\(3\left(2x-1\right)-5\left(x-3\right)=0\)
\(6x-3-5x+15=0\)
\(x+12=0\)
\(x=-12\)
a) \(3\left(2x-1\right)-5\left(x-3\right)=0\)
\(\Leftrightarrow6x-3-5x+15=0\)
\(\Leftrightarrow x+12=0\)
\(\Leftrightarrow x=-12\)
b) \(\left(x-3\right)\left(x-7\right)-x^2=1\)
\(\Leftrightarrow x^2-10x+21-x^2=1\)
\(\Leftrightarrow-10x=-20\Leftrightarrow x=2\)
c) \(5\left(x+3\right)-2x\left(3+x\right)=0\)
\(\Leftrightarrow5x+15-6x-2x^2=0\)
\(\Leftrightarrow-2x^2-x+15=0\)
Ta có: \(\Delta=1^2+4.2.5=41\)
pt có 2 nghiệm
\(x_1=\frac{1+\sqrt{41}}{-4}\);\(x_2=\frac{1-\sqrt{41}}{-4}\)
\(5\left(x+3\right)-2x\left(3+x\right)=0\)
\(5\left(x+3\right)+2x\left(x+3\right)=0\)
\(\left(5+2x\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5+2x=0\\x+3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=-3\end{cases}}\)
\(\left(x-3\right)\left(x-7\right)-x^2=1\)
\(x^2-7x-3x+21-x^2-1=0\)
\(-10x+22=0\)
\(-10x=-22\)
\(x=\frac{11}{5}\)
a, 3(2x-1) - 5(x-3) = 0
=> 3(2x-1) = 5(x-3)
=> 3(2x-1) = 5(x-3) hoặc 3(2x-1) = -5(x-3)
+) 3(2x-1) = 5 (x-3)
<=> 6x-3 = 5x-15
<=> 6x - 5x = -15+3
<=> x = -12
+) 3(2x-1) = -5(x-3)
<=> 6x-3 = -5x+15
<=> 6x+5x = 15+3
<=> 11x = 18
<=> x = 18/11
Vậy .....
a, 3 ( 2x - 1 ) - 5 ( x - 3 ) = 0
6x - 3 - 5x + 15 = 0
x + 12 = 0
x = - 12
Vậy ...
b, ( x - 3 ) ( x - 7 ) - x2 = 1
x2 - 7x - 3x + 21 - x2 = 1
21 - 10x = 1
10x = 20
x = 2
Vậy ...
c, 5 ( x + 3 ) - 2x ( x + 3 ) = 0
( x + 3 ) ( 5 - 2x ) = 0
\(\Rightarrow\orbr{\begin{cases}x+3=0\\5-2x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\2x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=\frac{5}{2}\end{cases}}}\)
Vậy...
