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\(a)\)
\(\left(x-1\right)\left(x^3+x^2+x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)\)
\(=x^4-1\)
\(b)\)
\(\left(x+1\right)\left(x^4-x^3+x^2-1x+1\right)\)
\(=x^5+1^5\)
\(=x^5+1\)
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\(\left(2x-y\right)^3-3x.\left(y-2x\right)^2-4x^2+4xy-y^2\)
\(=\left[\left(2x-y\right)^3-3x.\left(y-2x\right)^2\right]-\left[4x^2+4xy-y^2\right]\)
\(=\left[\left(2x-y\right)^2.\left(2x-y\right)-3x.\left(y-2x\right)^2\right]-\left(2x-y\right)^2\)
\(=\left[\left(y-2x\right)^2.\left(2x-y-3x\right)\right]-\left(2x-y\right)^2\)
=\(\left(y-2x\right)^2.\left(-x-y\right)-\left(2x-y\right)^2\)
\(=\left(y-2x\right)^2.\left(-x-y-1\right)\)
Chúc bạn học tốt
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phân tích đa thức
(2x – y)^3 – 3x(y – 2x)2 – 4x2 + 4xy – y ^2
= -(y-2x)^2(y+x+1)
đây nha bạn
\(\left(2x-y\right)^3-3x\left(y-2x\right)^2-4x^2+4xy-y^2\)
\(=\left(2x-y\right)^3-3x\left(2x-y\right)^2-\left(2x-y\right)^2\)
\(=\left(2x-y\right)^3-\left(2x-y\right)^2\left(3x+1\right)\)
\(=\left(2x-y\right)^2\left(2x-y-3x-1\right)\)
\(=\left(2x-y\right)^2\left(-x-y-1\right)\)
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\(\left(x-3\right)\left(4x-1\right)\left(6x-2\right)\left(2x+1\right)\)
\(\left(4x^2-12x-x+3\right)\left(12x^2-4x+6x-2\right)\)
\(\left(4x^2-13x+3\right)\left(12x^2+2x-2\right)\)
\(48x^4-13x^3+36x^2-156x^3-26x^2+26x-8x^2+26x-6\)
\(48x^4-169x^3+2x^2+52x-6\)
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\(\left(x-3\right)\left(4x-1\right)-\left(6x+2\right)\left(2x-1\right)\)
\(=4x^2-x-12x+3-\left(12x^2-6x+4x-2\right)\)
\(=4x^2-x-12x+3-12x^2+6x-4x+2\)
\(=-8x^2-11x+5\)
Đặt a - 2b + c = x
b - 2c + a = y
c - 2a + b = z
Ta có x + y + z = 0
=> x + y = - z
=> (x + y)3 = (-z)3
=> x3 + y3 + 3xy(x + y) = - z3
=> x3 + y3 + z3 = -3xy(x + y)
=> x3 + y3 + z3 = 3xyz
mà x3 + y3 + z3 = 0
=> 3xyz = 0
=> xyz = 0
=> x = 0 hoặc y = 0 hoặc z = 0
Khi x = 0 => a - 2b + c = 0
=> a + c = 2b
=> a + b + c = 3b \(⋮\)3
Khi y = 0 => b - 2c + a = 0
=> a + b = 2c
=> a + b + c = 3c \(⋮\)3
Khi z = 0 => c - 2a + b = 0
=> b + c = 2a
=> a + b + c = 3a \(⋮\)3
Vậy a + b + c \(⋮3\)