(x-2)(x+3)=0 (2x-3)(x^2+1)=0 3x(x-2)+4(x-2)=0 (3x-4)^2-(x-3)^2=0
2 câu trả lời
Đáp án:
↓
Giải thích các bước giải:
(x-2)(x+3)=0
⇒ [x−2=0x+3=0
⇒ [x=2x=−3
(2x-3)(x2+1)=0
(2x-3)(x2+1)=0
⇒ [2x−3=0x2+1=0
⇒ [x=32 ( lấy )x=−1 ( loại )
3x(x-2)+4(x-2)=0
⇒(3x+4)(x-2)=0
⇒ [3x+4=0x−2=0
⇒ [x=−43x=2
(3x-4)2-(x-3)2=0
⇒
=> (4x - 7)(2x - 1) = 0
=> \left[ \begin{array}{l}4x - 7 = 0\\2x - 1 = 0\end{array} \right.
=> \left[ \begin{array}{l}x=\dfrac{7}{4}\\x = \dfrac{1}{2}\end{array} \right.
#Sad
Đáp án+Giải thích các bước giải:
(x-2)(x+3)=0^{}
=> \left[ \begin{array}{l}x-2=0\\x+3=0\end{array} \right.
=> \left[ \begin{array}{l}x=2\\x=-3\end{array} \right.
Vậy x=2 hoặc x=-3
______________________
(2x-3)(x^2+1)=0^{}
=> \left[ \begin{array}{l}2x-3=0\\x^2+1=0\end{array} \right.
=> \left[ \begin{array}{l}2x=3\\x^2=-1\end{array} \right.
=> \left[ \begin{array}{l}x=\dfrac{3}{2}\\x^2=-1 (ko có giá trị của x)\end{array} \right.
Vậy x=\dfrac{3}{2}
_____________________
3x(x-2)+4(x-2)=0^{}
=> (x-2)(3x+4)=0^{}
=> \left[ \begin{array}{l}x-2=0\\3x+4=0\end{array} \right.
=> \left[ \begin{array}{l}x=2\\3x=-4\end{array} \right.
=> \left[ \begin{array}{l}x=2\\x=\dfrac{-4}{3}\end{array} \right.
Vậy x=2 hoặc x=\dfrac{-4}{3}
_______________________
(3x-4)^2-(x-3)^2=0^{}
=> [(3x-4)+(x-3)].[(3x-4-(x-3)]=0^{}
=> (3x-4+x-3)(3x-4-x+3)=0^{}
=> (4x-7)(2x-1)=0^{}
=> \left[ \begin{array}{l}4x-7=0\\2x-1=0\end{array} \right.
=> \left[ \begin{array}{l}4x=7\\2x=1\end{array} \right.
=> \left[ \begin{array}{l}x=\dfrac{7}{4}\\x=\dfrac{1}{2}\end{array} \right.
Vậy x=\dfrac{7}{4} hoặc x=\dfrac{1}{2}