phân tích đa thức sau thành nhân tử
a^5+a^3-a^2-1
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Bài 1:
a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(8-2x\right)\)
\(=2\left(4-x\right)\left(2x+1\right)\)
b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(=3\left(3x-2\right)\left(x-2\right)\)
Bài 2:
a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)
\(=\left(a-b\right)\left(2a-4b\right)\)
\(=2\left(a-b\right)\left(a-2b\right)\)
f: Ta có: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(a,x\left(a-b\right)+2a-2b=x\left(a-b\right)+2\left(a-b\right)=\left(a-b\right)\left(x+2\right)\\ b,Sửa:ax+ay+5x+5y=a\left(x+y\right)+5\left(x+y\right)=\left(a+5\right)\left(x+y\right)\)
\(a,=\left(x+2\right)\left(a-b\right)\\ b,Sửa:ax+ay+5x+5y\\ =a\left(x+y\right)+5\left(x+y\right)\\ =\left(a+5\right)\left(x+y\right)\)
\(a^3+a+30\)
\(=a^3+3a^2-3a^2-9a+10a+30\)
\(=\left(a+3\right)\left(a^2-3a+10\right)\)
\(x^3+x^2+100\)
\(=x^3+5x^2-4x^2-20x+20x+100\)
\(=\left(x+5\right)\left(x^2-4x+20\right)\)
\(a^3+4a^2+4a+3\)
\(=a^3+3a^2+a^2+3a+a+3\)
\(=a^2\left(a+3\right)+a\left(a+3\right)+\left(a+3\right)\)
\(=\left(a+3\right)\left(a^2+a+1\right)\)
\(a,=\left(a-5\right)^2-4b^2=\left(a-2b-5\right)\left(a+2b-5\right)\\ b,=ax^2+a-a^2x-x=ax\left(a-x\right)+\left(a-x\right)=\left(ax+1\right)\left(a-x\right)\)
a: \(=\left(a-5-2b\right)\left(a-5+2b\right)\)
b: \(ax^2+a-a^2x-x\)
\(=ax\left(x-a\right)-\left(x-a\right)\)
\(=\left(x-a\right)\left(ax-1\right)\)
a
\(8x^3-\dfrac{1}{125}y^3\\ =\left(2x\right)^3-\left(\dfrac{1}{5}y\right)^3\\ =\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\\ =\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)
b
\(-x^3+6x^2y-12xy^2+8y^3\\ =-\left(x^3-6x^2y+12xy^2-8y^3\right)\\ =-\left(x^3-3.2y.x^2+3.\left(2y\right)^2.x-\left(2y\right)^3\right)\\ =-\left(x-2y\right)^3\\ =-\left(x-2y\right)\left(x-2y\right)\left(x-2y\right)\)
a: 8x^3-1/125y^3
=(2x)^3-(1/5y)^3
=(2x-1/5y)(4x^2+2/5xy+1/25y^2)
b: =(2y-x)^3
\(a,=\left(x-1\right)^4-2\left(x-1\right)^2+1\\ =\left[\left(x-1\right)^2-1\right]^2\\ =\left(x^2-2x-2\right)^2\\ b,=\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]-4\\ =\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\\ =\left(x^2+6x\right)^2+13\left(x^2+6x\right)+36\\ =\left(x^2+6x+4\right)\left(x^2+6x+9\right)\\ =\left(x+3\right)^2\left(x^2+6x+4\right)\)
\(a^5+a^3-a^2-1=a^3\left(a^2+1\right)-\left(a^2+1\right)=\left(a^3-1\right)\left(a^2+1\right)\)
\(a^5+a^3-a^2-1=a^3\left(a^2+1\right)-\left(a^2+1\right)=\left(a^3-1\right)\left(a^2+1\right)=\left(a-1\right)\left(a^2+a+1\right)\left(a^2+1\right)\)