Toán Lớp 8: a)16x^3y+1/4yz^3
b)x^4-x^2+2x-1
c)x^4+2x^3-6x-9=0
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Toán Lớp 8: a)16x^3y+1/4yz^3
b)x^4-x^2+2x-1
c)x^4+2x^3-6x-9=0
gIÚP MÌNH NGAY VỚI Ạ
Comments ( 1 )
a)16{x^3}y + \frac{1}{4}y{z^3}\\
= 2y\left( {8{x^3} + \frac{1}{8}{z^3}} \right)\\
= 2y\left( {2x + \frac{1}{2}z} \right)\left( {4{x^2} – xz + \frac{1}{4}{z^2}} \right)\\
b){x^4} – {x^2} + 2x – 1\\
= {x^4} – \left( {{x^2} – 2x + 1} \right)\\
= {x^4} – {\left( {x – 1} \right)^2}\\
= \left( {{x^2} + x – 1} \right)\left( {{x^2} – x + 1} \right)\\
c){x^4} + 2{x^3} – 6x – 9 = 0\\
\Leftrightarrow {x^4} – 9 + 2{x^3} – 6x = 0\\
\Leftrightarrow \left( {{x^2} – 3} \right)\left( {{x^2} + 3} \right) – 2x\left( {{x^2} – 3} \right) = 0\\
\Leftrightarrow \left( {{x^2} – 3} \right)\left( {{x^2} + 3 – 2x} \right) = 0\\
\Leftrightarrow \left( {x – \sqrt 3 } \right)\left( {x + \sqrt 3 } \right)\left( {x – 1} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \sqrt 3 \\
x = – \sqrt 3 \\
x = 1\\
x = – 3
\end{array} \right.\\
Vậy\,x \in \left\{ { – 3; – \sqrt 3 ;1;\sqrt 3 } \right\}
\end{array}$