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Khi x = - 1; y = 1 thì xy = (-1).1= -1
Ta có: xy – x2y2 + x3y3 – x4y4 + x5y5 – x6.y6
= xy – (xy)2 + (xy)3 – (xy)4 + (xy)5 – (xy)6
= -1 – (-1)2 + (-1)3 – (-1)4 + (-1)5 - (-1)6
= -1 – 1 + (-1) – 1 + (-1) – 1
= - 6
Chọn đáp án D
1) \(x^3-1=x^3-1^3=\left(x-1\right)\left(x^2+x+1\right)\)
2) \(27x^3-64=\left(3x\right)^3-4^3=\left(3x-4\right)\left(9x^2+12x+4\right)\)
3) \(8x^3+1=\left(2x\right)^3+1^3=\left(2x+1\right)\left(4x^2-2x+1\right)\)
Lời giải:
$f(1)=a+b+c=6$
$f(2)=4a+2b+c=16$
$f(12)-f(-9)=(144a+12b+c)-(81a-9b+c)$
$=63a+21b=21(3a+b)$
$=21[(4a+2b+c)-(a+b+c)]=21(16-6)=21.10=210$
1) \(\left(3x-2a\right)^3\)
\(=\left(3x\right)^3-3\left(3x\right)^2\cdot2a+3\cdot3x\cdot\left(2a\right)^2-\left(2a\right)^3\)
\(=27x^3-3\cdot9x^2\cdot2a+3\cdot3x\cdot4a^2-8a^3\)
\(=27x^3-54ax^2+36a^2x-8a^3\)
2) \(\left(\dfrac{x+y}{3}\right)^3\)
\(=\dfrac{\left(x+y\right)^3}{27}\)
\(=\dfrac{x^3+3x^2y+3xy^2+y^3}{27}\)
3) \(\left(3x+\dfrac{y}{3}\right)^3\)
\(=\dfrac{\left(3x+y\right)^3}{27}\)
\(=\dfrac{27x^3+27x^2y+9xy^2+y^3}{27}\)
a) Ta có: \(\frac{1}{27}x^3-8y^6\)
\(=\left(\frac{1}{3}x\right)^3-\left(2y^2\right)^3\)
\(=\left(\frac{1}{3}x-2y^2\right)\left(\frac{1}{9}x^2+\frac{2}{3}xy^2+4y^4\right)\)
b) Ta có: \(t^2x^6-\frac{4}{9}y^4\)
\(=\left(tx^3\right)^2-\left(\frac{2}{3}y^2\right)^2\)
\(=\left(tx^3-\frac{2}{3}y^2\right)\left(tx^3+\frac{2}{3}y^2\right)\)
c) Ta có: \(64x^6+\frac{1}{27}y^3\)
\(=\left(4x^2\right)^3+\left(\frac{1}{3}y\right)^3\)
\(=\left(4x^2+\frac{1}{3}y\right)\left(8x^4-\frac{4}{3}x^2y+\frac{1}{9}y^2\right)\)
d) Ta có: \(\frac{1}{16}a^2x^6-y^4\)
\(=\left(\frac{1}{4}ax^3\right)^2-\left(y^2\right)^2\)
\(=\left(\frac{1}{4}ax^3-y^2\right)\left(\frac{1}{4}ax^3+y^2\right)\)
e) Ta có: \(m^4x^6-\frac{4}{25}y^2\)
\(=\left(m^2x^3\right)^2-\left(\frac{2}{5}y\right)^2\)
\(=\left(m^2x^3-\frac{2}{5}y\right)\left(m^2x^3+\frac{2}{5}y\right)\)
f) Ta có: \(27x^6-\frac{1}{64}y^3\)
\(=\left(3x^2\right)^3-\left(\frac{1}{4}y\right)^3\)
\(=\left(3x^2-\frac{1}{4}y\right)\left(9x^4+\frac{3}{4}x^2y+\frac{1}{16}y^2\right)\)