Toán Lớp 8: 1. Rút gọn các biểu thức sau:
a) ( x + y )^2 + (x – y )^2
b) ( 2x + 1 ) ( 2x – 1 ) – 4x^2
c) 2 ( x – y ) ( x + y ) + ( x + y )^2 + ( x – y )^2
d) ( x + y )^3 – ( x – y )^3 – 2y^3
bài 2: a) 5x ( x – 3y ) -10y ( x – 3y )
b) -3a^2 + 6ab – 3b^2
c) x^4 -x^3 -x + 1
d) x^2 – 4x + 3
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1)\\
a){\left( {x + y} \right)^2} + {\left( {x – y} \right)^2}\\
= {x^2} + 2xy + {y^2} + {x^2} – 2xy + {y^2}\\
= 2{x^2} + 2{y^2}\\
b)\left( {2x + 1} \right)\left( {2x – 1} \right) – 4{x^2}\\
= {\left( {2x} \right)^2} – 1 – 4{x^2}\\
= 4{x^2} – 1 – 4{x^2}\\
= – 1\\
c)2\left( {x – y} \right)\left( {x + y} \right) + {\left( {x + y} \right)^2} + {\left( {x – y} \right)^2}\\
= {\left( {x + y} \right)^2} + 2\left( {x – y} \right)\left( {x + y} \right) + {\left( {x – y} \right)^2}\\
= {\left( {x + y + x – y} \right)^2}\\
= {\left( {2x} \right)^2}\\
= 4{x^2}\\
d){\left( {x + y} \right)^3} – {\left( {x – y} \right)^3} – 2{y^3}\\
= {x^3} + 3{x^2}y + 3x{y^2} + {y^3}\\
– \left( {{x^3} – 3{x^2}y + 3x{y^2} – {y^3}} \right) – 2{y^3}\\
= 6{x^2}y + 2{y^3} – 2{y^3}\\
= 6{x^2}y\\
B2)\\
a)5x\left( {x – 3y} \right) – 10y\left( {x – 3y} \right)\\
= 5\left( {x – 3y} \right)\left( {x – 2y} \right)\\
b) – 3{a^2} + 6ab – 3{b^2}\\
= – 3\left( {{a^2} – 2ab + {b^2}} \right)\\
= – 3{\left( {a – b} \right)^2}\\
c){x^4} – {x^3} – x + 1\\
= {x^3}\left( {x – 1} \right) – \left( {x – 1} \right)\\
= \left( {x – 1} \right)\left( {{x^3} – 1} \right)\\
= \left( {x – 1} \right)\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)\\
= {\left( {x – 1} \right)^2}\left( {{x^2} + x + 1} \right)\\
d){x^2} – 4x + 3\\
= {x^2} – x – 3x + 3\\
= x\left( {x – 1} \right) – 3\left( {x – 1} \right)\\
= \left( {x – 1} \right)\left( {x – 3} \right)
\end{array}$