009^2/4
=> a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a+b+c) = 2009^2/4
Mà a+b+c = 0
=> a^2b^2 + b^2c^2 + c^2a^2 = 2009^2/4
Lại có : a^2 + b^2+ c^2 = 2009
=> (a^2 + b^2 + c^2)^2 = 2009^2
a^4 + b^4 + c^4 + 2(a^2b^2 + b^2c^2 + c^2a^2) = 2009^2
a^4 + b^4 + c^4 = 2009^2 – 2009^2/2
a^2 + b^4 + c^4 = 2009^2/2