lâu lâu ms làm,xin câu này :))
\(a,2x\left(x-3\right)-3\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(2x-3\right)=0\)
=>\(x-3=0\) hoặc \(2x-3=0\)
=>\(x=3\) hoặc \(x=\dfrac{3}{2}\)
Vậy....
\(b,3x\left(x-4\right)-x+4=0\)
=>\(\left(x-4\right)\left(3x-1\right)=0\)
=>....
=>\(x=4\) hoặc \(x=\dfrac{1}{3}\)
Vậy...
\(c,4x^2-1=0\)
=>\(4x^2=1\)
=>\(x^2=\dfrac{1}{4}\)
=>\(x=-\dfrac{1}{2}\) hoặc \(x=\dfrac{1}{2}\)
Vậy....
a. 2x(x-3) - 3(x-3) = 0
=> (2x-3)(x-3) = 0
=> 2x-3 = 0 hoặc x-3 = 0
=> x = 3/2 hoặc x=3
b.3x(x-4) - x + 4 = 0
=> 3x(x-4) - (x-4) = 0
=> (x-4)(3x-1) = 0
=> x-4 = 0 hoặc 3x-1 = 0
=> x = 4 hoặc x = 1/3
c. 4x2 - 1 = 0
=> 4x2 = 1
=> x2 = 1/4
=> x = 1/2 hoặc x = -1/2
a) 2x(x-3) - 3(x-3) =0
\(\Leftrightarrow\) (2x-3)(x-3)=0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b)3x(x-4)* x +4 =0
\(\Leftrightarrow\) 3x2 -12x*x +4=0
\(\Leftrightarrow\)3x2-12x2 +4=0
\(\Leftrightarrow\) -9x2 + 22 =0
\(\Leftrightarrow\)22 - (3x)2 =0
\(\Leftrightarrow\)( 2-3x)(2+ 3x) =0
\(\Leftrightarrow\left[{}\begin{matrix}2-3x=0\\2+3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{-2}{3}\end{matrix}\right.\)
c)4x2 - 1 = 0
\(\Leftrightarrow\)(2x)2 - 12= 0
\(\Leftrightarrow\)(2x -1)(2x +1)=0
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
