Toán Lớp 8: Tìm x, biết:
a)(x+3)^2-x(x-1)=0
b)(x-2)^2-3(2-x)=0
c)x^2-x=2
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Toán Lớp 8: Tìm x, biết:
a)(x+3)^2-x(x-1)=0
b)(x-2)^2-3(2-x)=0
c)x^2-x=2
Comments ( 1 )
a)x = – \dfrac{9}{7}\\
b)\left[ \begin{array}{l}
x = 2\\
x = – 1
\end{array} \right.\\
c)\left[ \begin{array}{l}
x = 2\\
x = – 1
\end{array} \right.
\end{array}\)
a){(x + 3)^2} – x(x – 1) = 0\\
\to {x^2} + 6x + 9 – {x^2} + x = 0\\
\to 7x + 9 = 0\\
\to x = – \dfrac{9}{7}\\
b){(x – 2)^2} – 3(2 – x) = 0\\
\to {(x – 2)^2} + 3\left( {x – 2} \right) = 0\\
\to \left( {x – 2} \right)\left( {x – 2 + 3} \right) = 0\\
\to \left[ \begin{array}{l}
x – 2 = 0\\
x + 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = – 1
\end{array} \right.\\
c){x^2} – x = 2\\
\to {x^2} – x – 2 = 0\\
\to {x^2} – 2x + x – 2 = 0\\
\to x\left( {x – 2} \right) + \left( {x – 2} \right) = 0\\
\to \left( {x – 2} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = – 1
\end{array} \right.
\end{array}\)