
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


a)2x2-6x=0
=>x(2x-6)=0
=>x=0 hoặc 2x-6=0
Với 2x-6=0 =>2x=6 <=>x=3

Answer:
\(3x^2-4x=0\)
\(\Rightarrow x\left(3x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)
\(\left(x^2-5x\right)+x-5=0\)
\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
\(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)
\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(5x\left(x-3\right)-x+3=0\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)
\(x^2-2x+5=0\)
\(\Rightarrow\left(x^2-2x+1\right)+4=0\)
\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)
Vậy không có giá trị \(x\) thoả mãn
\(x^2+x-6=0\)
\(\Rightarrow x^2+3x-2x-6=0\)
\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

(x - 4)(5x -2 -3)=0
Nên x - 4 = 0 nên x = 4
hoặc 5x - 2 - 3 = 0 nên x = 1
\(\Leftrightarrow5x^2-2x-20x+8-3x+12=0\)
\(\Leftrightarrow5x^2-25x+20=0\)
\(\Leftrightarrow5.\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow x^2-5x+4=0\)
\(=x^2-x-4x+4=0\)
\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left(x-4\right)=0\)
\(\Leftrightarrow\) \(x-1=0\) và \(x-4=0\)
\(\Leftrightarrow\) \(x=1\) và \(x=4\)

(x - 4)(5x - 2) - 3(x - 4) = 0
=> (x - 4)(5x - 2 - 3) = 0
=> (x - 4)(5x - 5) = 0
=> (x - 4).5.(x - 1) = 0
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}}\)
vậy

Bài 2: Tính giá trị của biểu thức sau:
\(16x^2-y^2=\left(4x+y\right)\left(4x-y\right)\)
Thay \(\hept{\begin{cases}x=87\\y=13\end{cases}}\)
\(\Rightarrow\left(4.87+13\right)\left(4.87-13\right)=361.335=120935\)
Bài 4: Tìm x
a) \(9x^2+x=0\)
\(\Rightarrow x\left(9x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\9x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{9}\end{cases}}\)
b) \(27x^3+x=0\)
\(\Rightarrow x\left(27x^2+1=0\right)\)
\(\Rightarrow\orbr{\begin{cases}x=0\\27x^2+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\27x^2=\left(-1\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=\frac{-1}{27}\end{cases}}\)
Ta có: \(\frac{-1}{27}\) loại vì \(x^2\ge0\forall x\)
Vậy \(x=0\)

#)Giải :
a)\(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}\)
\(\Leftrightarrow5x-3\left[4x-2\left(4x-15x+6\right)\right]\)
\(\Leftrightarrow5x-3\left(4x-8x+30x-12\right)\)
\(\Leftrightarrow5x-12x+24x-90x+36\)
\(\Leftrightarrow-73x+36=182\)
\(\Leftrightarrow-73x=146\)
\(\Leftrightarrow x=-2\)
Bài giải
\(\left(x-4\right)\left(5x-2\right)-3\left(x-4\right)=0\)
\(\left(x-4\right)\left(5x-2-3\right)=0\)
\(\left(x-4\right)\left(5x-5\right)=0\)
\(\left(x-4\right)x\left(x-1\right)=0\)
\(\Rightarrow\)Hoặc \(x-4=0\text{ }\Rightarrow\text{ }x=4\)
Hoặc \(x=0\)
Hoặc \(x-1=0\text{ }\Rightarrow\text{ }x=1\)
\(\Rightarrow\text{ }x\in\left\{4\text{ ; }0\text{ ; }1\right\}\)
( x - 4 )( 5x - 2 ) - 3( x - 4 ) = 0
⇔ 5x2 - 2x - 20x + 8 - 3x + 12 = 0
⇔ 5x2 - 25x + 20 = 0
⇔ 5x2 - 5x - 20x + 20 = 0
⇔ 5x( x - 1 ) - 20( x - 1 ) = 0
⇔ ( x - 1 )( 5x - 20 ) = 0
⇔ \(\orbr{\begin{cases}x-1=0\\5x-20=0\end{cases}\text{⇔}}\orbr{\begin{cases}x=1\\x=4\end{cases}}\)