Register Now

Login

Lost Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Tìm x biết : $\frac{4}{2x^{2}+6x}$ + $\frac{x^{2}+4}{x^{2}-9}$ – $\frac{1}{3-x}$ = 1

Toán Lớp 8: Tìm x biết :
$\frac{4}{2x^{2}+6x}$ + $\frac{x^{2}+4}{x^{2}-9}$ – $\frac{1}{3-x}$ = 1

Comments ( 1 )

  1. Giải đáp:
    \( \left[ \begin{array}{l}
    x =  – 9 + \sqrt {87} \\
    x =  – 9 – \sqrt {87} 
    \end{array} \right.\)
    Lời giải và giải thích chi tiết:
    \(\frac{4}{{2{x^2} + 6x}} + \frac{{{x^2} + 4}}{{{x^2} – 9}} – \frac{1}{{3 – x}} = 1\)
    Điều kiện: \(x \ne  \pm 3\)
    \(\begin{array}{l}\frac{4}{{2{x^2} + 6x}} + \frac{{{x^2} + 4}}{{{x^2} – 9}} – \frac{1}{{3 – x}} = 1\\ \Leftrightarrow \frac{4}{{2x\left( {x + 3} \right)}} + \frac{{{x^2} + 4}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} + \frac{1}{{x – 3}} = 1\\ \Leftrightarrow \frac{2}{{x\left( {x + 3} \right)}} + \frac{{{x^2} + 4}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} + \frac{1}{{x – 3}} = 1\\ \Rightarrow 2\left( {x – 3} \right) + x\left( {{x^2} + 4} \right) + x\left( {x + 3} \right) = x\left( {x + 3} \right)\left( {x – 3} \right)\\ \Leftrightarrow 2x – 6 + {x^3} + 4x + {x^2} + 3x = {x^3} – 9x\\ \Leftrightarrow {x^3} + {x^2} + 9x – 6 = {x^3} – 9x\\ \Leftrightarrow {x^2} + 18x – 6 = 0\\ \Leftrightarrow {x^2} + 18x + 81 = 6 + 81\\ \Leftrightarrow {\left( {x + 9} \right)^2} = 87\\ \Leftrightarrow \left[ \begin{array}{l}x + 9 = \sqrt {87} \\x – 9 =  – \sqrt {87} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  – 9 + \sqrt {87} \\x =  – 9 – \sqrt {87} \end{array} \right.\end{array}\)
    Vậy \(\left[ \begin{array}{l}x =  – 9 + \sqrt {87} \\x =  – 9 – \sqrt {87} \end{array} \right..\)

Leave a reply

222-9+11+12:2*14+14 = ? ( )