Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, (2x-5)3 = (2x)3 - 3.(2x)2.5 + 3.2x.52 - 53
= 8x3 - 60x2 + 150x - 125
Câu b, câu c làm tương tự như câu a
a) A = (x + 2y)(x^2 - 2xy + 4y^2) - 8(x^3 + y^3)
A = x(x^2 - 2xy + 4y^2) + 2y(x^2 - 2xy + 4y^2) - 8(x^3 + y^3)
A = x^3 - 2x^2y + 4xy^2 + 2x^2y - 4xy^2 + 8y^3 - 8x^3 - 8y^3
A = -7x^3
b) B = (2x + y)^3 - (8x^3 + y^3) - 2x^2y
B = (2x + y)[(2x)^2 + 2.2xy + y^2] - 8x^3 - y^3 - 2x^2y
B = 2x[(2x)^2 + 2.2xy + y^2] + y[(2x)^2 + 2.2xy + y^3] - 8x^3 - y^3 - 2x^2y
B = 8x^3 + 8x^2y + 2xy^2 + 4x^2y + y^3 - 8x^3 - y^3 - 2x^2y
B = 10x^2y + 6xy^2
1/
( a + b )3 + ( a - b )3 - 6ab2 < đã sửa >
= a3 + 3a2b + 3ab2 + b3 + a3 - 3a2b + 3ab2 - b3 - 6ab2
= 2a3
2/
A = x2 + y2 - 2x - 4y + 6 = ( x2 - 2x + 1 ) + ( y2 - 4y + 4 ) + 1 = ( x - 1 )2 + ( y - 2 )2 + 1 ≥ 1 ∀ x, y
Dấu "=" xảy ra khi x = 1 ; y = 2
=> MinA = 1 <=> x = 1 ; y = 2
B = 2x2 + 8x + 10 = 2( x2 + 4x + 4 ) + 2 = 2( x + 2 )2 + 2 ≥ 2 ∀ x
Dấu "=" xảy ra khi x = -2
=> MinB = 2 <=> x = -2
C = 25x2 + 3y2 - 10x + 11 = ( 25x2 - 10x + 1 ) + 3y2 + 10 = ( 5x - 1 )2 + 3y2 + 10 ≥ 10 ∀ x, y
Dấu "=" xảy ra khi x = 1/5 ; y = 0
=> MinC = 10 <=> x = 1/5 ; y = 0
D = ( x - 3 )2 + ( x - 11 )2
Đặt t = x - 7
D = ( t + 4 )2 + ( t - 4 )2
= t2 + 8t + 16 + t2 - 8t + 16
= t2 + 32 ≥ 32 ∀ t
Dấu "=" xảy ra khi t = 0
=> x - 7 = 0 => x = 7
=> MinD = 32 <=> x = 7
\(A=2x^4+4x^3-7x^3-14x^2+8x^2+16x\)
\(=2x^2\left(x^2+2x\right)-7x\left(x^2+2x\right)+8\left(x^2+2x\right)\)
\(=\left(2x^2-7x+8\right)\left(x^2+2x\right)\)
\(=x\left(x+2\right)\left(2x^2-7x+8\right)\)
\(B=2x\left(x^2-4x+4-y^2\right)\)
\(=2x\left(\left(x-2\right)^2-y^2\right)\)
\(=2x\left(x-y-2\right)\left(x+y-2\right)\)
\(C=x\left(8y^2+8xy+2x^2-z^2\right)\)
\(=x\left(2\left(4y^2+4xy+x^2\right)-z^2\right)\)
\(=x\left(2\left(x+2y\right)^2-z^2\right)\)
\(=x\left(\sqrt{2}x+2\sqrt{2}y-z\right)\left(\sqrt{2}x+2\sqrt{2}y+z\right)\)
\(D=4a^4+10a^3+6a^2-6a^2-15a-9\)
\(=2a^2\left(2a^2+5a+3\right)-3\left(2a^2+5a+3\right)\)
\(=\left(2a^2-3\right)\left(2a^2+5a+3\right)\)
\(E=4a^3-ab^2+2ab-4a^2\)
\(=a\left(4a^2-b^2\right)-2a\left(2a-b\right)\)
\(=a\left(2a+b\right)\left(2a-b\right)-2a\left(2a-b\right)\)
\(=\left(2a-b\right)\left(2a^2+ab-2a\right)\)
\(F=5a^2-10a-4a+8\)
\(=5a\left(a-2\right)-4\left(a-2\right)\)
\(=\left(5a-4\right)\left(a-2\right)\)
\(G=a\left(2x+3y\right)-\left(2x+3y\right)\)
\(=\left(a-1\right)\left(2x+3y\right)\)
a) \(\dfrac{6x^2y^3-2x^2y+6xy}{6xy}\)
\(=\dfrac{6x^2y^3}{6xy}-\dfrac{2x^2y}{6xy}+\dfrac{6xy}{6xy}\)
\(=xy^2-\dfrac{x}{3}+1\)
b) \(\dfrac{4\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{2\left(x+y\right).2\left(x+y\right)^2}{2\left(x+y\right)}\)
\(=2\left(x+y\right)^2\)
c) \(\dfrac{8x^3+27y^3}{2x+3y}\)
\(=\dfrac{\left(2x\right)^3+\left(3y\right)^3}{2x+3y}\)
\(=\dfrac{\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]}{2x+3y}\)
\(=4x^2-6xy+9y^2\)
d) \(\dfrac{48x^4y^3-12x^2y^5+6x^2y^2}{3x^2y^2}\)
\(=\dfrac{48x^4y^3}{3x^2y^2}-\dfrac{12x^2y^5}{3x^2y^2}+\dfrac{6x^2y^2}{3x^2y^2}\)
\(=16x^2y-4y^3+2\)
sao đã khai triển lại còn thu gọn ???