Toán Lớp 8: tìm x:
1.x^2+5x^2-3x=15
2.x^2-2/2x-3=2
3.(3x-1)(2x+7)+(x+1)(3-6x)=16
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Toán Lớp 8: tìm x:
1.x^2+5x^2-3x=15
2.x^2-2/2x-3=2
3.(3x-1)(2x+7)+(x+1)(3-6x)=16
Comments ( 1 )
1){x^3} + 5{x^2} – 3x = 15\\
\Leftrightarrow {x^2}\left( {x + 5} \right) – 3x – 15 = 0\\
\Leftrightarrow {x^2}\left( {x + 5} \right) – 3\left( {x + 5} \right) = 0\\
\Leftrightarrow \left( {x + 5} \right)\left( {{x^2} – 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 5\\
{x^2} = 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 5\\
x = \sqrt 3 ;x = – \sqrt 3
\end{array} \right.\\
Vậy\,x = – 5;x = – \sqrt 3 ;x = \sqrt 3 \\
2)Dkxd:x \ne \dfrac{3}{2}\\
\dfrac{{{x^2} – 2}}{{2x – 3}} = 2\\
\Leftrightarrow {x^2} – 2 = 4x – 6\\
\Leftrightarrow {x^2} – 4x + 4 = 0\\
\Leftrightarrow {\left( {x – 2} \right)^2} = 0\\
\Leftrightarrow x = 2\left( {tm} \right)\\
Vậy\,x = 2\\
3)\left( {3x – 1} \right)\left( {2x + 7} \right) + \left( {x + 1} \right)\left( {3 – 6x} \right) = 16\\
\Leftrightarrow 6{x^2} + 21x – 2x – 7 + 3x – 6{x^2} + 3 – 6x = 16\\
\Leftrightarrow 16x = 20\\
\Leftrightarrow x = \dfrac{5}{4}\\
Vậy\,x = \dfrac{5}{4}
\end{array}$