tìm x
5x2 - 15x = 0
5x2 -15x = 0
3 (x+5)-2x(x+5)=0
\(5x^2-15x=0\Leftrightarrow5x\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\\ 3\left(x+5\right)-2x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
1) \(5x^2-15x=0\)
\(\Rightarrow5x\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
2) \(3\left(x+5\right)-2x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(3-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(5x^2-15x=0\)
\(\Rightarrow x.\left(5x-15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\5x=15\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy x ∈ {0 ; 3}
\(3.\left(x+5\right)-2x.\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right).\left(3-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\\3-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\2x=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy x ∈ \(\left\{-5;\dfrac{3}{2}\right\}\)
2(x-3)(x2+1)+15x-5x2=0
\(\Leftrightarrow\left(x-3\right)\left(2x^2+2\right)+5x\left(3-x\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x^2-5x+2\right)=0\)
=>(x-3)(x-2)(2x-1)=0
=>x=3 hoặc x=2 hoặc x=1/2
\(2\left(x-3\right)\left(x^2+1\right)+15x-5x^2=0\\ \Leftrightarrow\left(x-3\right)\left(2x^2+2\right)-5x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(2x^2-5x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left[\left(2x^2-4x\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x-3\right)\left[2x\left(x-2\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x-3\right)\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(2\left(x-3\right)\left(x^2+1\right)+15x-5x^2=0\)
\(\Leftrightarrow2x^3+2x-6x^2-6+15x-5x^2=0\)
\(\Leftrightarrow2x^3-11x^2+17x-6=0\)
\(\Leftrightarrow2x^3-4x^2-7x^2+14x+3x-6=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-7x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-7x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-x-6x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-1\right)-3\left(2x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-1=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{2;\dfrac{1}{2};3\right\}\)
Tìm đa thức M biết:
a) x 3 - 5 x 2 +x - 5 = (x - 5).M;
b) ( x 2 - 4x - 3).M = 2 x 4 - 13 x 3 + 14 x 2 + 15x.
10(x+3) - 5x2-15x=0
thach ai lam dc :)
\(\Leftrightarrow\left(x+3\right)\left(10-5x\right)=0\)
=>x=-3 hoặc x=5
10(x+3)-5x(x+3)=0
(x+3)(10-5x)=0
x+3=0 hoặc 10-5x=0
x=-3. 2-x=0. x=2
Phân tích đa thức sau thành nhân tử: 5x2(x – 2y) – 15x(x – 2y)
5x2 (x – 2y)– 15x(x – 2y) = x.5x(x - 2y) - 3.5x(x - 2y)
= (x - 3).5x(x - 2y)
10(x+3) - 5x2-15x=0
cau cuoi cua de thi help em
=>(x+3)(10-5x)=0
=>x=-3 hoặc x=2
a) x2 – x;
b) 5x2(x – 2y) – 15x(x – 2y);
c) 3(x – y) – 5x(y – x).
\(a,x^2-x=x\left(x-1\right)\\ b,5x^2\left(x-2y\right)-15x\left(x-2y\right)=\left(5x^2-15x\right)\left(x-2y\right)=5x\left(x-3\right)\left(x-2y\right)\\ c,3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)=\left(5x+3\right)\left(x-y\right)\)
1/2x2 + 4x
2/15x3 + 5x2 - 10x
3) 5x2(x-2y) + 15x (x -2y)
4) 3(x - y) + 5x(y - x)
5/ 5x2 - 10x
6) 3x-6y
7)25x2+5x3+x2y
8)14x2y-21xy2+28x2y2
9)x(y-1)- y(y-1)
10) 10x(x-y)-8y(y-x)
giúp mình vs mn cảm ơn mn rất nhiều ạ
1, \(2x^2+4x=2x\left(x+2\right)\)
2, \(15x^3+5x^2-10x=5x\left(3x^2+x-2\right)=5x\left(x-\dfrac{2}{3}\right)\left(x+1\right)\)
3) \(5x^2\left(x-2y\right)+15x\left(x-2y\right)=\left(5x^2+15x\right)\left(x-2y\right)=5x\left(x+3\right)\left(x-2y\right)\)
4) \(3\left(x-y\right)+5x\left(y-x\right)=\left(x-y\right)\left(3-5x\right)\)
5) \(5x^2-10x=5x\left(x-2\right)\)
6) \(3x-6y=3\left(x-2y\right)\)
7) \(25x^2+5x^3+x^2y=x^2\left(25+5x+y\right)\)
8) \(14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)\)
9) \(x\left(y-1\right)-y\left(y-1\right)=\left(x-1\right)\left(y-1\right)\)
10) \(10x\left(x-y\right)-8y\left(y-x\right)=\left(10x+8y\right)\left(x-y\right)=2\left(5x+4y\right)\left(x-y\right)\)
\(1,=2x\left(x+2\right)\\ 2,=5x\left(3x^2+x-2\right)\\ 3,=\left(x-2y\right)\left(5x^2+15x\right)=5x\left(x+3\right)\left(x-2y\right)\\ 4,=\left(x-y\right)\left(3-5x\right)\\ 5,=5x\left(x-2\right)\\ 6,=3\left(x-2y\right)\\ 7,=5x^2\left(5+x+y\right)\\ 8,=7xy\left(2x-3y+4xy\right)\\ 9,=\left(y-1\right)\left(x-y\right)\\ 10,=\left(x-y\right)\left(10x+8y\right)=2\left(5x+4y\right)\left(x-y\right)\)
bài 3 phân tích đa thức sau thành nhân tử
a 4x2 -16 + (3x +12) (4-2x)
b x3 + X2Y -15x -15y
c 3(x+8) -x2 -8x
d x3 -3x2 + 1 -3x
e 5x2 -5y2 -20x + 20y
kkk =0)
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) \(5x^2-5y^2-20x+20y\)
\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y-4\right)\)