Ta có: $x^2+\dfrac{1}{x^2}=7$$\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-2=7$$\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2=9$$\Rightarrow x+\dfrac{1}{x}=3$ (vì $x>0$)$\Leftrightarrow\left(x+\dfrac{1}{x}\right)^3=27$$\Leftrightarrow x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)=27$$\Leftrightarrow x^3+\dfrac{1}{x^3}=18$Mặt khác: $x^4+\dfrac{1}{x^4}=\left(x^2+\dfrac{1}{x^2}\right)^2-2=7^2-2=47$Khi đó: $\left(x^3+\dfrac{1}{x^3}\right)\left(x^4+\dfrac{1}{x^4}\right)=18.47$$\Leftrightarrow x^7+\dfrac{1}{x^7}+x+\dfrac{1}{x}=846$$\Leftrightarrow x^7+\dfrac{1}{x^7}+3=846$$\Leftrightarrow x^7+\dfrac{1}{x^7}=843$hay $A=843$$\text{#}Toru$Câu trả lời: Ta có: $x^2+\dfrac{1}{x^2}=7$$\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-2=7$$\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2=9$$\Rightarrow x+\dfrac{1}{x}=3$ (vì $x>0$)$\Leftrightarrow\left(x+\dfrac{1}{x}\right)^3=27$$\Leftrightarrow x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)=27$$\Leftrightarrow x^3+\dfrac{1}{x^3}=18$Mặt khác: $x^4+\dfrac{1}{x^4}=\left(x^2+\dfrac{1}{x^2}\right)^2-2=7^2-2=47$Khi đó: $\left(x^3+\dfrac{1}{x^3}\right)\left(x^4+\dfrac{1}{x^4}\right)=18.47$$\Leftrightarrow x^7+\dfrac{1}{x^7}+x+\dfrac{1}{x}=846$$\Leftrightarrow x^7+\dfrac{1}{x^7}+3=846$$\Leftrightarrow x^7+\dfrac{1}{x^7}=843$hay $A=843$$\text{#}Toru$