\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

1:

a: =>3x=6

=>x=2

b: =>4x=16

=>x=4

c: =>4x-6=9-x

=>5x=15

=>x=3

d: =>7x-12=x+6

=>6x=18

=>x=3

2:

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4

a: Ta có: \(x^2+3x+4=0\)

\(\text{Δ}=3^2-4\cdot1\cdot4=9-16=-7< 0\)

Do đó: Phương trình vô nghiệm

Bài `1:`

`h)(3/4x-1)(5/3x+2)=0`

`=>[(3/4x-1=0),(5/3x+2=0):}=>[(x=4/3),(x=-6/5):}`

______________

Bài `2:`

`b)3x-15=2x(x-5)`

`<=>3(x-5)-2x(x-5)=0`

`<=>(x-5)(3-2x)=0<=>[(x=5),(x=3/2):}`

`d)x(x+6)-7x-42=0`

`<=>x(x+6)-7(x+6)=0`

`<=>(x+6)(x-7)=0<=>[(x=-6),(x=7):}`

`f)x^3-2x^2-(x-2)=0`

`<=>x^2(x-2)-(x-2)=0`

`<=>(x-2)(x^2-1)=0<=>[(x=2),(x^2=1<=>x=+-2):}`

`h)(3x-1)(6x+1)=(x+7)(3x-1)`

`<=>18x^2+3x-6x-1=3x^2-x+21x-7`

`<=>15x^2-23x+6=0<=>15x^2-5x-18x+6=0`

`<=>(3x-1)(5x-1)=0<=>[(x=1/3),(x=1/5):}`

`j)(2x-5)^2-(x+2)^2=0`

`<=>(2x-5-x-2)(2x-5+x+2)=0`

`<=>(x-7)(3x-3)=0<=>[(x=7),(x=1):}`

`w)x^2-x-12=0`

`<=>x^2-4x+3x-12=0`

`<=>(x-4)(x+3)=0<=>[(x=4),(x=-3):}`

`m)(1-x)(5x+3)=(3x-7)(x-1)`

`<=>(1-x)(5x+3)+(1-x)(3x-7)=0`

`<=>(1-x)(5x+3+3x-7)=0`

`<=>(1-x)(8x-4)=0<=>[(x=1),(x=1/2):}`

`p)(2x-1)^2-4=0`

`<=>(2x-1-2)(2x-1+2)=0`

`<=>(2x-3)(2x+1)=0<=>[(x=3/2),(x=-1/2):}`

`r)(2x-1)^2=49`

`<=>(2x-1-7)(2x-1+7)=0`

`<=>(2x-8)(2x+6)=0<=>[(x=4),(x=-3):}`

`t)(5x-3)^2-(4x-7)^2=0`

`<=>(5x-3-4x+7)(5x-3+4x-7)=0`

`<=>(x+4)(9x-10)=0<=>[(x=-4),(x=10/9):}`

`u)x^2-10x+16=0`

`<=>x^2-8x-2x+16=0`

`<=>(x-2)(x-8)=0<=>[(x=2),(x=8):}`

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

x4−3x3−2x2+6x+4=0x4−3x3−2x2+6x+4=0

⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0

⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0

⇔(x2−x−2)(x2−2x−2)=0⇔(x2−x−2)(x2−2x−2)=0

⇔(x+1)(x−2)(x−1−√3)(x−1+√3)=0⇔(x+1)(x−2)(x−1−3)(x−1+3)=0

⇔⎡⎢ ⎢ ⎢ ⎢⎣x=−1x=2x=1+√3x=1−√3

tl

x4−3x3−2x2+6x+4=0x4−3x3−2x2+6x+4=0

⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0

⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0

⇔(x2−x−2)(x2−2x−2)=0⇔(x2−x−2)(x2−2x−2)=0

⇔(x+1)(x−2)(x−1−√3)(x−1+√3)=0⇔(x+1)(x−2)(x−1−3)(x−1+3)=0

⇔⎡⎢ ⎢ ⎢ ⎢⎣x=−1x=2x=1+√3x=1−√3

^HT^

Ta thấy x=0 không là nghiệm của phương trình

chia cả 2 vế cho x^2 ta được:

PT <=> x^2-3x-6+3/x+1/(x^2)=0

       <=> (x^2-2+1/(x^2))-3(x-1/x)-4=0

      <=> (x-1/x)^2-3(x-1/x)-4=0

Đặt x-1/x=y

PT <=> y^2-3y-4=0

     <=> y=-4 hoặc y=1

Tại y=-4 , ta có x+1/x+4=0

                       <=> x^2+4x+1=0

                       <=> x=-2+ √3 hoăc x=-2-  √ 3

Tại y=1 ta có x^2-x-1=0

                 <=> x=(1- √  5)/2 hoặc x=(1+  √5)/2