Giải phương trình a, 4(x+2) – 1 = x-5 b, 2/x+1 – 1/x-2 = 3x -11/(x+1)(x-2) c, (x-2)(2x-1) = 5(x-2) Giải hộ tớ với ạ 07/09/2021 Bởi Nevaeh Giải phương trình a, 4(x+2) – 1 = x-5 b, 2/x+1 – 1/x-2 = 3x -11/(x+1)(x-2) c, (x-2)(2x-1) = 5(x-2) Giải hộ tớ với ạ
`a)4(x+2)-1=x-5` `<=>4x+8-1=x-5` `<=>3x=-12` `<=>x=-4` `b)2(x+1)-1/(x-2)=(3x-11)/((x+1)(x-2))` `=>2(x-2)-1(x+1)=3x-11` `<=>2x-4-x-1=3x-11` `<=>-2x=-6` `<=>x=3` `c)(x-2)(2x-1)=5(x-2)` `<=>(x-2)(2x-1)-5(x-2)=0` `<=>(x-2)(2x-1-5)=0` `<=>(x-2)(2x-6)=0` `<=>2(x-2)(x-3)=0` `<=>`\(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\) Bình luận
Đáp án+Giải thích các bước giải: `a)4(x+2)-1=x-5` `⇔4x+8-1=x-5` `⇔4x+7=x-5` `⇔4x-x=-5-7` `⇔3x=-12` `⇔x=-4` Vậy `S=\{-4\}` `b)ĐKXĐ:x\ne -1;x\ne 2` `2/(x+1)-1/(x-2)=(3x-11)/((x+1)(x-2))` `⇔(2(x-2))/((x+1)(x-2))-(1(x+1))/((x-2)(x+1))=(3x-11)/((x+1)(x-2))` `⇒2(x-2)-(x+1)=3x-11` `⇔2x-4-x-1=3x-11` `⇔x-5=3x-11` `⇔3x-x=-5+11` `⇔2x=6` `⇔x=3(TM)` Vậy `S=\{3\}` `c)(x-2)(2x-1)=5(x-2)` `⇔(x-2)(2x-1)-5(x-2)=0` `⇔(x-2)[(2x-1)-5]=0` `⇔(x-2)(2x-1-5)=0` `⇔(x-2)(2x-6)=0` \(⇔\left[ \begin{array}{l}x-2=0\\2x-6=0\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=2\\2x=6\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\) Vậy `S=\{2;3\}` Bình luận
`a)4(x+2)-1=x-5`
`<=>4x+8-1=x-5`
`<=>3x=-12`
`<=>x=-4`
`b)2(x+1)-1/(x-2)=(3x-11)/((x+1)(x-2))`
`=>2(x-2)-1(x+1)=3x-11`
`<=>2x-4-x-1=3x-11`
`<=>-2x=-6`
`<=>x=3`
`c)(x-2)(2x-1)=5(x-2)`
`<=>(x-2)(2x-1)-5(x-2)=0`
`<=>(x-2)(2x-1-5)=0`
`<=>(x-2)(2x-6)=0`
`<=>2(x-2)(x-3)=0`
`<=>`\(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\)
Đáp án+Giải thích các bước giải:
`a)4(x+2)-1=x-5`
`⇔4x+8-1=x-5`
`⇔4x+7=x-5`
`⇔4x-x=-5-7`
`⇔3x=-12`
`⇔x=-4`
Vậy `S=\{-4\}`
`b)ĐKXĐ:x\ne -1;x\ne 2`
`2/(x+1)-1/(x-2)=(3x-11)/((x+1)(x-2))`
`⇔(2(x-2))/((x+1)(x-2))-(1(x+1))/((x-2)(x+1))=(3x-11)/((x+1)(x-2))`
`⇒2(x-2)-(x+1)=3x-11`
`⇔2x-4-x-1=3x-11`
`⇔x-5=3x-11`
`⇔3x-x=-5+11`
`⇔2x=6`
`⇔x=3(TM)`
Vậy `S=\{3\}`
`c)(x-2)(2x-1)=5(x-2)`
`⇔(x-2)(2x-1)-5(x-2)=0`
`⇔(x-2)[(2x-1)-5]=0`
`⇔(x-2)(2x-1-5)=0`
`⇔(x-2)(2x-6)=0`
\(⇔\left[ \begin{array}{l}x-2=0\\2x-6=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=2\\2x=6\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\)
Vậy `S=\{2;3\}`