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Ta có
M = 3 x 2 ( x 2 + y 2 ) + 3 y 2 ( x 2 + y 2 ) – 5 ( y 2 + x 2 ) = ( x 2 + y 2 ) ( 3 x 2 + 3 y 2 – 5 ) = ( x 2 + y 2 ) [ 3 ( x 2 + y 2 ) – 5 ]
Mà x 2 + y 2 = 1 nên M = 1.(3.1 – 5) = -2. Vậy M = -2
Đáp án cần chọn là: D
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Chọn D.
(x − 5)(x + 3) = x(x + 3) – 5( x + 3) = x 2 + 3x - 5x - 15 = x 2 − 2x − 15
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P = 5x( x 2 – 3) + x 2 (7 – 5x) – 7 x 2
= 5x. x 2 +5x. (-3) + x 2 . 7 + x 2 . (- 5x) – 7 x 2
= 5 x 3 – 15x + 7 x 2 - 5 x 3 – 7 x 2
= ( 5 x 3 – 5 x 3 ) + ( 7 x 2 – 7 x 2 ) – 15x
= - 15x
Thay x = -5 vào P = -15x ta được: P = - 15.(-5) = 75
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2 x 2 (5 x 3 − 4 x 2 y − 7xy + 1)
= 2 x 2 . 5 x 3 – 2 x 2 .4 x 2 .y+ 2 x 2 .(– 7xy) + 2 x 2 .1
= 10 x 5 − 8 x 4 y − 14 x 3 y + 2 x 2
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a) (2 + xy)2 = 22 + 2.2.xy + (xy)2 = 4 + 4xy + x2y2
b) (5 – 3x)2 = 52 – 2.5.3x + (3x)2 = 25 – 30x + 9x2
c) (5 – x2)(5 + x2) = 52 – (x2)2 = 25 – x4
d) (5x – 1)3 = (5x)3 – 3.(5x)2.1 + 3.5x.12 – 13 = 125x3 – 75x2 + 15x – 1
e) (2x – y)(4x2 + 2xy + y2) = (2x – y)[(2x)2 + 2x.y + y2] = (2x)3 – y3 = 8x3 – y3
f) (x + 3)(x2 – 3x + 9) = (x + 3)(x2 – 3x + 32) = x3 + 33 = x3 + 27
a) (2 + xy)2 = 22 + 2.2.xy + (xy)2 = 4 + 4xy + x2y2
b) (5 – 3x)2 = 52 – 2.5.3x + (3x)2 = 25 – 30x + 9x2
c) (5 – x2)(5 + x2) = 52 – (x2)2 = 25 – x4 d) (5x – 1)3 = (5x)3 – 3.(5x)2.1 + 3.5x.12 – 13 = 125x3 – 75x2 + 15x – 1
e) (2x – y)(4x2 + 2xy + y2) = (2x – y)[(2x)2 + 2x.y + y2] = (2x)3 – y3 = 8x3 – y3
f) (x + 3)(x2 – 3x + 9) = (x + 3)(x2 – 3x + 32) = x3 + 33 = x3 + 27
(5 – x2)(5 + x2)
= 52 – (x2)2 (Áp dụng HĐT (3))
= 25 – x4