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a) 2x.(4x - 1)
câu b), c) mik ko biết
ko mong b cho mik
nhưng vẫn hi vọng b hoặc ai đó sẽ làm vậy
b) \(4x^4+1=4x^4+4x^2+1-4x^2\)
\(=\left(2x^2+1\right)^2-\left(2x\right)^2\)
\(=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
a) \(8x^2-2x-1=8x^2-4x+2x-1=4x.\left(2x-1\right)+\left(2x-1\right)=\left(2x-1\right)\left(4x+1\right)\)
b) \(4x^4+1=\left(2x^2\right)^2+4x^2+1-4x^2=\left(2x^2+1\right)^2-4x^2=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
c) \(\left(x^2-2x\right)\left(x^2-2x-1\right)-6=x^4-2x^3-x^2-2x^3+4x^2+2x-6\)
\(=x^4-4x^3+3x^2+2x-6=\left(x^4-3x^3\right)-\left(x^3-3x^2\right)+\left(2x-6\right)\)
\(=x^3.\left(x-3\right)-x^2.\left(x-3\right)+2.\left(x-3\right)=\left(x-3\right).\left(x^3-x^2+2\right)\)
\(=\left(x-3\right)\left[\left(x^3+x^2\right)+\left(-2x^2-2x\right)+\left(2x+2\right)\right]\)
\(=\left(x-3\right)\left[x^2\left(x+1\right)-2x.\left(x+1\right)+2.\left(x+1\right)\right]=\left(x-3\right)\left(x+1\right)\left(x^2-2x+2\right)\)
a, 8x^2-2x-1 = 8x2-4x+2x-1 = 4x ( 2x -1) + (2x-1) = (4x+1)(2x-1)
b) 4x4+1 = (2x2)2 + 4x2+ 1 - 4x2 = (2x2+1)2-(2x)2 = (2x2+1-2x)(2x2+1+2x)
\(=x^3+x^2+x^2+x+x+1=x^2\left(x+1\right)+x\left(x+1\right)+x+1\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
x3 + 2x2 + 2x + 1
= (x3 + 1) + (2x2 + 2x)
= (x + 1)(x2 + x + 1) + 2x(x + 1)
= (x + 1)(x2 + x + 1 + 2x)
= (x + 1)(x2 + 3x + 1)
Chúc bạn học tốt
\(x^6+2x^5+x^4-2x^3-2x^2+1=\left(x^3+x^2-1\right)^2\)
\(a)\left(x^2+2x\right)\left(x^2+2x+4\right)+3\)
Để đơn giản hơn cũng như là dễ nhìn hơn thì ta :
Đặt : \(x^2+2x=a\)
Do đó ta có đa thức :
\(a.\left(a+4\right)+3=a^2+4a+3\)
\(=a^2+a+3a+3\)
\(=a\left(a+1\right)+3\left(a+1\right)\)
\(=\left(a+1\right)\left(a+3\right)\)
\(=\left(x^2+2x+1\right)\left(x^2+2x+3\right)\)
\(=\left(x+1\right)^2.\left(x^2+2x+3\right)\)
Hoặc bạn có thể đặt \(x^2+2x+2=t\)
Thì \(P=\left(x^2+2x\right)\left(x^2+2x+4\right)+3\)
\(P=\left(t-2\right)\left(t+2\right)+3\)
\(P=t^2-4+3\)
\(P=t^2-1\)
\(P=\left(t-1\right)\left(t+1\right)\)
\(P=\left(x^2+2x+1\right)\left(x^2+2x+3\right)\)
\(P=\left(x+1\right)^2\left(x^2+2x+3\right)\)
#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
1) \(x\left(4x+1\right)\)
2) \(3\left(x-3y\right)\)
3) \(\left(2x+1\right)\left(2x+1+2\right)=\left(2x+1\right)\left(2x+3\right)\)
\(x^3+2x^2+2x+1\)
\(=\left(x^3+1\right)+2x^2+2x\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+2x\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
x3+2x2+2x+1= (x3+x2+x)+(x2+x+1)= x(x2+x+1)+(x2+x+1)= (x2+x+1)(x+1)