Toán Lớp 9: Thực hiện phép tính:
∛(20+14 √2) + ∛(20-14 √2)
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Toán Lớp 9: Thực hiện phép tính:
∛(20+14 √2) + ∛(20-14 √2)
Comments ( 2 )
Ta có:
$A^3=20+14\sqrt{2}+3(\sqrt[3]{20+14\sqrt{2}})^2.\sqrt[3]{20-14\sqrt{2}}+3\sqrt[3]{20+14\sqrt{2}}.(\sqrt[3]{20-14\sqrt{2}})^2+20-14\sqrt{2}$
$A^3=40+6\sqrt[3]{20+14\sqrt{2}}+6\sqrt[3]{20-14\sqrt{2}}$
$A^3=40+6(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}})$
$A^3=40+6A$
$⇔A^3-40-6A=0$
$⇔A^3-4A^2+4A^2-16A+10A-40=0$
$⇔A^2(A-4)+4A(A-4)+10(A-4)=0$
$⇔(A^2+4A+10)(A-4)=0(1)$
Do $A>0$ nên
$⇔A=4$
Vậy bt có giá trị là 4