Toán Lớp 8: Giải các phương trình sau:
a)x(x-1)(x+1)(x+2)=24
b)$x^{4}$ +$2x^{3}$- $2x^{2}$ +2x-3=0
c)$x^{3}$ -$6x^{2}$ +12x+19=0
d)$x^{3}$ +$5x^{2}$ -4x-20=0
e)$x^{4}$ -$5x^{3}$-$12x^{2}$ -5x+1=0
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 8: Giải các phương trình sau:
a)x(x-1)(x+1)(x+2)=24
b)$x^{4}$ +$2x^{3}$- $2x^{2}$ +2x-3=0
c)$x^{3}$ -$6x^{2}$ +12x+19=0
d)$x^{3}$ +$5x^{2}$ -4x-20=0
e)$x^{4}$ -$5x^{3}$-$12x^{2}$ -5x+1=0
Comments ( 1 )
a)x\left( {x – 1} \right)\left( {x + 1} \right)\left( {x + 2} \right) = 24\\
\Leftrightarrow x\left( {x + 1} \right)\left( {x – 1} \right)\left( {x + 2} \right) = 24\\
\Leftrightarrow \left( {{x^2} + x} \right)\left( {{x^2} + x – 2} \right) = 24\\
Dat:{x^2} + x = a\\
\Leftrightarrow a\left( {a – 2} \right) = 24\\
\Leftrightarrow {a^2} – 2a – 24 = 0\\
\Leftrightarrow {a^2} + 4a – 6a – 24 = 0\\
\Leftrightarrow \left( {a + 4} \right)\left( {a – 6} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
a + 4 = 0\\
a – 6 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + x + 4 = 0\left( {vn} \right)\\
{x^2} + x – 6 = 0
\end{array} \right.\\
\Leftrightarrow \left( {x – 2} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow x = 2;x = – 3\\
Vậy\,x = 2;x = – 3\\
b){x^4} + 2{x^3} – 2{x^2} + 2x – 3 = 0\\
\Leftrightarrow {x^4} – {x^3} + 3{x^3} – 3{x^2} + {x^2} – x + 3x – 3 = 0\\
\Leftrightarrow {x^3}\left( {x – 1} \right) + 3{x^2}\left( {x – 1} \right) + x\left( {x – 1} \right) + 3\left( {x – 1} \right) = 0\\
\Leftrightarrow \left( {x – 1} \right)\left( {{x^3} + 3{x^2} + x + 3} \right) = 0\\
\Leftrightarrow \left( {x – 1} \right)\left( {x + 3} \right)\left( {{x^2} + 1} \right) = 0\\
\Leftrightarrow x = 1;x = – 3\\
Vậy\,x = 1;x = – 3\\
c){x^3} – 6{x^2} + 12x + 19 = 0\\
\Leftrightarrow {x^3} + {x^2} – 7{x^2} – 7x + 19x + 19 = 0\\
\Leftrightarrow {x^2}\left( {x + 1} \right) – 7x\left( {x + 1} \right) + 19\left( {x + 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {{x^2} – 7x + 19} \right) = 0\\
\Leftrightarrow x + 1 = 0\\
\Leftrightarrow x = – 1\\
Vậy\,x = – 1\\
d){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 x = – 5;x = – 2;x = 2\\
Vậy\,x = – 5;x = – 2;x = 2\\
e){x^4} – 5{x^3} – 12{x^2} – 5x + 1 = 0\\
\Leftrightarrow {x^4} + {x^3} – 6{x^3} – 6{x^2} – 6{x^2} – 6x + x + 1 = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {{x^3} – 6{x^2} – 6x + 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {{x^3} + {x^2} – 7{x^2} – 7x + x + 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {x + 1} \right)\left( {{x^2} – 7x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 1\\
{x^2} – 2.x.\dfrac{7}{2} + \dfrac{{49}}{4} = \dfrac{{45}}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 1\\
{\left( {x – \dfrac{7}{2}} \right)^2} = {\left( {\dfrac{{3\sqrt 5 }}{2}} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 1\\
x = \dfrac{{7 \pm 3\sqrt 5 }}{2}
\end{array} \right.\\
Vậy\,x = – 1;x = \dfrac{{7 \pm 3\sqrt 5 }}{2}
\end{array}$