Cho phương trình x^2 - 10x - 8 = 0 có hai nghiệm x1; x2 Không giải phương trình hãy tính giá trị của các biểu thức

Giải chi tiết

Áp dụng định lí Vi-ét ta có: $\left\{\begin{array}{l}{x}_1+x_2=10\\ x_1{x}_2=-8\end{array}\right.$ 

Ta có ${x_1}^2+{x_2}^2={\left(x_1+x_2\right)}^2-2x_1{x}_2={10}^2-2\left(-8\right)=116$

Khi đó

$A=\frac{1}{{x_1}^2}+\frac{1}{{x_2}^2}=\frac{{x_1}^2+{x_2}^2}{{\left(x_1{x}_2\right)}^2}=\frac{116}{{\left(-8\right)}^2}=\frac{29}{16}$

 $C=\left(x_1-x_2\right)\left({x_1}^2-{x_2}^2\right)=\left(x_1-x_2\right)\left(x_1-x_2\right)\left(x_1+x_2\right)={\left(x_1-x_2\right)}^2\left(x_1+x_2\right)$

     $=\left({x_1}^2+{x_2}^2-2x_1{x}_2\right)\left(x_1+x_2\right)=\left[116-2.\left(-8\right).10\right]=1320$

$D=\left|x_1-x_2\right|\Rightarrow D^2={\left(x_1-x_2\right)}^2={x_1}^2+{x_2}^2-2x_1{x}_2\Rightarrow D^2=132\Rightarrow D=2\sqrt{33}$(Vì $D>0$)

$E={x_1}^4+{x_2}^4$

${x_1}^4+{x_2}^4+2x_1^2{x}_2^2-2x_1^2{x}_2^2={\left({x_1^2}_{}+{x_2^2}_{}\right)}^2-2{\left(x_1{x}_2\right)}^2={116}^2-2.{\left(-8\right)}^2=13328$         

$F={x_1}^5+{x_2}^5=x_1^3+x_2^3={\left({x_1}^{}+{x_2}^{}\right)}^3-3{x_1}^{}{x_2}^{}\left(x_{{}^1}+x_{{}^2}\right)={10}^3-3.10(-8)=1240$

$F={x_1}^5+{x_2}^5=\left({x_1}^2+{x_2}^2\right)\left({x_1}^3+{x_2}^3\right)-{x_1}^2{x_2}^3-{x_1}^3{x_2}^2$

$=\left({x_1}^2+{x_2}^2\right)\left({x_1}^3+{x_2}^3\right)-{x_1}^2{x_2}^2\left(x_1+x_2\right)=116.1240-{\left(-8\right)}^2.10=143200$