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`

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

`<=>6(2x-5)-10(x+3)=15(2-3x)-30x-60`

`<=>12x-30-10x-30=30-45x-30x-60`

`<=>2x-60=-30-75x`

`<=>77x=30`

`<=>x=30/77`

Vậy `S={30/77}`

`12)(x^2-3x)^2-2(x^2-3)=8`

`<=>x^4+9x^2-6x^3-2x^2+6-8=0`

`<=>x^4-6x^3+7x^2-2=0`

`<=>x^4-x^3-5x^3+5x^2+2x^2-2x+2x-2=0`

`<=>x^3(x-1)-5x^2(x-1)+2x(x-1)+2(x-1)=0`

`<=>(x-1)(x^3-5x^2+2x+2)=0`

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

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

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

`<=>(x-1)^2[(x-2)^2-6]=0`

`<=>(x-1)(x-2-\sqrt{6})(x-2+\sqrt{6})=0`

`<=>` $\left[ \begin{array}{l}x=1\\x=2-\sqrt{6}\\x=2+\sqrt{6}\end{array} \right.$

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

lệnh for...to...do:

a)program tinh_tong;

uses crt;

var i,s:byte;

begin

  clrscr;

  s:=0;

  for i:=1 to 9 do s:=s+i;

  write(s);

readln;

end.

b)

program tinh_tong;

uses crt;

var i,s:byte;

begin

  clrscr;

  s:=0;

  for i:=1 to 14 do

begin

if i mod 2=0 then

s:=s+i;

end;

  write(s);

readln;

end.

c)

program tinh_tong;

uses crt;

var i,s:byte;

begin

  clrscr;

  s:=0;

  for i:=1 to 15 do

begin

if i mod 2=1 then

s:=s+i;

end;

  write(s);

readln;

end.

lệnh while...do

a)program tinh_tong;

uses crt;

var i,s:byte;

begin

  clrscr;

  s:=0;

  i:=1;

while i<=9 do

begin

  s:=s+i;

i:=i+1;

end;

  write(s);

readln;

end.

b)program tinh_tong;

uses crt;

var i,s:byte;

begin

  clrscr;

  s:=0;

  i:=1;

while i<=14 do

begin

if i mod 2=0 then

  s:=s+i

else i:=i+1;

end;

  write(s);

readln;

end.

c)

program tinh_tong;

uses crt;

var i,s:byte;

begin

  clrscr;

  s:=0;

  i:=1;

while i<=15 do

begin

if i mod 2=1 then

  s:=s+i

else i:=i+1;

end;

  write(s);

readln;

end.