![](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)
\(a,x^3y^2-xy^2=xy^2\left(x^2-1\right)=xy^2\left(x-1\right)\left(x+1\right)\\ b,2x^3y^2+4x^2y^2+2xy^2=2xy^2\left(x^2+2x+1\right)=2xy^2\left(x+1\right)^2\\ c,3x^3y-12x^2y+12xy=2xy\left(x^2-4x+4\right)=2xy\left(x-2\right)^2\\ d,6x^3y+12x^2y^2+6xy^3=6xy\left(x^2+2xy+y^2\right)=6xy\left(x+y\right)^2\\ e,x^2\left(x-y\right)+y^2\left(y-x\right)=\left(x^2-y^2\right)\left(x-y\right)=\left(x-y\right)^2\left(x+y\right)\\ f,9x^2\left(x-2\right)-4y^2\left(x-2\right)=\left(9x^2-4y^2\right)\left(x-2\right)=\left(3x-2y\right)\left(3x+2y\right)\left(x-2\right)\)
Tick plz
a: \(x^3y^2-xy^2=xy^2\left(x^2-1\right)=xy^2\left(x-1\right)\left(x+1\right)\)
b: \(2x^3y^2+4x^2y^2+2xy^2=2xy^2\left(x^2+2x+1\right)=2xy^2\cdot\left(x+1\right)^2\)
c: \(3x^3y-12x^2y+12xy=3xy\left(x^2-4x+4\right)=3xy\cdot\left(x-2\right)^2\)
d: \(6x^3y+12x^2y^2+6xy^3=6xy\left(x^2+2xy+y^2\right)=6xy\cdot\left(x+y\right)^2\)
e: \(x^2\left(x-y\right)+y^2\left(y-x\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)
f: \(9x^2\left(x-2\right)-4y^2\left(x-2\right)=\left(x-2\right)\left(3x-2y\right)\left(3x+2y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
13x2 + 9y2 - 30x + 12xy + 25 = 0
<=> (9y2 + 12xy + 4y2) + (9x2 - 30x + 25) = 0
<=> (3y + 2x)2 + (3x - 5)2 = 0
Dễ thấy \(\left(3y+2x\right)^2\ge0;\left(3x-5\right)^2\ge0\forall x,y\)
nên \(\left(3y+2x\right)^2+\left(3x-5\right)^2\ge0\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}3y+2x=0\\3x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{10}{9}\\x=\dfrac{5}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(36x^2-25\right)-\left(6x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(6x+5\right)\left(6x-5\right)-\left(6x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(6x+5\right)\left(6x-5-x-1\right)=0\)
\(\Leftrightarrow\left(6x+5\right)\left(5x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{6}\\x=\frac{6}{5}\end{cases}}\)
\(\left(36x^2-25\right)-\left(6x+5\right)\left(x+1\right)=0\Leftrightarrow\left(6x-5\right)\left(6x+5\right)-\left(6x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(6x+5\right)\left(5x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{6}\\x=\frac{6}{5}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
11: \(2x^2-12xy+18y^2\)
\(=2\left(x^2-6xy+9y^2\right)\)
\(=2\left(x-3y\right)^2\)
12: \(\left(x^2+x\right)^2+3\left(x^2+x\right)+2\)
\(=\left(x^2+x+2\right)\left(x^2+x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,(x^2+9)^2-36x^2=0
(x^2+9-6x)(x^2+9+6x=0
=>x^2+9-6x=0 hoac x^2+9+6x=0
+,x^2+9-6x=0
x^2-6x+9=0
( x-3)^2 =0
=>x-3 =0
x =3
+,x^2+9+6x=0
x^2+6x+9=0
(x+3)^2 =0
=>x+3 =0
x =-3
Ý dưới cũng tương tự..
Dấu trừ ở trước thì bạn phải đổi dấu trog ngoặc