Giải các phương trình sau:
(5x-2)/6+(3-4x)/2=2-(x+7)/3
Bài 4. Giải các phương trình sau:
a) (x – 5)(x + 3) = 0; b) 3x (x – 5) + 6(x – 5) = 0;
c) 2(x – 3) – 4(3 – x) = 0 d) x2 + 5x + 6 = 0.
Đáp án:
`↓↓`
Giải thích các bước giải:
$\dfrac{5x-2}{6}+$ $\dfrac{3-4x}{2}=2-$ $\dfrac{x+7}{3}$
`⇔\frac{5x-2}{6}+\frac{3(3-4x)}{6}=\frac{12}{6}-\frac{2(x+7)}{6}`
`⇔5x-2+9-12x=12-2x-14`
`⇔-7x+7=-2x-2`
`⇔5x=9`
`⇔x=9/5`
Vậy `S=\{9/5\}`
Bài 4:
`a)(x-5)(x+3)=0`
\(⇔\left[ \begin{array}{l}x-5=0\\x+3=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=5\\x=-3\end{array} \right.\)
Vậy `S=\{5;-3\}`
`b)3x(x-5)+6(x-5)=0`
`⇔(3x+6)(x-5)=0`
`⇔3(x+2)(x-5)=0`
\(⇔\left[ \begin{array}{l}x+2=0\\x-5=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=-2\\x=5\end{array} \right.\)
Vậy `S=\{-2;5\}`
`c)2(x-3)-4(3-x)=0`
`⇔2(x-3)+4(x-3)=0`
`⇔6(x-3)=0`
`⇔x-3=0`
`⇔x=3`
Vậy `S=\{3\}`
`d)x^2+5x+6=0`
`⇔x^2+2x+3x+6=0`
`⇔x(x+2)+3(x+2)=0`
`⇔(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 `S=\{-2;-3\}`
Đáp án + Giải thích các bước giải:
`(5x-2)/(6)+(3-4x)/(2)=2-(x+7)/(3)`
`⇔(5x-2)/(6)+(3(3-4x))/(6)=(12)/(6)-(2(x+7))/(6)`
`⇒5x-2+3(3-4x)=12-2(x+7)`
`⇔5x-2+9-12x=12-2x-14`
`⇔5x-12x+2x=2-9+12-14`
`⇔-5x=-9`
`⇔x=(9)/(5)`
Vậy `S={(9)/(5)}`
Bài `4:`
`a//(x-5)(x+3)=0`
`⇔` \(\left[ \begin{array}{l}x-5=0\\x+3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=5\\x=-3\end{array} \right.\)
Vậy `S={5;-3}`
`b//3x(x-5)+6(x-5)=0`
`⇔(x-5)(3x+6)=0`
`⇔` \(\left[ \begin{array}{l}x-5=0\\3x+6=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=5\\x=-2\end{array} \right.\)
Vậy `S={5;-2}`
`c//2(x-3)-4(3-x)=0`
`⇔2(x-3)+4(x-3)=0`
`⇔(x-3)(2+4)=0`
`⇔6(x-3)=0`
`⇔x-3=0`
`⇔x=3`
Vậy `S={3}`
`d//x^{2}+5x+6=0`
`⇔(x^{2}+2x)+(3x+6)=0`
`⇔x(x+2)+3(x+2)=0`
`⇔(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 `S={-2;-3}`