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\(x^6-x^4+2x^3+2x^2=x^2\left(x^4-x^2+2x+2\right)\)
\(=x^2\left(\left(x^4+2x^3+x^2\right)+\left(-2x^3-4x^2-2x\right)+\left(2x^2+4x+2\right)\right)\)
\(=x^2\left(x^2+2x+1\right)\left(x^2-2x+2\right)\)
\(=x^2\left(x+1\right)^2\left(x^2-2x+2\right)\)
xin lỗi nhưng mình chưa hiểu lắm, bannj có thể viết rõ hơn không
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Đặt x^2 + 2x = a ta có:
a^2 - 9a + 20 = (a - 4)(a - 5)
Thay ngược lại ta có: (x^2 + 2x - 4)(x^2 + 2x - 5)
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(x2 - x)2 - 2 * (x2 - x) - 15
đặt x2 - x = a
có: a2 - 2a - 15 = (a2 - 2a + 1) - 16 = (a - 1)2 - 16 = (a - 5) (a + 3)
thay vào đc: (x2 - x - 5) (x2 - x +3)
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(x^2+x)^2-2(x^2+x)-15
=(x2+x)2-2(x2+x)+1-16
=(x2+x-1)2-16
=(x2+x-1+4)(x2+x-1-4)
=(x2+x+3)(x2+x-5)
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Ta có : \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15=\left[\left(x^2+x\right)-2\left(x^2+x\right)+1\right]-16=\left(x^2+x-1\right)^2-4^2\)
\(=\left(x^2+x-5\right)\left(x^2+x+3\right)\)
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Đặt \(t=x^2+8x+11\) và \(A=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\): \(\Rightarrow A=\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1=\left(t-1\right)\left(t+1\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)
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đặt x^2+x = y
=> y^2 - 2y - 15
= y^2 - 2y + 1 - 16
= ( y - 1 )^2 - 16
= ( y - 1 )^2 - 4^2
= ( y - 1 - 4 ) x ( y-1+4)
=(y -5) (y+3)
= (x^2 +x-5) (x^2+x+3)
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(x^2 + x)^2 - 2.(x^2 + x) - 15
= (x2 + x)2 - 2(x2 + x) + 1 - 16
= (x2 + x + 1)2 - 16
= (x2 + x + 1 - 4)(x2 + x + 1 + 4)
= (x2 + x - 3)(x2 + x + 5)
\(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=\left(\left(x^2+x\right)^2-2\left(x^2+x\right)+1\right)-16\)
\(=\left(x^2+x-1\right)^2-4^2\)
\(=\left(x^2+x-1-4\right)\left(x^2+x-1+4\right)\)
\(=\left(x^2+x-5\right)\left(x^2+x+3\right)\)