Đáp án:
\(\begin{array}{l}
a,\\
9.\left( {5x + 1} \right)\left( {x - 1} \right)\\
b,\\
4.\left( {2y - 17} \right).\left( {5y - 11} \right)\\
c,\\
{\left( {\dfrac{1}{2}x - 5y} \right)^2}\\
d,\\
\left( {y + z} \right).\left( {x + 3} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{\left( {7x - 4} \right)^2} - {\left( {2x - 5} \right)^2}\\
= \left[ {\left( {7x - 4} \right) - \left( {2x - 5} \right)} \right].\left[ {\left( {7x - 4} \right) + \left( {2x - 5} \right)} \right]\\
= \left( {7x - 4 - 2x + 5} \right).\left( {7x - 4 + 2x - 5} \right)\\
= \left( {5x + 1} \right).\left( {9x - 9} \right)\\
= 9.\left( {5x + 1} \right)\left( {x - 1} \right)\\
b,\\
49{\left( {y - 4} \right)^2} - 9.{\left( {y + 2} \right)^2}\\
= {7^2}.{\left( {y - 4} \right)^2} - {3^2}.{\left( {y + 2} \right)^2}\\
= {\left[ {7.\left( {y - 4} \right)} \right]^2} - {\left[ {3.\left( {y + 2} \right)} \right]^2}\\
= {\left( {7y - 28} \right)^2} - {\left( {3y + 6} \right)^2}\\
= \left[ {\left( {7y - 28} \right) - \left( {3y + 6} \right)} \right].\left[ {\left( {7y - 28} \right) + \left( {3y + 6} \right)} \right]\\
= \left( {7y - 28 - 3y - 6} \right).\left( {7y - 28 + 3y + 6} \right)\\
= \left( {4y - 34} \right).\left( {10y - 22} \right)\\
= 2.\left( {2y - 17} \right).2.\left( {5y - 11} \right)\\
= 4.\left( {2y - 17} \right).\left( {5y - 11} \right)\\
c,\\
\dfrac{1}{4}{x^2} + 25{y^2} - 5xy\\
= \dfrac{1}{4}{x^2} - 5xy + 25{y^2}\\
= {\left( {\dfrac{1}{2}x} \right)^2} - 2.\dfrac{1}{2}x.5y + {\left( {5y} \right)^2}\\
= {\left( {\dfrac{1}{2}x - 5y} \right)^2}\\
d,\\
xy + xz + 3y + 3z\\
= \left( {xy + xz} \right) + \left( {3y + 3z} \right)\\
= x.\left( {y + z} \right) + 3.\left( {y + z} \right)\\
= \left( {y + z} \right).\left( {x + 3} \right)
\end{array}\)
