Toán Lớp 8: X ³ – 5x^2 + 8x – 4
4x ² -25 – (2x – 5) (2x + 3)
9x ² y ² – ( x ² + y ² – z ²) ²
(X ² + x) ² + 4x ² +4x -12
-14x ² + 39x -10
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Toán Lớp 8: X ³ – 5x^2 + 8x – 4
4x ² -25 – (2x – 5) (2x + 3)
9x ² y ² – ( x ² + y ² – z ²) ²
(X ² + x) ² + 4x ² +4x -12
-14x ² + 39x -10
Comments ( 1 )
a){x^3} – 5{x^2} + 8x – 4\\
= {x^3} – 2{x^2} – 3{x^2} + 6x + 2x – 4\\
= {x^2}\left( {x – 2} \right) – 3x\left( {x – 2} \right) + 2\left( {x – 2} \right)\\
= \left( {x – 2} \right)\left( {{x^2} – 3x + 2} \right)\\
= \left( {x – 2} \right).\left( {{x^2} – 2x – x + 2} \right)\\
= \left( {x – 2} \right)\left( {x – 2} \right)\left( {x – 1} \right)\\
= {\left( {x – 2} \right)^2}.\left( {x – 1} \right)\\
b)4{x^2} – 25 – \left( {2x – 5} \right)\left( {2x + 3} \right)\\
= \left( {2x – 5} \right)\left( {2x + 5} \right) – \left( {2x – 5} \right)\left( {2x + 3} \right)\\
= \left( {2x – 5} \right)\left( {2x + 5 – 2x – 3} \right)\\
= 2\left( {2x + 5} \right)\\
c){\left( {{x^2} + x} \right)^2} + 4{x^2} + 4x – 12\\
= {\left( {{x^2} + x} \right)^2} + 4\left( {{x^2} + x} \right) – 12\\
Đặt:\left( {{x^2} + x} \right) = a\\
\Leftrightarrow {a^2} + 4a – 12\\
= {a^2} + 6a – 2a – 12\\
= \left( {a + 6} \right)\left( {a – 2} \right)\\
= \left( {{x^2} + x + 6} \right)\left( {{x^2} + x – 2} \right)\\
= \left( {{x^2} + x + 6} \right)\left( {x – 1} \right)\left( {x + 2} \right)\\
e) – 14{x^2} + 39x – 10\\
= – 14{x^2} + 4x + 35x – 10\\
= – 2x\left( {7x – 2} \right) + 5\left( {7x – 2} \right)\\
= \left( {7x – 2} \right)\left( { – 2x + 5} \right)
\end{array}$