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a) \(6x^2-11x+3\)
\(=6x^2-9x-2x+3\)
\(=3x\left(2x-3\right)-\left(2x-3\right)\)
\(=\left(3x-1\right)\left(2x-3\right)\)
b) \(2x^2+3x-27\)
\(=2x^2-6x+9x-27\)
\(=2x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(2x+9\right)\left(x-3\right)\)
Ta có:
N x3 – 3x2 + 3x – 1 = x3 – 3.x2.1 + 3.x.12 – 13 = (x – 1)3
U 16 + 8x + x2 = 42 + 2.4.x + x2 = (4 + x)2 = (x + 4)2
H 3x2 + 3x + 1 + x3 = x3 + 3x2 + 3x + 1 = (x + 1)3 = (1 + x)3
 1 – 2y + y2 = 12 – 2.1.y + y2 = (1 – y)2 = (y – 1)2
Điền vào bảng như sau:
| (x – 1)3 | (x + 1)3 | (y – 1)2 | (x – 1)3 | (1 + x)3 | (1 – y)2 | (x + 4)2 |
| N | H | Â | N | H | Â | U |
Vậy: Đức tính đáng quý là "NHÂN HẬU"
(Chú ý: Bạn có thể làm theo cách ngược lại, tức là khai triển các biểu thức (x – 1)3, (x + 1)3, (y – 1)2, (x + 4)2 … để tìm xem kết quả ứng với chữ nào và điền vào bảng.)
\(1,\\ 12x^6y^3:4x^3y=3x^3y^2\\ \left(x+1\right)\left(x^2-x+1\right)=x^3+1\\ 2x^2y\left(x^2+3xy\right)=3x^4y+6x^3y^2\\ 2,\\ a,=2xy\left(2x+3y-4\right)\\ b,=\left(x-3\right)\left(x+y\right)\\ c,=\left(x-2\right)\left(x+2\right)+y\left(x-2\right)=\left(x+y+2\right)\left(x-2\right)\\ d,=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\\ 3,\\ a,\Leftrightarrow x^2-x^2+2x=2\\ \Leftrightarrow2x=2\Leftrightarrow x=1\\ b,\Leftrightarrow\left(x-2\right)\left(x-2+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) \(\dfrac{1}{x^3-8}=\dfrac{1}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
\(\dfrac{3}{4-2x}=\dfrac{-3}{2\left(x-2\right)}=\dfrac{-3\left(x^2+2x+4\right)}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
b) \(\dfrac{x}{x^2-1}=\dfrac{x}{\left(x+1\right)\left(x-1\right)}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2\left(x-1\right)}\)
\(\dfrac{1}{x^2+2x+1}=\dfrac{1}{\left(x+1\right)^2}=\dfrac{x-1}{\left(x+1\right)^2\left(x-1\right)}\)
c) \(\dfrac{1}{x+2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{5}{2-x}=\dfrac{-5}{x-2}=\dfrac{-5\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)^2}\)
d) \(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{\left(x-y\right)^2}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{2x}{x^2-y^2}=\dfrac{2x}{\left(x+y\right)\left(x-y\right)}=\dfrac{6x\left(x-y\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{x^2-xy+y^2}{x^2-2xy+y^2}=\dfrac{x^2-xy+y^2}{\left(x-y\right)^2}=\dfrac{3\left(x^2-xy+y^2\right)\left(x+y\right)}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{3\left(x^3+y^3\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
em c.ơn trước
\(5,\dfrac{4}{x-2}+\dfrac{x}{x+1}-\dfrac{x^2-2}{\left(x-2\right)\left(x+1\right)}=0\left(dkxd:x\ne2;-1\right)\)
\(\Rightarrow4\left(x+1\right)+x\left(x-2\right)-x^2-2=0\)
\(\Rightarrow4x+4+x^2-2x-x^2-2=0\)
\(\Rightarrow2x+2=0\)
\(\Rightarrow x=-1\left(loai\right)\)
Vậy \(S=\varnothing\)
