Câu 1:

`a)5(x+2)=3x-10`

`⇔5x+10=3x-10`

`⇔5x-3x=-10-10`

`⇔2x=-20`

`⇔x=(-20):2`

`⇔x=-10`

Vậy `S={-10}`

`b)(2x+5)(3x-1)=(2x+5)(x-3)`

`⇔(2x+5)(3x-1)-(2x+5)(x-3)=0`

`⇔(2x+5)[(3x-1)-(x-3)]=0`

`⇔(2x+5)(3x-1-x+3)=0`

`⇔(2x+5)(2x+2)=0`

`⇔`$\left[\begin{matrix} 2x+5=0\\ 2x+2=0\end{matrix}\right.$

`⇔`$\left[\begin{matrix} 2x=-5\\ 2x=-2\end{matrix}\right.$

`⇔`$\left[\begin{matrix} x=-\dfrac{5}{2}\\ x=-1\end{matrix}\right.$

Vậy `S={-5/2;-1}`

`c)2/(x²+2x)-1/x=(x-2)/(x+2)(ĐKXĐ:x`$\neq$ `0,x`$\neq$ `-2)`

`⇔2/[x(x+2)]-1/x=(x-2)/(x+2)`

`⇔2/[x(x+2)]-(x+2)/[x(x+2)]=[x(x-2)]/[x(x+2)]`

`⇒2-(x+2)=x(x-2)`

`⇔2-x-2=x²-2x`

`⇔2-x-2-x²+2x=0`

`⇔-x²+(-x+2x)+(2-2)=0`

`⇔-x²+x=0`

`⇔-x(x-1)=0`

`⇔`$\left[\begin{matrix} -x=0\\ x-1=0\end{matrix}\right.$

`⇔`$\left[\begin{matrix} x=0(Ko TM ĐKXĐ)\\ x=1(TM ĐKXĐ)\end{matrix}\right.$

Vậy `S={1}`

`d)|2x-1|+5=6`

`⇔|2x-1|=6-5`

`⇔|2x-1|=1`

`⇔`$\left[\begin{matrix} 2x-1=1\\ 2x-1=-1\end{matrix}\right.$

`⇔`$\left[\begin{matrix} 2x=2\\ 2x=0\end{matrix}\right.$

`⇔`$\left[\begin{matrix} x=1\\ x=0\end{matrix}\right.$

Vậy `S={0;1}`

`e)5(x-3)=3x+10`

`⇔5x-15=3x+10`

`⇔5x-3x=10+15`

`⇔2x=25`

`⇔x=25/2`

Vậy `S={25/2}`

`f)(2x-5)(3x+1)=(2x-5)(x+3)`

`⇔(2x-5)(3x+1)-(2x-5)(x+3)=0`

`⇔(2x-5)[(3x+1)-(x+3)]=0`

`⇔(2x-5)(3x+1-x-3)=0`

`⇔(2x-5)(2x-2)=0`

`⇔`$\left[\begin{matrix} 2x-5=0\\ 2x-2=0\end{matrix}\right.$

`⇔`$\left[\begin{matrix} 2x=5\\ 2x=2\end{matrix}\right.$

`⇔`$\left[\begin{matrix} x=\dfrac{5}{2}\\ x=1\end{matrix}\right.$

Vậy `S={5/2;1}`

`g)2/(x²-2x)+1/x=(x+2)/(x-2)(ĐKXĐ:x`$\neq$ `0,x`$\neq$ `2)`

`⇔2/[x(x-2)]+1/x=(x+2)/(x-2)`

`⇔2/[x(x-2)]+(x-2)/[x(x-2)]=[x(x+2)]/[x(x-2)]`

`⇒2+x-2=x(x+2)`

`⇔2+x-2=x²+2x`

`⇔2+x-2-x²-2x=0`

`⇔-x²+(x-2x)+(-2+2)=0`

`⇔-x²-x=0`

`⇔-x(x+1)=0`

`⇔`$\left[\begin{matrix} -x=0\\ x+1=0\end{matrix}\right.$

`⇔`$\left[\begin{matrix} x=0(Ko TM ĐKXĐ)\\ x=-1(TM ĐKXĐ)\end{matrix}\right.$

Vậy `S={-1}`

`h)|2x-1|-5=4`

`⇔|2x-1|=4+5`

`⇔|2x-1|=9`

`⇔`$\left[\begin{matrix} 2x-1=9\\ 2x-1=-9\end{matrix}\right.$

`⇔`$\left[\begin{matrix} 2x=10\\ 2x=-8\end{matrix}\right.$

`⇔`$\left[\begin{matrix} x=5\\ x=-4\end{matrix}\right.$

Vậy `S={5;-4}`

Câu 2:

`a)3-2x≥4`

`⇔-2x≥4-3`

`⇔-2x≥1`

`⇔x≤-1/2`

Vậy `S={x|x≤-1/2}`

Biểu diễn trên trục số:`(ảnh 1)`

`b)(4x-5)/3>(7-x)/5`

`⇔[5(4x-5)]/15>[3(7-x)]/15`

`⇔5(4x-5)>3(7-x)`

`⇔20x-25>21-3x`

`⇔20x+3x>21+25`

`⇔23x>46`

`⇔x>46:23`

`⇔x>2`

Vậy `S={x|x>2}`

Biểu diễn trên trục số:`(ảnh 1)`

`c)4x-3≥2x+4`

`⇔4x-2x≥4+3`

`⇔2x≥7`

`⇔x≥7/2`

Vậy `S={x|x≥7/2}`

Biểu diễn trên trục số:`(ảnh 1)`

`d)(-2x-3)/4>(x-4)/3`

`⇔[3(-2x-3)]/12>[4(x-4)]/12`

`⇔3(-2x-3)>4(x-4)`

`⇔-6x-9>4x-16`

`⇔-6x-4x>``-16+9`

`⇔-10x>``-7`

`⇔x<7/10`

Vậy `S={x|x<7/10}`

Biểu diễn trên trục số:`(ảnh 1)`

Giải thích các bước giải:

1/. 

a/. 5(x + 2) = 3x - 10

⇔ 5x + 10 - 3x + 10 = 0

⇔ (5x - 3x) + (+10 + 10) = 0

⇔ 2x  + 20 = 0

⇔ 2x  = - 20 

⇒ x = - `(20)/5`

x = - 10

Vậy S = { - 10 }

b/. (2x + 5)(3x - 1) = (2x + 5)(x - 3)

⇔ (2x + 5)(3x - 1) - (2x + 5)(x - 3) = 0

⇔ (2x + 5)[3x - 1 - (x - 3)] = 0

⇔ (2x + 5)(3x - 1 - x + 3) = 0

⇔ (2x + 5)[(3x - x) + (- 1 + 3)] = 0

⇔ (2x + 5)(2x + 2) = 0

⇒ 2x + 5 = 0 hay 2x + 2 = 0

⇒ 2x = - 5 hay 2x = - 2

⇒ x = -`5/2` hay x = -1

Vậy S = { -`5/2`; - 1}}

c/. `2/(x² + 2x)`- `1/x` = `(x-2)/(x+2)`

ĐKXĐ:  x # 0; x # - 2

MTC: x(x + 2)

⇔ `2/(x(x+2)`- `(x+2)/(x(x+2)` = `(x(x-2))/[x(x+2)]`

⇔ 2 - x - 2 = x² - 2x

⇔  x² - 2x + x = 0

⇔  x² - x = 0

⇔ x(x - 1) = 0

⇒ x = 0 (loại) hay x - 1 = 0

⇒ x = 0 (loại) hay x = 1 (nhận)

Vậy S = { 1}

d/. |2x - 1| + 5 = 6

⇔ |2x - 1|  = 6 - 5

⇔  |2x - 1|  = 1

⇒ 2x - 1 = 1 hay 2x - 1 = - 1

⇒ 2x = 2 hay 2x = 0

⇒ x = 1 hay x = 0

Vậy S = { 1; 0}

e/. 5(x - 3)= 3x + 10

⇔ 5x - 15 - 3x - 10 = 0

⇔ (5x - 3x) + (- 15 - 10) = 0

⇔ 2x - 25 = 0

⇒ 2x = 25

⇒ x = `(25)/2`

Vậy S = {`(25)/2`}

f/. (2x - 5)(3x + 1) = (2x - 5)(x + 3)

⇔ (2x - 5)(3x + 1) - (2x - 5)(x + 3) = 0

⇔ (2x - 5)[3x + 1 - (x + 3)] = 0

⇔ (2x - 5)(3x + 1 - x - 3) = 0

⇔ (2x - 5)(2x - 2) = 0

⇒ 2x - 5 = 0 hay 2x - 2 = 0

⇒ 2x = 5 hay 2x = 2

⇒ x = `5/2` hay x = 1

Vậy S = { `5/2` ; 1}

g/. `2/(x² - 2x)`+ `1/x` = `(x+2)/(x-2)`

ĐKXĐ:  x # 0; x # 2

MTC: x(x - 2)

`2/[x(x-2)]`+ `(x-2)/[x(x-2)]` = `[x(x+2)]/[x(x - 2)]`

⇔ 2 + x - 2 = x² + 2x

⇔  x² + 2x - x = 0

⇔ x² + x = 0

⇔ x(x + 1) = 0

⇒ x = 0 (loại) hay x + 1 = 0

⇒ x = 0 (loại) hay x = - 1  (nhận)

Vậy S = { - 1}

h/. |2x - 1| - 5 = 4

⇔ |2x - 1| = 4 + 5 

⇔ |2x - 1| = 9

⇒ 2x - 1 = 9 hay 2x - 1 = - 9

⇒ 2x = 10 hay 2x = - 8

⇒ x = 5 hay x = - 4

Vậy S = { 5; - 4}

2/.

a/. 3 - 2x ≥ 4

⇔ - 2x + 3 - 4 ≥ 0

⇔ - 2x - 1 ≥ 0

⇔ - 2x ≥ 1

⇒ x ≤ -`1/2`

Vậy S = {x| x ≤ -`1/2` }

b/. `(4x 5)/3`> `(7- x)/5`

⇔ `[5(4x-5)]/(15)`> `[3(7-x)]/15`

⇔ 20x - 25 > 21 - 3x

⇔ 20x + 3x - 25 - 21 > 0

⇔ 23x - 46 > 0

⇔ 23x > 46

⇒ x > `(46)/(23)`

⇒ x > 2

Vậy S = { x| x > 2}

c/. 4x - 3 ≥ 2x + 4

⇔ 4x - 2x - 3 -  4 ≥ 0

⇔ 2x - 7 ≥ 0

⇔ 2x ≥ 7

⇒ x ≥ `7/2`

Vậy S = { x| x ≥ `7/2`}

d/. `(-2x-3)/4`> `(x-4)/3`

⇔ `[3(-2x-3)]/(12)`> `[4(x-4)]/(12)`

⇔ - 6x - 9 > 4x - 16

⇔ - 6x - 9 - 4x + 16 > 0

⇔ - 10x + 7 > 0

⇔  - 10x  > - 7

⇔   x < `(-7)/(10)`

Vậy S = { x| x < `(-7)/(10)`}

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