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\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
x4 - 4x3 - 8x2 + 8x
= x(x3 - 4x2 - 8x + 8)
= x[x3 + 8 - 4x(x + 2)]
= x[(x + 2)(x2 - 2x + 4) - 4x(x + 2)]
= x(x + 2)(x2 - 6x + 4)
= x(x + 2)(x2 - 6x + 9 - 5)
= \(x\left(x+2\right)\left[\left(x-3\right)^2-5\right]=x\left(x+2\right)\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)\)
\(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left(x^3-6x^2+2x^2+4x-12x+8\right)\)
\(=x\left[\left(x^3-6x^2+4x\right)+\left(2x^2-12x+8\right)\right]\)
\(=x\left[x\left(x^2-6x+4\right)+2\left(x^2-6x+4\right)\right]\)
\(=x\left(x^2-6x+4\right)\left(x+2\right)\)
\(=x\left[\left(x-3\right)^2-\left(\sqrt{5}\right)^2\right]\left(x+2\right)\)
\(=x\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\left(x+2\right)\)
\(=\left(x^2+x+4\right)^2+3x\left(x^2+x+4\right)+5x\left(x^2+x+4\right)+15x^2\\ =\left(x^2+x+4\right)\left(x^2+x+4+3x\right)+5x\left(x^2+x+4+3x\right)\\ =\left(x^2+x+4+3x\right)\left(x^2+x+4+5x\right)\\ =\left(x^2+4x+4\right)\left(x^2+6x+4\right)\\ =\left(x+2\right)^2\left(x^2+6x+4\right)\)
Trả lời:
a, x4 + 3x3 + x2 + 3x
= ( x4 + 3x3 ) + ( x2 + 3x )
= x3 ( x + 3 ) + x ( x + 3 )
= ( x3 + x ) ( x + 3 )
= x ( x2 + 1 ) ( x + 3 )
b, Sửa đề: x4 - x2 + 8x - 8
= ( x4 - x2 ) + ( 8x - 8 )
= x2 ( x2 - 1 ) + 8 ( x - 1 )
= x2 ( x - 1 ) ( x + 1 ) + 8 ( x - 1 )
= ( x - 1 ) [ x2 ( x + 1 ) + 8 ]
= ( x - 1 ) ( x3 + x2 + 8 )
\(x^4+8x^3+28x^2+48x-13\)
\(=x^4+4x^3+13x^2+4x^3+16x^2+52x-x^2-4x-13\)
\(=x^2\left(x^2+4x+13\right)+4x\left(x^2+4x+13\right)-\left(x^2+4x+13\right)\)
\(=\left(x^2+4x-1\right)\left(x^2+4x+13\right)\)
= \(x^4-2x^3-6x^3+12x^2-x^2+2x+6x-12\)
= \(x^3\left(x-2\right)-6x^2\left(x-2\right)-x\left(x-2\right)+6\left(x-2\right)\)
= \(\left(x-2\right)\left(x^3-6x^2-x+6\right)\)
= \(\left(x-2\right)\left(x^2\left(x-6\right)-\left(x-6\right)\right)\)
= \(\left(x-2\right)\left(x-6\right)\left(x-1\right)\left(x+1\right)\)
x4 - 8x3 + 11x2 + 8x - 12
= (x3 - 7x2 + 4x + 12)(x - 1)
= (x3 - 8x + 12)(x + 1)(x - 1)
= (x - 6)(x - 2)(x + 1)(x - 1)
Bài 1:
\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)
Bài 2:
\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)
Bài 3:
\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)
\(x^4-8x=x\left(x^3-8\right)=x\left(x-2\right)\left(x^2+2x+4\right)\)
x4 - 8x = x.(x3 - 8) = x. (x3 - 23) = x.(x - 2).(x2 + 2x +4)