Toán Lớp 8: tìm x
a, 2x(x + 3) + 3x – 9 = 0
b, x^2 – ( x+ 5)( x- 2) = 18
c, ( 7x – 9)^2 – ( 5x + 4) ^2 = 0
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Toán Lớp 8: tìm x
a, 2x(x + 3) + 3x – 9 = 0
b, x^2 – ( x+ 5)( x- 2) = 18
c, ( 7x – 9)^2 – ( 5x + 4) ^2 = 0
Comments ( 1 )
a)\left[ \begin{array}{l}
x = \dfrac{{ – 9 + 3\sqrt {17} }}{4}\\
x = \dfrac{{ – 9 – 3\sqrt {17} }}{4}
\end{array} \right.\\
b)x = – \dfrac{8}{3}\\
c)\left[ \begin{array}{l}
x = \dfrac{{13}}{2}\\
x = \dfrac{5}{{12}}
\end{array} \right.
\end{array}\)
a)2x\left( {x + 3} \right) + 3x – 9 = 0\\
\to 2{x^2} + 6x + 3x – 9 = 0\\
\to 2{x^2} + 9x – 9 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{ – 9 + 3\sqrt {17} }}{4}\\
x = \dfrac{{ – 9 – 3\sqrt {17} }}{4}
\end{array} \right.\\
b){x^2} – (x + 5)(x – 2) = 18\\
\to {x^2} – {x^2} – 3x + 10 = 18\\
\to – 3x – 8 = 0\\
\to x = – \dfrac{8}{3}\\
c){(7x – 9)^2} – {(5x + 4)^2} = 0\\
\to {(7x – 9)^2} = {(5x + 4)^2}\\
\to \left| {7x – 9} \right| = \left| {5x + 4} \right|\\
\to \left[ \begin{array}{l}
7x – 9 = 5x + 4\left( {DK:x \ge \dfrac{9}{7}} \right)\\
7x – 9 = – 5x – 4\left( {DK:x < \dfrac{9}{7}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 13\\
12x = 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{13}}{2}\\
x = \dfrac{5}{{12}}
\end{array} \right.
\end{array}\)