Toán Lớp 8: Tìm x biết :
a) 36-9(x+1)²=0
b) 36(x-2)²-4(x+1)²=0
c) (x+2)²-(x-2)(x+2)=0
d) x³-8=(x-2)³
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Toán Lớp 8: Tìm x biết :
a) 36-9(x+1)²=0
b) 36(x-2)²-4(x+1)²=0
c) (x+2)²-(x-2)(x+2)=0
d) x³-8=(x-2)³
Comments ( 1 )
a,\\
\left[ \begin{array}{l}
x = 1\\
x = – 3
\end{array} \right.\\
b,\\
\left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x = \dfrac{5}{4}
\end{array} \right.\\
c,\\
x = – 2\\
d,\\
\left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.
\end{array}\)
a,\\
36 – 9{\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow 9{\left( {x + 1} \right)^2} = 36\\
\Leftrightarrow {\left( {x + 1} \right)^2} = 4\\
\Leftrightarrow {\left( {x + 1} \right)^2} – 4 = 0\\
\Leftrightarrow {\left( {x + 1} \right)^2} – {2^2} = 0\\
\Leftrightarrow \left( {x + 1 – 2} \right)\left( {x + 1 + 2} \right) = 0\\
\Leftrightarrow \left( {x – 1} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 1 = 0\\
x + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = – 3
\end{array} \right.\\
b,\\
36{\left( {x – 2} \right)^2} – 4.{\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow 4.\left[ {9{{\left( {x – 2} \right)}^2} – {{\left( {x + 1} \right)}^2}} \right] = 0\\
\Leftrightarrow 9{\left( {x – 2} \right)^2} – {\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow {3^2}.{\left( {x – 2} \right)^2} – {\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow {\left[ {3.\left( {x – 2} \right)} \right]^2} – {\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow {\left( {3x – 6} \right)^2} – {\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow \left[ {\left( {3x – 6} \right) – \left( {x + 1} \right)} \right].\left[ {\left( {3x – 6} \right) + \left( {x + 1} \right)} \right] = 0\\
\Leftrightarrow \left( {2x – 7} \right).\left( {4x – 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x – 7 = 0\\
4x – 5 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x = \dfrac{5}{4}
\end{array} \right.\\
c,\\
{\left( {x + 2} \right)^2} – \left( {x – 2} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left( {x + 2} \right).\left[ {\left( {x + 2} \right) – \left( {x – 2} \right)} \right] = 0\\
\Leftrightarrow \left( {x + 2} \right).4 = 0\\
\Leftrightarrow x + 2 = 0\\
\Leftrightarrow x = – 2\\
d,\\
{x^3} – 8 = {\left( {x – 2} \right)^3}\\
\Leftrightarrow {x^3} – {2^3} = {\left( {x – 2} \right)^3}\\
\Leftrightarrow \left( {x – 2} \right).\left( {{x^2} + x.2 + {2^2}} \right) – {\left( {x – 2} \right)^3} = 0\\
\Leftrightarrow \left( {x – 2} \right).\left( {{x^2} + 2x + 4} \right) – {\left( {x – 2} \right)^3} = 0\\
\Leftrightarrow \left( {x – 2} \right).\left[ {\left( {{x^2} + 2x + 4} \right) – {{\left( {x – 2} \right)}^2}} \right] = 0\\
\Leftrightarrow \left( {x – 2} \right).\left[ {\left( {{x^2} + 2x + 4} \right) – \left( {{x^2} – 2.x.2 + {2^2}} \right)} \right] = 0\\
\Leftrightarrow \left( {x – 2} \right).\left[ {\left( {{x^2} + 2x + 4} \right) – \left( {{x^2} – 4x + 4} \right)} \right] = 0\\
\Leftrightarrow \left( {x – 2} \right).6x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 2 = 0\\
6x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.
\end{array}\)