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( x2 + 8x + 7 ) ( x2 + 8x + 15 ) + 15
Đặt x2 + 8x + 7 = y ta có:
y ( y + 8 ) + 15
= y2 + 8y + 15
= ( y + 3 ) ( y + 5 )
= ( x2 + 8x + 10 ) ( x2 + 8x + 12 )
= ( x2 + 8x + 10 ) ( x + 2 ) ( x + 6 )
Đặt x2 + 8x + 7 = y ta có:
y ( y + 8 ) + 15
= y2 + 8y + 15
= ( y + 3 ) ( y + 5 )
= ( x2 + 8x + 10 ) ( x2 + 8x + 12 )
= ( x2 + 8x + 10 ) ( x + 2 ) ( x + 6 )
\(x^4+8x=x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)
Ta có:\(\left(x^2+8x-34\right)^2-\left(3x^2-8x+2\right)^2\)
\(=\left(x^2+8x-34+3x^2-8x+2\right)\left[x^2+8x-34-\left(3x^2-8x+2\right)\right]\)
\(=\left(4x^2-32\right)\left(x^2+8x-34-3x^2+8x-2\right)\)
\(=\left(4x^2-32\right)\left(-2x^2+16x-36\right)\)
\(=-2\left(4x^2-32\right)\left(x^2-8x+18\right)\)
cảm ơn.nhưng sao bạn không rút 4 trong \(4x^2-32\)
=> \(\left(4\left(x^2-8\right)\right)\cdot\left(-2\left(x^2-8x+18\right)\right)\)
=\(-8\left(x^2-8\right)\left(x^2-8x+18\right)\)
\(4x^4-8x^3+3x^2-8x+4\)
\(=\left(4x^4-8x^3\right)+\left(3x^2-6x\right)-\left(2x-4\right)\)
\(=4x^3\left(x-2\right)+3x\left(x-2\right)-2\left(x-2\right)\)
\(=\left(x-2\right)\left(4x^3+3x-2\right)\)
x4 - 4x3 - 8x2 + 8x
= x(x3 - 4x2 - 8x + 8)
= x[x3 + 8 - 4x(x + 2)]
= x[(x + 2)(x2 - 2x + 4) - 4x(x + 2)]
= x(x + 2)(x2 - 6x + 4)
= x(x + 2)(x2 - 6x + 9 - 5)
= \(x\left(x+2\right)\left[\left(x-3\right)^2-5\right]=x\left(x+2\right)\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)\)
\(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left(x^3-6x^2+2x^2+4x-12x+8\right)\)
\(=x\left[\left(x^3-6x^2+4x\right)+\left(2x^2-12x+8\right)\right]\)
\(=x\left[x\left(x^2-6x+4\right)+2\left(x^2-6x+4\right)\right]\)
\(=x\left(x^2-6x+4\right)\left(x+2\right)\)
\(=x\left[\left(x-3\right)^2-\left(\sqrt{5}\right)^2\right]\left(x+2\right)\)
\(=x\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\left(x+2\right)\)
\(x^4-8x+63\)
\(=x^4+4x^3+9x^2-4x^3-16x^2-36x+7x^2+28x+63\)
\(=\left(x^4+4x^3+9x^2\right)-\left(4x^3+16x^2+36x\right)+\left(7x^2+28x+63\right)\)
\(=x^2\left(x^2+4x+9\right)-4x\left(x^2+4x+9\right)+7\left(x^2+4x+9\right)\)
\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)
=(x^2+8x)^2+23(x^2+8x)+135
Cái này ko phân tích được nha bạn
\(2x^3-8x\)
\(=2x\left(x^2-4\right)\)
\(=2x\left(x+2\right)\left(x-2\right)\)
2x3-8x = 2x(x2- 4)
=2x(x+2)(x-2)