Bài 2:Giải các phương trình sau: a)4x-150=-x+50 b)x/x-1-2x/1-x^2=0 c)5-|x-12|=2x d)(x^2+x)^2+4(x^2+x)=12 16/11/2021 Bởi Josephine Bài 2:Giải các phương trình sau: a)4x-150=-x+50 b)x/x-1-2x/1-x^2=0 c)5-|x-12|=2x d)(x^2+x)^2+4(x^2+x)=12
Đáp án: d. \(\left[ \begin{array}{l}x = 1\\x = – 2\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}a.4x – 150 = – x + 50\\ \to 5x = 200\\ \to x = 40\\b.DK:x \ne \pm 1\\\frac{x}{{x – 1}} – \frac{{2x}}{{1 – {x^2}}} = 0\\ \to \frac{{x\left( {x + 1} \right) + 2x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} = 0\\ \to {x^2} + x + 2x = 0\\ \to \left[ \begin{array}{l}x = 0\\x + 3 = 0\end{array} \right.\\ \to \left[ \begin{array}{l}x = 0\\x = – 3\end{array} \right.\\c.5 – \left| {x – 12} \right| = 2x\\ \to \left| {x – 12} \right| = 5 – 2x\\ \to \left[ \begin{array}{l}x – 12 = 5 – 2x\\x – 12 = – 5 + 2x\end{array} \right.\\ \to \left[ \begin{array}{l}3x = 17\\x = – 7\end{array} \right.\\ \to \left[ \begin{array}{l}x = \frac{{17}}{3}\\x = – 7\end{array} \right.\\d.{({x^2} + x)^2} + 4({x^2} + x) = 12\left( 1 \right)\\Đặt:({x^2} + x) = t\\\left( 1 \right) \to {t^2} + 4t – 12 = 0\\ \to {t^2} – 2t + 6t – 12 = 0\\ \to t\left( {t – 2} \right) + 6\left( {t – 2} \right) = 0\\ \to \left[ \begin{array}{l}t = 2\\t = – 6\end{array} \right.\\ \to \left[ \begin{array}{l}({x^2} + x) = 2\\({x^2} + x) = – 6\end{array} \right.\\ \to \left[ \begin{array}{l}{x^2} – x + 2x – 2 = 0\\{x^2} + x + 6 = 0\left( 2 \right)\end{array} \right.\\ \to \left[ \begin{array}{l}x – 1 = 0\\x + 2 = 0\end{array} \right.\\ \to \left[ \begin{array}{l}x = 1\\x = – 2\end{array} \right.\end{array}\) Phương trình (2) vô nghiệm do Δ<0 Bình luận
Đáp án:
d. \(\left[ \begin{array}{l}
x = 1\\
x = – 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.4x – 150 = – x + 50\\
\to 5x = 200\\
\to x = 40\\
b.DK:x \ne \pm 1\\
\frac{x}{{x – 1}} – \frac{{2x}}{{1 – {x^2}}} = 0\\
\to \frac{{x\left( {x + 1} \right) + 2x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} = 0\\
\to {x^2} + x + 2x = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x + 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = – 3
\end{array} \right.\\
c.5 – \left| {x – 12} \right| = 2x\\
\to \left| {x – 12} \right| = 5 – 2x\\
\to \left[ \begin{array}{l}
x – 12 = 5 – 2x\\
x – 12 = – 5 + 2x
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = 17\\
x = – 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \frac{{17}}{3}\\
x = – 7
\end{array} \right.\\
d.{({x^2} + x)^2} + 4({x^2} + x) = 12\left( 1 \right)\\
Đặt:({x^2} + x) = t\\
\left( 1 \right) \to {t^2} + 4t – 12 = 0\\
\to {t^2} – 2t + 6t – 12 = 0\\
\to t\left( {t – 2} \right) + 6\left( {t – 2} \right) = 0\\
\to \left[ \begin{array}{l}
t = 2\\
t = – 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
({x^2} + x) = 2\\
({x^2} + x) = – 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} – x + 2x – 2 = 0\\
{x^2} + x + 6 = 0\left( 2 \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x – 1 = 0\\
x + 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = – 2
\end{array} \right.
\end{array}\)
Phương trình (2) vô nghiệm do Δ<0
a/ 4x-150=-x+50
<=> 4x+x=50+150
<=> 4x=200
<=> x=50