a)4x2-9=0
⇔ (2x-3)(2x+3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b)(x+5)2-(x-1)2=0
⇔ (x+5-x+1)(x+5+x-1)=0
⇔ 12(x+2)=0
⇔ x=-2
c)x2-6x-7=0
⇔ x2-7x+x-7=0
⇔ x(x-7)+(x-7)=0
⇔ (x-7)(x+1)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
d)(x+1)2-(2x-1)2=0
⇔ (x+1-2x+1)(x+1+2x-1)=0
⇔3x(2-x)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
a, 4x2 - 9 = 0
<=> 4x2 = 9
<=> x2 = \(\dfrac{9}{4}\) => x = \(\sqrt{\dfrac{9}{4}}\)
b, (x + 5 )2 - ( x - 1 )2 = 0
<=> ( x+5-x+1 )(x+5+x-1) = 0
<=> 6(2x+4) = 0
<=> 12x+24=0
<=> 12x = -24
<=> x = -2
c, x2-6x-7=0
<=> x2+x-7x-7=0
<=> x(x+1)-7(x+1)=0
<=> (x-7)(x+1)=0
=> x+7=0 hoặc x+1=0
+ x-7=0 => x=7
+ x+1=0 => x=-1
d, \(\left(x+1\right)^2-\left(2x-1\right)^2=0\)
<=> \(\left(x+1-2x+1\right)\left(x+1+2x-1\right)=0\)
<=> (-x+2).3x=0
=> x=0 hoặc (-x+2).3=0
+ (-x+2).3=0 => -3x+6=0 => x=-2
b) (x +5)2 -(x -1)2=0
<=> [(x +5) -(x -1)][(x +5) +(x -1)]=0
<=> (x +5 -x +1)(x +5 +x -1)=0
<=> 6(2x+4)=0 <=>12(x +2)=0
=> x +2=0=> x=-2
vậy x= -2
c) x2 -6x -7=0
<=> x2 -7x +x -7=0
<=> (x2 +x)( -7x -7)=0
<=> x(x +1).-7(x +1)=0
<=> (x +1)(x -7)=0
<=> \(\left\{{}\begin{matrix}x+1=0\\x-7=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=-1\\x=7\end{matrix}\right.\)
Vậy S={-1; 7}
d) (x +1)2 -(2x -1)2=0
<=> [(x -1)-(2x -1)][(x -1)+(2x -1)]=0
<=> (x -1 -2x +1)(x -1 +2x -1)=0
<=> (x -2x)(3x -2)<=> -x(3x -2)=0
<=> \(\left\{{}\begin{matrix}-x=0\\3x-2=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy S={0; \(\dfrac{2}{3}\)}
a) 4x2 -9=0
<=> (2x2)2 -32=0
<=> (2x2 -3)(2x2 +3)=0
<=>\(\left\{{}\begin{matrix}2x^2-3=0\\2x^2+3=0\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}2x^2=3\\2x^2=-3\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x^2=1,5\\x^2=-1,5\left(L\right)\end{matrix}\right.\)
=> x=1,5
Vậy x=1,5
