Question

Nghiệm không nguyên của phương trình:

\(8\left(x^2+\dfrac{1}{x^2}\right)-34\left(x+\dfrac{1}{x}\right)+51=0\)

Answer 1

Đặt \(t=x+\dfrac{1}{x}\rightarrow t^2=x^2+\dfrac{1}{x^2}+2\Rightarrow x^2+\dfrac{1}{x^2}=t^2-2\)

PT \(\Leftrightarrow8\left(t^2-2\right)-34t+51=0\Leftrightarrow8t^2-34t+35=0\Leftrightarrow8t^2-14t-20t+35=0\Leftrightarrow2t\left(4t-7\right)-5\left(4t-7\right)=0\)\(\Leftrightarrow\left(4t-7\right)\left(2t-5\right)=0\Leftrightarrow\left[\begin{matrix}t=\dfrac{4}{7}\\t=\dfrac{5}{2}\end{matrix}\right.\)

Với \(t=\dfrac{4}{7}\Rightarrow\dfrac{4}{7}=x+\dfrac{1}{x}\Leftrightarrow7x^2-4x+7=0\Leftrightarrow x^2-4x+4+3x^2+7=0\Leftrightarrow\left(x-2\right)^2+3x^2+7=0\left(vn\right)\)Với \(t=\dfrac{5}{2}\Leftrightarrow x+\dfrac{1}{x}=\dfrac{5}{2}\Leftrightarrow2x^2-5x+2=0\Leftrightarrow2x^2-4x-x+2=0\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)=0\) \(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}x=2\left(l\right)\\x=\dfrac{1}{2}\left(n\right)\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{2}\)

Answer 2

Chương mê cc

Answer 3

ủa lộn con ngân mới là ghệ chương chứ

Answer 4

Ta có :

\(8\left(x^2+\frac{1}{x^2}\right)-34\left(x+\frac{1}{x}\right)+51=0\)

=>\(8\left(x+\frac{1}{x}\right)^2-34\left(x+\frac{1}{x}\right)+51=0\)

=>\(\left(x+\frac{1}{x}\right)\left(8x+\frac{8}{x}-34\right)+51=0\)

=>\(\left(x+\frac{1}{x}\right)\left(\frac{8x^2+8}{x}-34\right)+51=0\)

=>\(\frac{8x^3+8x}{x}-34x+\frac{8x^2+8}{x^2}+\frac{-34}{x}+51=0\)

=>\(\frac{8x^2+8}{x^2}+\frac{8x^3-34x^2+8x-34}{x}+51=0\)

=>\(\frac{8x^2+8+8x^4-34x^3+8x^2-34x+51x^2}{x^2}=0\)

=>\(\frac{8x^4-34x^3+67x^2-34x+8}{x^2}=0\)

=>\(8x^4-34x^3+67x^2-34x+8=0\)

Phương trình không có nghiệm nếu nghiệm không nguyên

Answer 5

đặt \(x+\dfrac{1}{x}=t\Rightarrow\left|t\right|\ge2\\ \)\(\Rightarrow x^2+\dfrac{1}{x^2}=t^2-2\)

\(\Leftrightarrow8t^2-16-34t+51=0\Leftrightarrow8t^2-34t+35=0\)

\(\Leftrightarrow\left(8t^2-14t\right)-\left(20t-35\right)=2\left(4t-7\right)-5\left(4t-7\right)=\left(4t-7\right)\left(2t-5\right)=0\)\(\left[\begin{matrix}t=\dfrac{7}{4}< 2\Rightarrow loai\\t=\dfrac{5}{2}>2\Rightarrow\left(nhan\right)\end{matrix}\right.\)

\(x+\dfrac{1}{x}=\dfrac{5}{2}\Leftrightarrow x^2-\dfrac{5}{2}x+1=0\Leftrightarrow\left(x-\dfrac{5}{4}\right)^2=\dfrac{9}{16}\)

\(\left[\begin{matrix}x=\dfrac{5-3}{4}=\dfrac{1}{2}\\x=\dfrac{5+3}{4}=2\left(loai\right)\end{matrix}\right.\)

Kết luận: x=1/2