K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
19 tháng 10 2021
5x2-15x=0
5x(x-3) =0
TH1: 5x=0 TH2: x-3=0
=>x=0 => x=3
Vậy x thuộc {0;3}
29 tháng 1 2022
\(\Leftrightarrow\left(x-3\right)\left(2x^2+2\right)+5x\left(3-x\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x^2-5x+2\right)=0\)
=>(x-3)(x-2)(2x-1)=0
=>x=3 hoặc x=2 hoặc x=1/2
29 tháng 1 2022
\(2\left(x-3\right)\left(x^2+1\right)+15x-5x^2=0\\ \Leftrightarrow\left(x-3\right)\left(2x^2+2\right)-5x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(2x^2-5x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left[\left(2x^2-4x\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x-3\right)\left[2x\left(x-2\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x-3\right)\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
19 tháng 10 2021
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) \(10x\left(x-y\right)-8y\left(y-x\right)=\left(10x+8y\right)\left(x-y\right)=2\left(5x+4y\right)\left(x-y\right)\)
19 tháng 10 2021
\(1,=2x\left(x+2\right)\\ 2,=5x\left(3x^2+x-2\right)\\ 3,=\left(x-2y\right)\left(5x^2+15x\right)=5x\left(x+3\right)\left(x-2y\right)\\ 4,=\left(x-y\right)\left(3-5x\right)\\ 5,=5x\left(x-2\right)\\ 6,=3\left(x-2y\right)\\ 7,=5x^2\left(5+x+y\right)\\ 8,=7xy\left(2x-3y+4xy\right)\\ 9,=\left(y-1\right)\left(x-y\right)\\ 10,=\left(x-y\right)\left(10x+8y\right)=2\left(5x+4y\right)\left(x-y\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 8 2021
1.
\(\left(x-5\right)^2+3\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-5+3\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
2.
\(\left(x^2-9\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 8 2021
3.
\(\left(2x+1\right)^2+\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1+x-1\right)=0\)
\(\Leftrightarrow\left(2x+1\right).3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\2x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
4.
\(\left(x-1\right)\left(x+3\right)+\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1+x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
C
1
L
14 tháng 8 2021
a) \(x^2-x+x=4\)
\(x^2=4\)
\(x=\pm2\)
b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)
\(\left(x-5\right)\left(3x-2\right)=0\)
\(\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
c) Ta có: \(a+b+c=5-3-2=0\)
\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)
d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :
\(t^2-11t+18=0\)
\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)
\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)
\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)
3D
Giải phương trình
6(2x-5)-6x2+15x=0
\(\dfrac{x+2}{x}\)=\(\dfrac{x^2+5x+4}{x^2+2x}\)+\(\dfrac{x}{x+2}\)
1
7 tháng 4 2022
a: =>6(2x-5)-3x(2x-5)=0
=>(2x-5)(6-3x)=0
=>x=5/2 hoặc x=2
b: \(\Leftrightarrow x^2+5x+4+x^2=x^2+4x+4\)
=>x2+x=0
=>x(x+1)=0
=>x=0(loại) hoặc x=-1(nhận)
\(5x^2-15x=0\Leftrightarrow5x\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\\ 3\left(x+5\right)-2x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
1) \(5x^2-15x=0\)
\(\Rightarrow5x\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
2) \(3\left(x+5\right)-2x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(3-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)