by Rút gọn biểu thức:
\(a)A = \left( {{x^2} – 3x + 9} \right)\left( {x + 3} \right) – \left( {54 + {x^3}} \right)\)
\(b)B = {\left( {x – 1} \right)^2} + 2\left( {x – 1} \right)\left( {x + 1} \right) + {\left( {x + 1} \right)^2} + 5\)
\(c)C = 9{x^2} – 2xy + \frac{1}{9}{y^2} – 2\left( {3x – \frac{1}{3}y} \right)\left( {3x + \frac{1}{3}y} \right) + {\left( {3x + \frac{1}{3}y} \right)^2}\)
A. \( a) 27\)
\( b) 4{x^2} + 8\)
\( c) \frac{4}{9}{y^2}\)
B. \( a) -27\)
\( b) 4{x^2} + 5\)
\( c) \frac{4}{9}{y^2}\)
C. \( a) 81\)
\( b) 4{x^2} + 5\)
\( c) -\frac{2}{3}{y^2}\)
D. \( a) -27\)
\( b) 4{x^2} + 5\)
\( c) \frac{2}{3}{y^2}\)
Hướng dẫn Chọn đáp án là: B
Lời giải chi tiết:
Hướng dẫn giải chi tiết
\(\begin{array}{l}a)\;A = \left( {{x^2} – 3x + 9} \right)\left( {x + 3} \right) – \left( {54 + {x^3}} \right)\\A = \left( {{x^2} – 3x + {3^2}} \right)\left( {x + 3} \right) – \left( {54 + {x^3}} \right)\\A = {x^3} + {3^3} – 54 – {x^3}\\A = 27 – 54 = – 27\end{array}\)
\(\begin{array}{l}b)\;B = {\left( {x – 1} \right)^2} + 2\left( {x – 1} \right)\left( {x + 1} \right) + {\left( {x + 1} \right)^2} + 5\\B = {\left( {\left( {x – 1} \right) + \left( {x + 1} \right)} \right)^2} + 5\\B = {\left( {x – 1 + x + 1} \right)^2} + 5\\B = {\left( {2x} \right)^2} + 5 = 4{x^2} + 5\end{array}\)
\(\begin{array}{l}c)\;C = 9{x^2} – 2xy + \frac{1}{9}{y^2} – 2\left( {3x + \frac{1}{3}y} \right)\left( {3x – \frac{1}{3}y} \right) + {\left( {3x + \frac{1}{3}y} \right)^2}\\C = {\left( {3x} \right)^2} – 2.3x.\frac{1}{3}y + {\left( {\frac{1}{3}y} \right)^2} – 2\left( {3x + \frac{1}{3}y} \right)\left( {3x – \frac{1}{3}y} \right) + {\left( {3x + \frac{1}{3}y} \right)^2}\\C = {\left( {3x – \frac{1}{3}y} \right)^2} – 2\left( {3x + \frac{1}{3}y} \right)\left( {3x – \frac{1}{3}y} \right) + {\left( {3x + \frac{1}{3}y} \right)^2}\\C = {\left( {\left( {3x – \frac{1}{3}y} \right) – \left( {3x + \frac{1}{3}y} \right)} \right)^2}\\C = {\left( {3x – \frac{1}{3}y – 3x – \frac{1}{3}y} \right)^2}\\C = {\left( { – \frac{2}{3}y} \right)^2} = \frac{4}{9}{y^2}\end{array}\)
