Phân tích thành nhân tử
x\(^3\)-x\(^2\)-4
4x\(^4\)-3x\(^2\)+1
x\(^8\)+x+1
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\(x^5+x+1\)
\(=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
Câu 1:
\(=x^2-\left(y-4\right)^2\)
\(=\left(x-y+4\right)\cdot\left(x+y-4\right)\)
\(x^3\left(2+x\right)^2-\left(x+2\right)^2+1-x^3\\ =\left(x+2\right)^2\left(x^3-1\right)-\left(x^3-1\right)\\ =\left[\left(x+2\right)^2-1\right]\left(x^3-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x^2+4x+3\right)=\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x^2+x+1\right)\)
\(x^4+x^3+2x^2+x+1=\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+1\right)\left(x^2+x+1\right)\)
Dễ thấy \(x^2+1>0\); \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) nên ta không thể phân tích thêm được nữa.
Vậy \(x^4+x^3+2x^2+x+1=\left(x^2+1\right)\left(x^2+x+1\right)\).
\(=x^2\left(x+3\right)-4\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-4\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(x+2\right)\)
\(4.\left(2x+3\right)\left(2x-1\right)\left(x-3\right)\left(4x+1\right)+44x^2\)
\(=4.\left(4x^2+4x-3\right)\left(4x^2-11x-3\right)+44x^2\)
Đặt \(4x^2+4x-3=t\)
\(\Rightarrow4.\left(2x+3\right)\left(2x-1\right)\left(x-3\right)\left(4x+1\right)+44x^2\)
\(=4.t.\left(t-15x\right)+44x^2\)
\(=4t^2-60tx+44x^2\)
\(=4.\left(t^2-15tx+11x^2\right)\)
Tự lm nốt nhé~
\(x^2\left(x-3\right)-4x+12=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
=x²(x-3)-4x+3.4
=x²(x-3)-4(x+3)
=x²(x-3)+4(x-3)
=(x-3)(x²+4)
=(x-3)(x²+2²)
=(x-3)(x-2)(x+2)
a: \(=x^2\left(x-2\right)\)
b: \(=\left(y+1\right)^2-x^2=\left(y+1+x\right)\left(y+1-x\right)\)
c: =(x-3)(x+2)