## 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

Question

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

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1 giờ 2021-09-07T10:35:21+00:00 2 Answers 0 views 0

1. 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

<=>$$\left[ \begin{array}{l}x=2\\x=3\end{array} \right.$$

2. 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\}