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\(x^2+5x-6=x^2-6x+x-6=x\left(x-6\right)+\left(x-6\right)=\left(x+1\right)\left(x-6\right)\)
\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\)
\(7x-6x^2-2=-6x^2+7x-2=-6\left(x^2-\frac{7}{6}x+\frac{1}{3}\right)=-6\left(x^2-\frac{7}{6}x+\frac{49}{144}-\frac{1}{144}\right)=-6\left[\left(x-\frac{7}{12}\right)^2-\frac{1}{144}\right]\)
1) x2-x+6x-6 = x(x-1)+6(x-1)=(x+6)(x-1)
2) 5x(x+y)-(x+y) =(5x-1)(x+y)
3) -6x2+7x-2 = -6x2+3x+4x-2= -6x(x-\(\frac{1}{2}\)) +4(x-\(\frac{1}{2}\)) =(-3x+2)(2x-1)
x3 + 7x - 6=x2 . x + 7x - 22 + 2 = (x2 - 22) + (x+7x)+2=(x-2) . (x+2) + 8x + 2
x3 - 5x + 8x - 4=x2 . x -5x + 8x -22 = (x2 - 22) . (x -5x + 8x )=(x-2) . (x+2) . 4x
x3 - 9x2 + 6x + 16=x2 . x - 9x2 + 6x + 16 = (x2 - 9x2) . (x+6x) + 16=(x-9x) . (x+9x) . 7x + 16
k mk nha
a) \(x^2+5x-6\)
\(=x^2-2x+3x-6\\ =\left(x^2-2x\right)+\left(3x-6\right)\\ =x\left(x-2\right)+3\left(x-2\right)\\ =\left(x-2\right)\left(x+3\right)\)
b) \(5x^2+5xy-x-y\)
\(=\left(5x^2+5xy\right)-\left(x+y\right)\\ =5x\left(x+y\right)-\left(x+y\right)\\ =\left(x+y\right)\left(5x+1\right)\)
c)\(7x-6x^2-2\)
\(=3x+4x-6x^2-2\\ =\left(3x-6x^2\right)+\left(4x-2\right)\\ =3x\left(1-2x\right)+2\left(2x-1\right)\\ =3x\left(1-2x\right)-2\left(1-2x\right)\\ =\left(1-2x\right)\left(3x-2\right)\)
\(1.x^4+6x^3+11x^2+6x+1\)
\(=x^4+6x^3+9x^2+2x^2+6x+1\)
\(=x^4+9x^2+1+6x^3+2x^2+6x\)
\(=\left(x^2\right)^2+\left(3x\right)^2+1^2+2.x^2.3x+2.x^2.1+2.3x.1\)
\(=\left(x^2+3x+1\right)^2\)
\(2,6x^4+5x^3-38x^2+5x+6\)
\(=6x^4+6x^3+2x^3-3x^3-36x^2+2x^2-3x^2-x^2-12x+18x-x+6\)
\(=\left(6x^4+2x^3\right)+\left(6x^3+2x^2\right)-\left(3x^3+x^2\right)-\left(36x^2+12x\right)+\left(18x+6\right)-\left(3x^2+x\right)\)
\(=2x^3\left(3x+1\right)+2x^2\left(3x+1\right)-x^2\left(3x+1\right)-12x\left(3x+1\right)+6\left(3x+1\right)-x\left(3x+1\right)\)
\(=\left(3x+1\right)\left(2x^3+2x^2-x^2-12x+6-x\right)\)
\(=\left(3x+1\right)\left[\left(2x^3-x^2\right)+\left(2x^2-x\right)-\left(12x-6\right)\right]\)
\(=\left(3x+1\right)\left[x^2\left(2x-1\right)+x\left(2x-1\right)-6\left(2x-1\right)\right]\)
\(=\left(3x+1\right)\left(2x-1\right)\left(x^2+x-6\right)\)
\(=\left(3x+1\right)\left(2x-1\right)\left(x^2+3x-2x-6\right)\)
\(=\left(3x+1\right)\left(2x-1\right)\left[\left(x^2+3x\right)-\left(2x+6\right)\right]\)
\(=\left(3x+1\right)\left(2x-1\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]\)
\(=\left(3x+1\right)\left(2x-1\right)\left(x+3\right)\left(x-2\right)\)
1. \(x^4+6x^3+11x^2+6x+1\)
\(=\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2+2x^2+6x+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
3. \(x^4-7x^3+14x^2-7x+1\)
\(=x^2\left(x^2-7x+14-\dfrac{7}{x}+\dfrac{1}{x^2}\right)\)
\(=x^2\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(7x+\dfrac{7}{x}\right)+14\right]\)
\(=x^2\left[\left(x+\dfrac{1}{x}\right)^2-7\left(x+\dfrac{1}{x}\right)+12\right]\)
\(=x^2\left[\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right).\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{1}{4}\right]\)
\(=x^2\left[\left(x+\dfrac{1}{x}-\dfrac{7}{2}\right)^2-\dfrac{1}{4}\right]\)
\(=\left(x^2+1-\dfrac{7}{2}x\right)^2-\left(\dfrac{1}{2}x\right)^2\)
\(=\left(x^2-3x+1\right)\left(x^2-4x+1\right)\)
Có thể phân tích thành HĐT tiếp hoặc không.
b: \(=x^4+x^2+36-2x^3+12x^2-12x+x^2-6x+9\)
\(=x^4-2x^3+14x^2-18x+45\)
\(=x^4+9x^2-2x^3-18x+5x^2+45\)
\(=\left(x^2+9\right)\left(x^2-2x+5\right)\)
d: \(=2x^4+2x^3+6x^2-x^3-x^2-3x+x^2+x+3\)
\(=\left(x^2+x+3\right)\left(2x^2-x+1\right)\)
e: \(=3x^4-3x^3-3x^2-2x^3+2x^2+2x+2x^2-2x-2\)
\(=\left(x^2-x-1\right)\left(3x^2-2x+1\right)\)
\(6x^4+5x^3-38x^2+5x+6=0\)
\(6x^4-12x^3+17x^3-34x^2-4x^2+8x-3x+6=0\)
\(6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)=0\)
\(\left(x-2\right)\left(6x^3+17x^2-4x-3\right)=0\)
\(\left(x-2\right)\left[6x^3-3x^2+20x^2-10x+6x-3\right]=0\)
\(\left(x-2\right)\left[6x^2\left(x-\dfrac{1}{2}\right)+20x\left(x-\dfrac{1}{2}\right)+6\left(x-\dfrac{1}{2}\right)\right]=0\)
\(\left(x-2\right)\left(x-\dfrac{1}{2}\right)\left(6x^2+20x+6\right)=0\)
=> \(\left[{}\begin{matrix}x-2=0\\x-\dfrac{1}{2}=0\\6x^2+20x+6=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\\left(3x+1\right)\left(x+3\right)=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\x=-3\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(6x^4+5x^3-38x^2+5x+6\)
= \(6x^4-15x^3+6x^2+20x^3-50x^2+20x+6x^2-15x+6\)
= \(3x^2\left(2x^2-5x+2\right)+10x\left(2x^2-5x+2\right)+3\left(2x^2-5x+2\right)\)
= \(\left(2x^2-5x+2\right)\left(3x^2+10x+3\right)\)
= \(\left(2x^2-x-4x+2\right)\left[3x^2+x+9x+3\right]\)
= \(\left[x\left(2x-1\right)-2\left(2x-1\right)\right]\left[x\left(3x+1\right)+3\left(3x+1\right)\right]\)
= \(\left(x-2\right)\left(2x-1\right)\left(3x+1\right)\left(x+3\right)\)
Chúc bạn học tốt !!!