Giải phương trình: a, x ² (x + 4,5) = 13,5 b, 4 (x + 1) ² - 9 (x - 1) ² = 0 mấy bạn chuyên toán giúp mình với:< nãy mới bị mấy bạn spam:<<
2 câu trả lời
#andy
\[\begin{array}{l}
a){x^2}\left( {x + 4,5} \right) = 13,5\\
\Leftrightarrow {x^3} + 4,5{x^2} = 13,5\\
\Leftrightarrow {x^3} + 3{x^2} + 1,5{x^2} + 4,5x - 4,5x - 13,5 = 0\\
\Leftrightarrow {x^2}\left( {x + 3} \right) + 1,5x\left( {x + 3} \right) - 4,5\left( {x + 3} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {{x^2} + 1,5x - 4,5} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {{x^2} + 3x - 1,5x - 4,5} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left[ {x\left( {x + 3} \right) - 1,5\left( {x + 3} \right)} \right] = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {x + 3} \right)\left( {x - 1,5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x - 1,5 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 1,5
\end{array} \right.\\
\Rightarrow S = \left\{ { - 3;1,5} \right\}\\
b)4{\left( {x + 1} \right)^2} - 9{\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow \left[ {2\left( {x + 1} \right){\rm{ }} - {\rm{ }}3\left( {x - 1} \right)} \right]{\rm{ }}\left[ {2\left( {x + 1} \right){\rm{ }} + {\rm{ }}3\left( {x - 1} \right)} \right] = 0\\
\Leftrightarrow \left( {2x + 2{\rm{ }} - {\rm{ }}3x{\rm{ }} + 3} \right){\rm{ }}\left( {2x + 2{\rm{ }} + {\rm{ }}3x - 3} \right) = 0\\
\Leftrightarrow \left( { - x{\rm{ }} + 5} \right){\rm{ }}\left( {5x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
- x + 5 = 0\\
5x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = \dfrac{1}{5}
\end{array} \right.\\
\Rightarrow S = \left\{ {5;\dfrac{1}{5}} \right\}\\
\end{array}\]
`a, x ² (x + 4,5) = 13,5`
`<=>x^3+4,5x^2-13,5=0`
`<=>x^3+3x^2+1,5x^2+4,5x-4,5x-13,5=0`
`<=>(x^3+3x^2)+(1,5x^2+4,5x)-(4,5x+13,5)=0`
`<=>x^2(x+3)+1,5x(x+3)-4,5(x+3)=0`
`<=>(x+3)(x^2+1,5-4,5)=0`
`<=>(x+3)(x+3)(x-1,5)=0`
`<=>(x+3)^2.(x-1,5)=0`
$⇔\left[\begin{matrix} (x+3)^2=0\\ x-1,5=0\end{matrix}\right.$
$⇔\left[\begin{matrix} x+3=0\\ x-1,5=0\end{matrix}\right.$
$⇔\left[\begin{matrix} x=-3\\ x=1,5\end{matrix}\right.$
Vậy `x in {1,5;-3}`
`b, 4 (x + 1) ² - 9 (x - 1) ² = 0`
`<=> 4 (x + 1) ² = 9 (x - 1) ²`
`<=>[2(x+1)]^2=[3(x-1)]^2`
`<=>(2x+2)^2=(3x-3)^2`
$⇔\left[\begin{matrix} 2x+2=3x-3\\ 2x+2=3-3x\end{matrix}\right.$
$⇔\left[\begin{matrix} -x=-5\\ 5x=1\end{matrix}\right.$
$⇔\left[\begin{matrix} x=5\\ x=\dfrac{1}{5}\end{matrix}\right.$
Vậy `x in {5;1/5}`