1) Giải các phương trình sau :
a) (x+2)(x-5) = 0;
b) (3x – 2)(4x + 5)=0;
c) (4x + 2)(x ² + 1)=0;
d) (x + 3).(2x – 5)=0.
1) Giải các phương trình sau :
a) (x+2)(x-5) = 0;
b) (3x – 2)(4x + 5)=0;
c) (4x + 2)(x ² + 1)=0;
d) (x + 3).(2x – 5)=0.
`\text{Đáp án: + Giải thích các bước giải:}`
`a ) ( x + 2 ) ( x – 5 ) = 0`
`⇔` \(\left[ \begin{array}{l}x+2=0\\x-5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-2\\x=5\end{array} \right.\)
`\text{Vậy S = { -2 ; 5}}`
`b ) (3x – 2)(4x + 5)=0`
`⇔` \(\left[ \begin{array}{l}3x-2=0\\4x + 5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}3x=2\\4x=-5\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\frac{2}{3}\\x=\frac{-5}{4}\end{array} \right.\)
`\text{Vậy S =}` `{ 2/3 ; (-5)/4 }`
`c ) ( 4x + 2 )( x^2 + 1 ) = 0`
`⇔` \(\left[ \begin{array}{l}4x+2=0\\x(x+1)=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}4x=-2\\x=0\\x +1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\frac{-1}{4}\\x=0\\x=-1\end{array} \right.\)
`\text{Vậy S =}` `{(-1)/4 ; -1;0}`
`d ) (x + 3)(2x – 5)=0`
`⇔` \(\left[ \begin{array}{l}x+3=0\\2x-5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-3\\x=\frac{5}{2}\end{array} \right.\)
`\text{Vậy S =}` `{-3;5/2}`
a) (x+2) (x-5) = 0
⇔ x+2 = 0 hay x-5=0
⇔ x = -2 hay x = 5
Vậy S={-2;5}
b) (3x – 2)(4x + 5)=0
⇔3x – 2 = 0 hay 4x + 5 = 0
⇔3x = 2 hay 4x = -5
⇔ x = $\frac{2}{3}$ hay x = $-\frac{5}{4}$
Vậy S={$\frac{2}{3}$; $-\frac{5}{4}$}
c)(4x + 2)(x ² + 1)=0
⇔4x + 2 = 0 hay x ² + 1 = 0
⇔4x = -2 hay x² = -1
⇔x = $-\frac{1}{2}$ hay x = ±$1$
Vậy S={$-\frac{1}{2}$; ±$1$}
d) (x + 3).(2x – 5)=0
⇔x + 3 = 0 hay 2x – 5 = 0
⇔x = -3 hay 2x = 5
⇔x = -3 hay x = $\frac{5}{2}$
Vậy S={-3; $\frac{5}{2}$}
$chucbanhoktot$
$chomikxinctlhnnha$