Question

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Answer 1

`a,` `12/5x^2y^2 - 9x^4 -4/25y^4`

`= -9x^4 + 12/5x^2y^2 -4/25y^4`

`= (-3x^2)^2 + 2 . (-3) x^2 . (-2/5)y^2 + (-2/5y)^2`

`= (-3x^2 - 2/5y)^2`

`@` Áp dụng HĐT `(1): (a + b)^2 = a^2 + 2ab + b^2`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`b,` ` a^2y^2 + b^2x^2 - 2axby`

`= a^2y^2 - 2axby + b^2x^2`

`= (ay)^2 - 2aybx + (bx)^2`

`= (ay - bx)^2`

`@` Áp dụng HĐT `(2): (a - b)^2 = a^2 - 2ab + b^2`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`c,` `64x^2 - (8a + b)^2`

`= (8x)^2 - (8a + b)^2`

`= (8x - 8a - b).(8x + 8a + b)`

`@` Áp dụng HĐT `(3): a^2 - b^2 = (a - b).(a + b)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`d,` `100 - (3x - y)^2`

`= 10^2 - (3x - y)^2`

`= (10 - 3x + y).(10 + 3x - y)`

`@` Áp dụng HĐT `(3): a^2 - b^2 = (a - b).(a + b)`

Answer 2

a)

$\frac{12}{5} x^{2} y^{2} - 9 x^{4} - \frac{4}{25} y^{4} = - [{(3 x^{2})}^{2} - 2.3 x^{2} . \frac{2}{5} y^{2} + {(\frac{2}{5} y^{2})}^{2}] = - {(3 x^{2} - \frac{2 y^{2}}{5})}^{2}$

b)

$a^{2} y^{2} + b^{2} x^{2} - 2 a x b y = {(b x)}^{2} - 2 . b x . a y + {(a y)}^{2} = {(b x - a y)}^{2}$

c)

$64 x^{2} - {(8 a + b)}^{2} = {(8 x)}^{2} - {(8 a + b)}^{2} = (8 x + 8 a + b) . (8 x - 8 a - b) d) 100 - {(3 x - y)}^{2} = 10^{2} - {(3 x - y)}^{2} = (10 + 3 x - y) . (10 - 3 x + y) e) 27 x^{3} - a^{3} b^{3} = {(3 x)}^{3} - {(a b)}^{3} = (3 x - a b) (9 x^{2} + 3 a b . x + a^{2} b^{2}) f) 27 x^{3} - 27 x^{2} + 3 x + 1 = {(3 x)}^{3} - 3 . {(3 x)}^{2} . 1 + 3.3 x . 1^{2} - 1 - 6 x + 2 = {(3 x - 1)}^{3} - 2 . (3 x - 1) = (3 x - 1) . [{(3 x - 1)}^{2} - 2] = (3 x - 1) . [{(3 x - 1)}^{2} - {(\sqrt{2})}^{2}] = (3 x - 1) . (3 x - 1 + \sqrt{2}) . (3 x - 1 - \sqrt{2}) g) x^{3} - 3 x^{2} + 3 x - 1 = {(x - 1)}^{3} h) \frac{1}{27} + x^{3} = {(\frac{1}{3})}^{3} + x^{3} = (\frac{1}{3} + x) . (\frac{1}{9} - \frac{1}{3} x + x^{2}) i) 0, 001 - 1000 x^{3} = {(0, 1)}^{3} - {(10 x)}^{3} = (0, 1 - 10 x) (0, 01 + x + 100 x^{2})$

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Answer 3

a)

$\frac{12}{5} x^{2} y^{2} - 9 x^{4} - \frac{4}{25} y^{4} = - [{(3 x^{2})}^{2} - 2.3 x^{2} . \frac{2}{5} y^{2} + {(\frac{2}{5} y^{2})}^{2}] = - {(3 x^{2} - \frac{2 y^{2}}{5})}^{2}$

b)

$a^{2} y^{2} + b^{2} x^{2} - 2 a x b y = {(b x)}^{2} - 2 . b x . a y + {(a y)}^{2} = {(b x - a y)}^{2}$

c)

$64 x^{2} - {(8 a + b)}^{2} = {(8 x)}^{2} - {(8 a + b)}^{2} = (8 x + 8 a + b) . (8 x - 8 a - b) d) 100 - {(3 x - y)}^{2} = 10^{2} - {(3 x - y)}^{2} = (10 + 3 x - y) . (10 - 3 x + y) e) 27 x^{3} - a^{3} b^{3} = {(3 x)}^{3} - {(a b)}^{3} = (3 x - a b) (9 x^{2} + 3 a b . x + a^{2} b^{2}) f) 27 x^{3} - 27 x^{2} + 3 x + 1 = {(3 x)}^{3} - 3 . {(3 x)}^{2} . 1 + 3.3 x . 1^{2} - 1 - 6 x + 2 = {(3 x - 1)}^{3} - 2 . (3 x - 1) = (3 x - 1) . [{(3 x - 1)}^{2} - 2] = (3 x - 1) . [{(3 x - 1)}^{2} - {(\sqrt{2})}^{2}] = (3 x - 1) . (3 x - 1 + \sqrt{2}) . (3 x - 1 - \sqrt{2}) g) x^{3} - 3 x^{2} + 3 x - 1 = {(x - 1)}^{3} h) \frac{1}{27} + x^{3} = {(\frac{1}{3})}^{3} + x^{3} = (\frac{1}{3} + x) . (\frac{1}{9} - \frac{1}{3} x + x^{2}) i) 0, 001 - 1000 x^{3} = {(0, 1)}^{3} - {(10 x)}^{3} = (0, 1 - 10 x) (0, 01 + x + 100 x^{2})$

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