các cậu giúp tớ các bài vè tìm x nha: a)x^3-8=(x-2^3) b)x^3+5x^2-4x-20=0 c)x^3-4x^2+4x=0 d)x^2-25+2(x+5)=0 e)x^2(x-2)+7x=14 g)x^2-25=6x-9 h)5x(x-3)-2x+6=0
các cậu giúp tớ các bài vè tìm x nha: a)x^3-8=(x-2^3) b)x^3+5x^2-4x-20=0 c)x^3-4x^2+4x=0 d)x^2-25+2(x+5)=0 e)x^2(x-2)+7x=14 g)x^2-25=6x-9 h)5x(x-3)-2x+6=0
Đáp án:
\(\begin{array}{l}
a)\,\,\,\left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right. & & & b)\,\,\left[ \begin{array}{l}
x = – 5\\
x = 2\\
x = – 2
\end{array} \right.\\
c)\,\,\left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right. & & & d)\,\,\left[ \begin{array}{l}
x = – 5\\
x = 3
\end{array} \right.\\
e)\,\,x = 2 & & & g)\,\,\left[ \begin{array}{l}
x = 8\\
x = – 2
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\,\,\,{x^3} – 8 = {\left( {x – 2} \right)^3}\\
\Leftrightarrow \left( {x – 2} \right)\left( {{x^2} + 2x + 4} \right) – \left( {x – 2} \right)\left( {{x^2} – 4x + 4} \right) = 0\\
\Leftrightarrow \left( {x – 2} \right)\left( {{x^2} + 2x + 4 – {x^2} + 4x – 4} \right) = 0\\
\Leftrightarrow \left( {x – 2} \right)6x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
6x = 0\\
x – 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right..\\
b)\,\,{x^3} + 5{x^2} – 4x – 20 = 0\\
\Leftrightarrow {x^2}\left( {x + 5} \right) – 4\left( {x + 5} \right) = 0\\
\Leftrightarrow \left( {x + 5} \right)\left( {{x^2} – 4} \right) = 0\\
\Leftrightarrow \left( {x + 5} \right)\left( {x – 2} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 5 = 0\\
x – 2 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – 5\\
x = 2\\
x = – 2
\end{array} \right..\\
c)\,\,{x^3} – 4{x^2} + 4x = 0\\
\Leftrightarrow x\left( {{x^2} – 4x + 4} \right) = 0\\
\Leftrightarrow x{\left( {x – 2} \right)^2} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x – 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right..\\
d)\,\,{x^2} – 25 + 2\left( {x + 5} \right) = 0\\
\Leftrightarrow \left( {x – 5} \right)\left( {x + 5} \right) + 2\left( {x + 5} \right) = 0\\
\Leftrightarrow \left( {x + 5} \right)\left( {x – 5 + 2} \right) = 0\\
\Leftrightarrow \left( {x + 5} \right)\left( {x – 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 5 = 0\\
x – 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – 5\\
x = 3
\end{array} \right..\\
e)\,\,\,{x^2}\left( {x – 2} \right) + 7x = 14\\
\Leftrightarrow {x^2}\left( {x – 2} \right) + 7x – 14 = 0\\
\Leftrightarrow {x^2}\left( {x – 2} \right) + 7\left( {x – 2} \right) = 0\\
\Leftrightarrow \left( {x – 2} \right)\left( {{x^2} + 7} \right) = 0\\
\Leftrightarrow x – 2 = 0\,\,\,\,\,\left( {{x^2} + 7 > 0\,\,\,\forall x} \right)\\
\Leftrightarrow x = 2.\\
g)\,\,{x^2} – 25 = 6x – 9\\
\Leftrightarrow {x^2} – 6x + 9 – 25 = 0\\
\Leftrightarrow {\left( {x – 3} \right)^2} – {5^2} = 0\\
\Leftrightarrow \left( {x – 3 – 5} \right)\left( {x – 3 + 5} \right) = 0\\
\Leftrightarrow \left( {x – 8} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 8 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x = – 2
\end{array} \right..
\end{array}\)