Toán 9 Cho $x=\sqrt[3]{7+5\sqrt{2}}-\frac{1}{\sqrt[3]{7+5 \sqrt{2}}}$.Tính $F=x^{3}+3x-14$

Nguyệt Băng

Học sinh mới
Thành viên
2 Tháng bảy 2018
38
39
6
Hà Nội
...
Cho x = [tex]\sqrt[3]{7+5\sqrt{2}}-\frac{1}{\sqrt[3]{7+5\sqrt{2}}}[/tex].Tính F=[tex]x^{3}+3x-14[/tex]
C1: $x=\sqrt[3]{(\sqrt 2+1)^3}-\dfrac1{\sqrt[3]{(\sqrt 2+1)^3}}=\sqrt 2+1-\dfrac1{\sqrt 2+1}=\sqrt 2+1-(\sqrt 2-1)=2\Rightarrow x^3+3x-14=0$
C2: $x^3=7+5\sqrt 2 - \dfrac1{7+5\sqrt 2}-3\sqrt[3]{7+5\sqrt 2}.\dfrac1{\sqrt[3]{7+5\sqrt 2}}.(\sqrt[3]{7+5\sqrt 2}-\dfrac1{\sqrt[3]{7+5\sqrt 2}})$
$\Leftrightarrow x^3=7+5\sqrt 2-\dfrac{7-5\sqrt 2}{(7+5\sqrt 2)(7-5\sqrt 2)}-3x$
$\Leftrightarrow x^3=7+5\sqrt 2-(5\sqrt 2-7)-3x$
$\Leftrightarrow x^3=14-3x\Leftrightarrow x^3+3x-14=0$
 

misoluto04@gmail.com

Banned
Banned
Thành viên
19 Tháng sáu 2018
895
462
101
19
Hà Nội
Good bye là xin chào...
C1: $x=\sqrt[3]{(\sqrt 2+1)^3}-\dfrac1{\sqrt[3]{(\sqrt 2+1)^3}}=\sqrt 2+1-\dfrac1{\sqrt 2+1}=\sqrt 2+1-(\sqrt 2-1)=2\Rightarrow x^3+3x-14=0$
C2: $x^3=7+5\sqrt 2 - \dfrac1{7+5\sqrt 2}-3\sqrt[3]{7+5\sqrt 2}.\dfrac1{\sqrt[3]{7+5\sqrt 2}}.(\sqrt[3]{7+5\sqrt 2}-\dfrac1{\sqrt[3]{7+5\sqrt 2}})$
$\Leftrightarrow x^3=7+5\sqrt 2-\dfrac{7-5\sqrt 2}{(7+5\sqrt 2)(7-5\sqrt 2)}-3x$
$\Leftrightarrow x^3=7+5\sqrt 2-(5\sqrt 2-7)-3x$
$\Leftrightarrow x^3=14-3x\Leftrightarrow x^3+3x-14=0$
Bạn làm sai rồi...
 
  • Like
Reactions: Cứu mạng@@
Top Bottom