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mavar
@mavar
July 2022
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СРОЧНО вычислите x1^4×x2+x1×x2^4 где х1 и х2 корни уравнения х^2+1,5х-2=0 СРОЧНО
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mmb1
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X1^4*x2+x2^4*x1=x1x2(x1^3+x2^3)=x1x2(x1+x2)(x1^2+x2^2-x1x2)=x1x2(x1+x2)((x1+x2)^2-3x1x2)
х^2+1,5х-2=0
по теореме виета
x1+x2=-3/2
x1x2=-2
-2*(-3/2)*((-3/2)^2-3*(-2))=3(9/4+6)=99/4
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Answers & Comments
Verified answer
X1^4*x2+x2^4*x1=x1x2(x1^3+x2^3)=x1x2(x1+x2)(x1^2+x2^2-x1x2)=x1x2(x1+x2)((x1+x2)^2-3x1x2)х^2+1,5х-2=0
по теореме виета
x1+x2=-3/2
x1x2=-2
-2*(-3/2)*((-3/2)^2-3*(-2))=3(9/4+6)=99/4