Toán Lớp 8: Phương pháp tách hạng tử:
1. x$x^{2}$+x-2
2. x$x^{2}$-5xy-14y$x^{2}$
3. x$x^{4}$-10x$x^{2}$+9
4. x$x^{2}$+7x+12
5.x$x^{2}$-4x+8
6.x$x^{4}$+x$x^{2}$-2
7.2x$x^{2}$+2xy-15y$x^{2}$
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Toán Lớp 8: Phương pháp tách hạng tử:
1. x$x^{2}$+x-2
2. x$x^{2}$-5xy-14y$x^{2}$
3. x$x^{4}$-10x$x^{2}$+9
4. x$x^{2}$+7x+12
5.x$x^{2}$-4x+8
6.x$x^{4}$+x$x^{2}$-2
7.2x$x^{2}$+2xy-15y$x^{2}$
.
.
Comments ( 1 )
1){x^2} + x – 2\\
= {x^2} + 2x – x – 2\\
= x\left( {x + 2} \right) – \left( {x + 2} \right)\\
= \left( {x + 2} \right)\left( {x – 1} \right)\\
2){x^2} – 5xy – 14{y^2}\\
= {x^2} + 2xy – 7xy – 14{y^2}\\
= x\left( {x + 2y} \right) – 7y\left( {x + 2y} \right)\\
= \left( {x + 2y} \right)\left( {x – 7y} \right)\\
3){x^4} – 10{x^2} + 9\\
= {x^4} – {x^2} – 9{x^2} + 9\\
= \left( {{x^2} – 1} \right)\left( {{x^2} – 9} \right)\\
= \left( {x – 1} \right)\left( {x + 1} \right)\left( {x – 3} \right)\left( {x + 3} \right)\\
4){x^2} + 7x + 12\\
= {x^2} + 3x + 4x + 12\\
= \left( {x + 3} \right)\left( {x + 4} \right)\\
5){x^2} – 4x + 8\\
= {x^2} – 4x + 4 + 4\\
= {\left( {x – 2} \right)^2} + 4\\
6){x^4} + {x^2} – 2\\
= {x^4} + 2{x^2} – {x^2} – 2\\
= \left( {{x^2} + 2} \right)\left( {{x^2} – 1} \right)\\
= \left( {{x^2} + 2} \right)\left( {x + 1} \right)\left( {x – 1} \right)\\
7)2{x^2} + xy – 15{y^2}\\
= 2{x^2} + 6xy – 5xy – 15{y^2}\\
= \left( {x + 3y} \right)\left( {2x – 5y} \right)
\end{array}$