Question

giải phương trình:

a) \(\frac{4}{x-5}+\frac{9}{x+5}=\frac{1}{2}\)

b) \(x^4-4x^2+3=0\)

Mình đang cần gấp ạ

Answer 1

a)

\(\frac{4}{x-5}+\frac{9}{x+5}=\frac{1}{2}\\ \Leftrightarrow\frac{4x+20}{2\cdot\left(x^2-25\right)}+\frac{9x-40}{2\cdot\left(x^2-25\right)}-\frac{x^2-25}{2\cdot\left(x^2-25\right)}=0\\ \Leftrightarrow\frac{4x+20+9x-40-x^2+25}{2\cdot\left(x^2-25\right)}=0\\ \Leftrightarrow\frac{13x+5-x^2}{2\cdot\left(x^2-25\right)}=0\\ \Rightarrow13x+5-x^2=0\\ \Rightarrow x=\left[{}\begin{matrix}-\frac{-13+3\sqrt{21}}{2}\\-\frac{-13-3\sqrt{21}}{2}\end{matrix}\right.\)

b)

\(x^4-4x^2+3=0\\ \Leftrightarrow\left(x^2-3\right)\cdot\left(x^2-1\right)=0\\ \Rightarrow x=\left[{}\begin{matrix}\sqrt{3}\\-\sqrt{3}\\1\\-1\end{matrix}\right.\)

Answer 2

Bài làm

a) \(\frac{4}{x-5}+\frac{9}{x+5}=\frac{1}{2}\) ĐKXĐ: x khác 5, x khác -5

\(\Rightarrow2.4.\left(x+5\right)+9.2\left(x-5\right)=2\left(x+5\right)\left(x-5\right)\)

\(\Leftrightarrow8x+40+18x-90=2x^2-50\)

\(\Leftrightarrow-2x^2+26x=0\)

\(\Leftrightarrow2x\left(-x+13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\-x+13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=13\end{matrix}\right.\)( tmđkxđ )

Vậy S = { 0; 13 }

b) x⁴ - 4x² + 3 = 0

<=> x⁴ - x² - 3x² + 3 = 0

<=> x²( x² - 1 ) - 3( x² - 1 ) = 0

<=> ( x² - 3 )( x² - 1 ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{3}\\x=\pm1\end{matrix}\right.\)

Vậy S = { 3; -3; 1; -1 }