![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)
b: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
c:\(-6x^2+7x-2\)
\(=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(-3x+2\right)\)
1.
a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
2.
a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)
3.
b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)
4.
a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(4x^2-16+\left(3x+12\right)\left(4-2x\right)\)
\(=\left(2x-4\right)\left(2x+4\right)-3\left(x+4\right)\left(2x-4\right)\)
\(=\left(2x-4\right)\left(2x+4-3x-12\right)\)
\(=-\left(2x-4\right)\left(x+8\right)\)
b) \(x^3+x^2y-15x-15y\)
\(=x^2\left(x+y\right)-15\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-15\right)\)
c) \(3\left(x+8\right)-x^2-8x\)
\(=3\left(x+8\right)-x\left(x+8\right)\)
\(=\left(x+8\right)\left(3-x\right)\)
d) \(x^3-3x^2+1-3x\)
\(=x^3+1-3x^2-3x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
d) \(5x^2-5y^2-20x+20y\)
\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y-4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
45 + x 3 - 5 x 2 - 9 x = x 3 - 5 x 2 - 9 x - 45 = x 2 x - 5 - 9 x - 5 = x - 5 x 2 - 9 = x - 5 x - 3 x + 3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{x^3 – 5x^2 + 8x – 4 }\)
\(\text{= x^3 – 4x^2 + 4x – x^2 + 4x – 4}\)
\(\text{= x( x^2 – 4x + 4 ) – ( x^2 – 4x + 4 )}\)
\(\text{= ( x – 1 ) ( x – 2 )^2}\)
\(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x-4\\ =x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\\ =\left(x^2-4x+4\right)\left(x-1\right)\\ =\left(x-2\right)^2\left(x-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$2x^2+y^2-2xy+2x-4y+9$
$=(x^2+y^2-2xy)+4(x-y)+(x^2-2x+1)+8$
$=(x-y)^2+4(x-y)+4+(x-1)^2+4$
$=(x-y+2)^2+(x-1)^2+4$
Này chỉ tính được min thôi chứ không phân tích được thành nhân tử bạn nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
giải bài toán: cho tam giác MNP, NTlà phân giác của góc N biết MN=4cm, NT=10cm, MP=8cm:TínhTM, TP?
\(x^3-5x^2+3x+9\)
Gỉa sử \(x^3-5x^2+3x+9=\left(x+a\right)\left(x^2+bx+c\right)\)
\(=x^3+bx^2+cx+ax^2+abx+ac\)
\(=x^3+\left(a+b\right)x^2+\left(c+ab\right)x+ac\)
Từ \(x^3-5x^2+3x+9=x^3+\left(a+b\right)x^2+\left(c+ab\right)x+ac\) Hợp nhất ta được :
\(\hept{\begin{cases}a+b=-5\\c+ab=3\\ac=9\end{cases}}\)
Dự đoán \(a=-3\Rightarrow\hept{\begin{cases}b=-2\\c=-3\end{cases}}\)
\(\Rightarrow\left(x+a\right)\left(x^2+bx+c\right)=\left(x-3\right)\left(x^2-2x-3\right)=\left(x-3\right)^2\left(x+1\right)\)
Vậy \(x^3-5x^2+3x+9=\left(x-3\right)^2\left(x+1\right)\)
DỄ ẸT MÀK KO BT LÀM CÒN LÊN ĐÂY HỎI ...LP 8 ĐỂ LÀM J DANH HIỆU HỞ~~~~~CÁI NÀY THÌ LP 5 CX LÀM ĐC NHÉ!