\(\frac{x^2\left(x-1\right)\left(x+1\right)}{\left(>0\right)}\le0\Rightarrow\left(x-1\right)\left(x+1\right)\le0\Rightarrow-1\le x\le1\)

bất kì số nào cũng viết dc như vậy

a = 1+1+1+......+1 + 0 + ..+0

\(PT:ax^2+bx+c=0\) (1) có 2 nghiệm pb  có dúng 1 nghiệm dương(x1)  => ac<0 ; \(\sqrt{\Delta}=b^2-4ac>0\)

\(PT:ct^2+bt+a=0\) (2) có ac<0 => \(\sqrt{\Delta}=b^2-4ac>0\) (theo trên) => (2) cũng có 2 nghiệm pb ,trái dấu ( 1 dương = t1 )

ta có :  x1>0 ; t1 >0  nên : 

          +   \(x_1.t_1=\frac{-b+\sqrt{\Delta}}{2a}.\frac{-b-\sqrt{\Delta}}{2c}=\frac{4ac}{4ac}=1\left(Neusa>0;c<0\right)\)

           +  \(x_1.t_1=\frac{-b-\sqrt{\Delta}}{2a}.\frac{-b+\sqrt{\Delta}}{2c}=\frac{4ac}{4ac}=1\left(Neusa<0;c>0\right)\)

=> \(x_1+t_1\ge2\sqrt{x_1.t_1}=2\)

+ m \(\ne\)-2

 \(\Delta'=\left(m-1\right)^2-\left(m+2\right)\left(3-m\right)=m^2-2m+1+m^2-m-6=2m^2-3m-5=\left(2m-5\right)\left(m+1\right)\)

\(m\ge\frac{5}{2};m\le-1\)

\(\int\limits^{x_1+x_2=\frac{2\left(m-1\right)}{\left(m+2\right)}=2-\frac{4}{m+2}}_{x_1x_2=\frac{3-m}{m+2}=-1+\frac{5}{m+2}}\)=>\(\int\limits^{5\left(S\right)=10-\frac{20}{m+2}}_{4x_1x_2=-4+\frac{20}{m+2}}\)=>5S+ 4P = 6

làm tiếp 

\(VT=\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}=\frac{1}{2}\left[\left(\frac{y}{x}+\frac{x}{y}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)+\left(\frac{z}{y}+\frac{y}{z}\right)-3\right]\ge\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)

=> dpcm

a) 8x - 75 = 5x +21

  8x - 5x = 21 + 75

 3x = 96

x =32

b) 9x + 25 = -2x + 58

   9x + 2x = 58 - 25

  11x = 33

  x =3

\(1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)

\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{99.100}+\frac{1}{50}=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)

  \(=2\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}=1\)

\(\left(2+\left(-4\right)\right)+\left(6+\left(-8\right)\right)+....+\left(x-4+\left(-\left(x-2\right)\right)\right)+x=2000\)

\(-2-2............-2+x=2000\) có ( x -2 ):4 số -2

-2 ( x -2) :4 +x = 2000

- ( x -2) + 2x = 4000

x =4000 -2 

x = 3998

x(3y + 5 ) = - 2 - 4y  => \(x=-\frac{4y+2}{3y+5}=-1-\frac{y-3}{3y+5}\in Z\)=> y -3 =0  hoặc y-3 = 3y+5 hoặc y -3 = -( 3y+5) hoặc 3y +5 = 1 hoặc 3y+5 =-1

=> y thuộc {3; -4 ; -2 } => x  tương ứng { -1; -2 ; -6}

Vậy (x;y) = (-1;3) ; ( -2; -4) ; ( -6; -2)