Giúp mik vs ạ! a) 3(x – 2)=7(10 – 4x) b) 2x^3+5x^2+3x=0 c) x+5/4 – 2x-3/3 = 6x-1/2+2x-1/12 d) 3/x-3 + 1/x+3 = 6/x^2-9 e) (x-1)(x-3)(x+5)(x+7)-297=0

Đáp án:

e) \(\left[ \begin{array}{l}
x = 4\\
x =  – 8
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a)3x – 6 = 70 – 28x\\
 \to 31x = 76\\
 \to x = \dfrac{{76}}{{31}}\\
b)2{x^3} + 5{x^2} + 3x = 0\\
 \to x\left( {2{x^2} + 5x + 3} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 0\\
2{x^2} + 5x + 3 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 0\\
\left( {x + 1} \right)\left( {2x + 3} \right) = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 0\\
x =  – 1\\
x =  – \dfrac{3}{2}
\end{array} \right.\\
c)\dfrac{{x + 5}}{4} – \dfrac{{2x – 3}}{3} = \dfrac{{6x – 1}}{2} + \dfrac{{2x – 1}}{{12}}\\
 \to \dfrac{{3\left( {x + 5} \right) – 4\left( {2x – 3} \right) – 6\left( {6x – 1} \right) – 2x + 1}}{{12}} = 0\\
 \to 3x + 15 – 8x + 12 – 36x + 6 – 2x + 1 = 0\\
 \to  – 43x + 34 = 0\\
 \to x = \dfrac{{34}}{{43}}\\
d)DK:x \ne  \pm 3\\
\dfrac{{3\left( {x + 3} \right) + x – 3 – 6}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} = 0\\
 \to 3x + 9 + x – 9 = 0\\
 \to x = 0\\
e)\left( {x – 1} \right)\left( {x + 5} \right)\left( {x – 3} \right)\left( {x + 7} \right) – 297 = 0\\
 \to \left( {{x^2} + 4x – 5} \right)\left( {{x^2} + 4x – 21} \right) – 297 = 0\\
Đặt:{x^2} + 4x – 5 = t\\
Pt \to t\left( {t – 16} \right) – 297 = 0\\
 \to {t^2} – 16t – 297 = 0\\
 \to \left[ \begin{array}{l}
t = 27\\
t =  – 11
\end{array} \right.\\
 \to \left[ \begin{array}{l}
{x^2} + 4x – 5 = 27\\
{x^2} + 4x – 5 =  – 11\left( {vô nghiệm} \right)
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x =  – 8
\end{array} \right.
\end{array}\)