Toán Lớp 8: câu 1:
a. (x+y)^2-(x-y)^2
b. 5x.(x-2y)+2.(2y-x)^2
c. (4x-8).(x^2+6)-(4x-8).(x+7)+9.(8-4x)
d. 7x.(y-4)^2-(4-y)^3
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Toán Lớp 8: câu 1:
a. (x+y)^2-(x-y)^2
b. 5x.(x-2y)+2.(2y-x)^2
c. (4x-8).(x^2+6)-(4x-8).(x+7)+9.(8-4x)
d. 7x.(y-4)^2-(4-y)^3
Comments ( 1 )
a){\left( {x + y} \right)^2} – {\left( {x – y} \right)^2}\\
= \left( {x + y + x – y} \right)\left( {x + y – x + y} \right)\\
= 2x.2y\\
= 4xy\\
b)5x\left( {x – 2y} \right) + 2{\left( {2y – x} \right)^2}\\
= 5x\left( {x – 2y} \right) + 2{\left( {x – 2y} \right)^2}\\
= \left( {x – 2y} \right)\left( {5x + 2.\left( {x – 2y} \right)} \right)\\
= \left( {x – 2y} \right)\left( {7x – 4y} \right)\\
c)\left( {4x – 8} \right)\left( {{x^2} + 6} \right) – \left( {4x – 8} \right)\left( {x + 7} \right)\\
+ 9\left( {8 – 4x} \right)\\
= \left( {4x – 8} \right)\left( {{x^2} + 6 – x – 7 – 9} \right)\\
= \left( {4x – 8} \right)\left( {{x^2} – x – 10} \right)\\
= 2\left( {x – 2} \right)\left( {{x^2} – x – 10} \right)\\
d)7x{\left( {y – 4} \right)^2} – {\left( {4 – y} \right)^3}\\
= 7x{\left( {y – 4} \right)^2} + {\left( {y – 4} \right)^3}\\
= {\left( {y – 4} \right)^2}\left( {7x + y – 4} \right)
\end{array}$