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a. \(=4x^3-12x^2-x^2+3x+6x-18=\left(x-3\right)\left(4x^2-x+6\right)\)
b. \(=-x^3+x^2-7x^2+7x-x+1=\left(x-1\right)\left(-x^2-7x-1\right)\)
c. \(=x^3+2x^2-6x^2-12x+4x+8=\left(x+2\right)\left(x^2-6x+4\right)\)
a, Cách 1 : \(x^2+5x+6=x^2+2x+3x+6=\left(x+2\right)\left(x+3\right)\)
Cách 2 : \(x^2+5x+6=x^2+2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}+6\)
\(=\left(x+\frac{5}{2}\right)^2-\frac{1}{4}=\left(x+2\right)\left(x+3\right)\)
b, Cách 1 : \(x^2-x-6=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\)
Cách 2 : \(x^2-x-6=x^2-x+\frac{1}{4}-\frac{1}{4}-6=\left(x-\frac{1}{2}\right)^2-\frac{25}{4}=\left(x-3\right)\left(x+2\right)\)
c, Cách 1 : \(x^2+6x+8=x^2+4x+2x+8=\left(x+2\right)\left(x+4\right)\)
Cách 2 : \(x^2+6x+8=x^2+6x+9-1=\left(x+3\right)^2-1=\left(x+2\right)\left(x+4\right)\)
d, Cách 1 : \(x^2-2x-8=x^2+2x-4x-8=\left(x-4\right)\left(x+2\right)\)
Cách 2 : \(x^2-2x-8=x^2-2x+1-9=\left(x-1\right)^2-9=\left(x-4\right)\left(x+2\right)\)
a) \(=x^2-2x-4x+8\)
\(=x\left(x-2\right)-4\left(x-2\right)\)
\(=\left(x-2\right)\left(x-4\right)\)
c) \(=x^3-x-6x-6\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\)
\(=x\left(x+1\right)\left(x-1-6\right)\)
\(=x\left(x+1\right)\left(x-7\right)\)
\(3,\)Nhẩm nghiệm của đa thức trên ta đc : -1
Ta có lược đồ sau :
| 1 | 1 | -4 | -4 | |
| -1 | 1 | 0 | -4 | 0 |
Phân tích thành nhân tử ta có :\(\left(x+1\right)\left(x^2-4\right)\)
1) \(3x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(3x+5\right)\)
2) \(4x(x-2y)-8y(2y-x)\)
\(=4x\left(x-2y\right)+8y\left(x-2y\right)\)
\(=\left(4x+8y\right)\left(x-2y\right)\)
\(=4\left(x+2y\right)\left(x-2y\right)\)
3) \(a^2\left(x-1\right)+b^2\left(1-x\right)\)
\(=a^2\left(x-1\right)-b^2\left(x-1\right)\)
\(=\left(a^2-b^2\right)\left(x-1\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(x-1\right)\)
4) \(3x\left(x-a\right)+4a\left(a-x\right)\)
\(=3x\left(x-a\right)-4a\left(x-a\right)\)
\(=\left(x-a\right)\left(3x-4a\right)\)
5) \(5x\left(x-y\right)^2+10y^2\left(y-x\right)^2\)
\(=5x\left(x-y\right)^2+10y^2\left(x-y\right)^2\)
\(=\left(5x+10y^2\right)\left(x-y\right)^2\)
\(=5\left(x+2y^2\right)\left(x-y\right)^2\)
6) \(3x\left(x-3\right)^2+9\left(3-x\right)^2\)
\(=3x\left(x-3\right)^2+9\left(x-3\right)^2\)
\(=\left(3x+9\right)\left(x-3\right)^2\)
\(=3\left(x+3\right)\left(x-3\right)^2\)
7) \(x\left(m-a\right)^2-y\left(a-m\right)^2\)
\(=x\left(a-m\right)^2-y\left(a-m\right)^2\)
\(=\left(x-y\right)\left(a-m\right)^2\)
8) \(6y^2\left(x-1\right)^2+9y\left(1-x\right)^2\)
\(=6y^2\left(x-1\right)^2+9y\left(x-1\right)^2\)
\(=\left(6y^2+9x\right)\left(x-1\right)^2\)
\(=3\left(2y^2+3x\right)\left(x-1\right)^2\)
#Ayumu
a)
\(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\left(ĐKXĐ:x\ne4;x\ne2\right)\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{-x^2+6x-8}\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{-x^2+4x+2x-8}\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{\left(-x^2+4x\right)+\left(2x-8\right)}\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{-x.\left(x-4\right)+2.\left(x-4\right)}\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{\left(x-4\right).\left(2-x\right)}\)
\(\Leftrightarrow\frac{x+3}{x-4}-\frac{x-1}{2-x}=\frac{2}{\left(x-4\right).\left(2-x\right)}\)
\(\Leftrightarrow\frac{\left(x+3\right).\left(2-x\right)}{\left(x-4\right).\left(2-x\right)}-\frac{\left(x-1\right).\left(x-4\right)}{\left(x-4\right).\left(2-x\right)}=\frac{2}{\left(x-4\right).\left(2-x\right)}\)
\(\Rightarrow\left(x+3\right).\left(2-x\right)-\left(x-1\right).\left(x-4\right)=2\)
\(\Leftrightarrow2x-x^2+6-3x-\left(x^2-4x-x+4\right)=2\)
\(\Leftrightarrow2x-x^2+6-3x-x^2+4x+x-4=2\)
\(\Leftrightarrow4x-2x^2+2=2\)
\(\Leftrightarrow4x-2x^2+2-2=0\)
\(\Leftrightarrow4x-2x^2=0\)
\(\Leftrightarrow2x.\left(2-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0:2\\x=2-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=2\left(KTM\right)\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{0\right\}.\)
Chúc bạn học tốt!