Toán Lớp 8: Tìm x:
a)2(x-1)+x=3
b)(2x-1)^2-25=0
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Toán Lớp 8: Tìm x:
a)2(x-1)+x=3
b)(2x-1)^2-25=0
Comments ( 2 )
a, 2(x-1) + x =3
2x – 2 + x = 3
3x – 2 = 3
=> 3x = 5
=> x = 5/3
Vậy x= 5/3
b, (2x-1)^2 – 25 = 0
( 2x-1)^2 – 5^2 =0
(2x-1+5)(2x-1-5)=0
(2x+4)(2x-6)=0
=>\(\left[ \begin{array}{l}2x+4=0\\2x-6=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}2x=-4\\2x=6\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
Vậy x ∈ {-2;3}
– Dùng HĐT số 3: a^2 – b^2 = (a+b)(a-b)
$a)2(x-1)+x=3$
$⇒ 2x-2+x=3
$⇒ 3x-2=3$
$⇒ 3x=5$
$⇒ x=\dfrac{5}{3}$
Vậy $x=\dfrac{5}{3}$
$b)(2x-1)^2-25=0$
$⇒ (2x-1)^2-5^2=0$
$⇒ (2x-1-5)(2x-1+5)=0$
$⇒ (2x-6)(2x+4)=0$
$⇒$ \(\left[ \begin{array}{l}2x-6=0\\2x+4=0\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}2x=6\\2x=-4\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=3\\x=-2\end{array} \right.\)
Vậy $x$ ∈ {$3;-2$}