`x(x-1)(x+1)(x+2)=24`

`<=>[x(x+1)][(x-1)(x+2)]=24`

`<=>(x^2+x)(x^2+x-2)=24`

`<=>(x^2+x-1)^2=25`

`+)x^2+x-1=5`

`<=>x^2+x-6=0`

`<=>x^2-2x+3x-6=0`

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

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

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

`+)x^2+x-1=-5`

`<=>x^2+x+4=0`

`<=>(x+1/2)^2+15/4=0` vô lý

Vậy `S={2,3}`

`11)4x^2+4-8x=9(x-2)^2`

`<=>4(x^2-2x+1)=9(x-2)^2`

`<=>(2x-2)^2=(3x-6)^2`

`<=>(3x-6-2x+2)(3x-6+2x-2)=0`

`<=>(x-4)(5x-8)=0`

`<=>` $\left[ \begin{array}{l}x=4\\x=\dfrac{8}{5}\end{array} \right.$

Vậy `S={4,5/8}`

2) Ta có: \(19-\left(x-5\right)^3=x\left(3-x^2\right)-24\left(x-6\right)\)

\(\Leftrightarrow19-\left(x^3-15x^2+75x-125\right)=3x-x^3-24x+144\)

\(\Leftrightarrow19-x^3+15x^2-75x+125=-x^3-21x+144\)

\(\Leftrightarrow-x^3+15x^2-75x+144+x^3+21x-144=0\)

\(\Leftrightarrow15x^2-54x=0\)

\(\Leftrightarrow x\left(15x-54\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\15x-54=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\15x=54\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{18}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{0;\dfrac{18}{5}\right\}\)

3) Ta có: \(x\left(5-x\right)\left(x+5\right)-4x\left(x+5\right)=2x+1-\left(2x-1\right)^2\)

\(\Leftrightarrow x\left(5-x\right)\left(5+x\right)-4x\left(x+5\right)=2x+1-\left(4x^2-4x+1\right)\)

\(\Leftrightarrow x\left(25-x^2\right)-4x^2-20x=2x+1-4x^2+4x-1\)

\(\Leftrightarrow25x-x^3-4x^2-20x-2x-1+4x^2-4x+1=0\)

\(\Leftrightarrow-x^3-x=0\)

\(\Leftrightarrow x\left(x^2+1\right)=0\)

mà \(x^2+1>0\forall x\)

nên x=0

Vậy: S={0}

7) Ta có: \(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)

\(\Leftrightarrow\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1=\dfrac{x+5}{61}+1+\dfrac{x+7}{59}+1\)

\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}=\dfrac{x+66}{61}+\dfrac{x+66}{59}\)

\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\)

\(\Leftrightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)

mà \(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\ne0\)

nên x+66=0

hay x=-66

Vậy: S={-66}

`10)x^2-11x+24=0`

`<=>x^2-3x-8x+24=0`

`<=>x(x-3)-8(x-3)=0`

`<=>(x-3)(x-8)=0`

`<=>` $\left[ \begin{array}{l}x=3\\x=8\end{array} \right.$

`8,(x+1)^3-4(x+1)=0`

`<=>(x+1)[(x+1)^2-4]=0`

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

`<=>(x+1)(x-2)(x+3)=0`

`<=>` $\left[ \begin{array}{l}x=2\\x=-1\\x=-3\end{array} \right.$

4) Ta có: \(\dfrac{2x-5}{5}-\dfrac{x+3}{3}=\dfrac{2-3x}{2}-x-2\)

\(\Leftrightarrow\dfrac{6\left(2x-5\right)}{30}-\dfrac{10\left(x+3\right)}{30}=\dfrac{15\left(2-3x\right)}{30}-\dfrac{30\left(x+2\right)}{30}\)

\(\Leftrightarrow12x-30-10x-30=30-45x-30x-60\)

\(\Leftrightarrow-22x-60=-75x-30\)

\(\Leftrightarrow-22x+75x=-30+60\)

\(\Leftrightarrow53x=30\)

\(\Leftrightarrow x=\dfrac{30}{53}\)

Vậy: \(S=\left\{\dfrac{30}{53}\right\}\)

5) Ta có: \(\dfrac{5x-3}{6}-\dfrac{7x-1}{4}=5\)

\(\Leftrightarrow\dfrac{2\left(5x-3\right)}{12}-\dfrac{3\left(7x-1\right)}{12}=\dfrac{60}{12}\)

\(\Leftrightarrow10x-6-21x+3=60\)

\(\Leftrightarrow-11x-3=60\)

\(\Leftrightarrow-11x=63\)

\(\Leftrightarrow x=-\dfrac{63}{11}\)

Vậy: \(S=\left\{-\dfrac{63}{11}\right\}\)

`9,x^3+x^2-2=0`

`x^3-x^2+2x^2-2=0`

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

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

`<=>x=1`

`14,x^2-2x+1=0`

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

`<=>x-1=0`

`<=>x=1`

`15,x^3+3x^2+3x+1=0`

`<=>(x+1)^3=0`

`<=>x+1=0`

`<=>x=-1`

Bài 6: 

1) Ta có: \(2x\left(x-5\right)-\left(x+3\right)^2=3x-x\left(5-x\right)\)

\(\Leftrightarrow2x^2-10x-\left(x^2+6x+9\right)=3x-5x+x^2\)

\(\Leftrightarrow2x^2-10x-x^2-6x-9-3x+5x-x^2=0\)

\(\Leftrightarrow-14x-9=0\)

\(\Leftrightarrow-14x=9\)

\(\Leftrightarrow x=-\dfrac{9}{14}\)

Vậy: \(S=\left\{-\dfrac{9}{14}\right\}\)

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

`<=>2x^2-10x-x^2-6x-9=3x-5x+x^2`

`<=>x^2-16x-9=x^2-2x`

`<=>14x=-9`

`<=>x=-9/14`

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) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

(3-2i)z+(4+5i)=7+3i

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