\(x^3-4x=0\)

\(\Rightarrow x\left(x^2-4\right)=0\)

\(\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow\)x = 0 , x - 2 = 0 hoặc x + 2 = 0

\(\Rightarrow\)\(x=0,x=2,x=-2\)

\(x^3-4x=0\)

\(\Leftrightarrow x\left(x^2-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)

Vậy \(x\in\left\{-2;0;2\right\}\)

          \(\frac{x-3}{2}+2< \frac{2x+1}{3}\)

\(\Leftrightarrow\)\(\frac{3\left(x-3\right)}{6}+\frac{12}{6}< \frac{2\left(2x+1\right)}{6}\)

\(\Leftrightarrow\)\(3x-9+12< 4x+2\)

\(\Leftrightarrow\)\(3x-4x< 2+9-12\)

\(\Leftrightarrow\)\(-x< -1\)

\(\Leftrightarrow\)\(x>1\)

Vậy :....................................

\(\frac{x-3}{2}+2< \frac{2x+1}{3}\)

\(\Rightarrow\frac{3\left(x-3\right)}{6}+\frac{12}{6}< \frac{2\left(2x+1\right)}{6}\)

\(\Leftrightarrow3x-9+12< 4x+2\)

\(\Leftrightarrow3x-4x< 2+9-12\)

\(\Leftrightarrow-x< -1\)

\(\Leftrightarrow x>1\)

Vây.....

\(2\frac{2}{x-1}=1+\frac{2x}{x+2}\)              \(\left(x\ne1;x\ne-2\right)\)

\(\Rightarrow\frac{2\left(x-1\right)+2}{x-1}=\frac{\left(x+2\right)+2x}{x+2}\)\(\Rightarrow2x^2+4x=3x^2+2x-3x+2\)

\(\Rightarrow\frac{2x-2+2}{x-1}=\frac{x+2+2x}{x+2}\)

\(\Rightarrow\frac{2x}{x-1}=\frac{3x+2}{x+2}\)

\(\Rightarrow2x\left(x+2\right)=\left(x-1\right)\left(3x+2\right)\)

\(\Rightarrow2x^2+4x=x\left(3x+2\right)-1\left(3x+2\right)\)

\(\Rightarrow2x^2+4x=x\left(3x+2\right)-1\left(3x+2\right)\)

\(2\frac{2}{x-1}=1+\frac{2x}{x+2}\) ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x\ne-2\end{cases}}\)

=> \(\frac{2\left(x-1\right)+2}{x-1}=\frac{x+2+2x}{x+2}\)

=> \(\frac{2\left(x-1+1\right)}{x-1}=\frac{x+2\left(x+1\right)}{x+2}\)

=> \(\frac{2x}{x-1}=\frac{x+2\left(x+1\right)}{x+2}\)

=> \(2x\left(x+2\right)=x+2\left(x+1\right)\left(x-1\right)\)

=> \(2x^2+4x=x+2\left(x^2-1\right)\)

=> \(2x^2+4x=x+2x^2-2\)

=> \(2x^2+4x-x-2x^2+2=0\)

=> \(3x+2=0\)

=> \(3x=-2\)

=> \(x=-\frac{2}{3}\)

Đặt \(a=\frac{x+y}{2};b=\frac{y+z}{2};c=\frac{z+x}{2}\)

Thì \(\Rightarrow a+b+c=\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=\frac{x+y+y+z+z+x}{2}=\)\(x+y+z=1\)

Bất đẳng thức đã tương đương với \(x+2y+z\ge4\left(x+y\right).\left(y+z\right).\left(z+x\right)\)

\(\Rightarrow a+b\ge16abc\)

Ta có: \(\left(a+b\right).\left(a+b+c\right)^2\ge4\left(a+b\right).4c\left(a+b\right)\ge16abc\left(đpcm\right).\)

cảm ơn bn

                 \(a^2x-b^2x+ab+b^2=0\)

\(\Leftrightarrow\)\(\left(a^2x-b^2x\right)+ab+b^2=0\)

\(\Leftrightarrow\)\(x\left(a^2-b^2\right)+b\left(a+b\right)=0\)

\(\Leftrightarrow\)\(x\left(a-b\right)\left(a+b\right)+b\left(a+b\right)=0\)

Với : \(a\ne\pm b\)\(\Rightarrow\)phương trình có 1 nghiệm duy nhất : \(x=\frac{-b\left(a+b\right)}{\left(a+b\right)\left(a-b\right)}=\frac{-b}{a-b}\)

 Với : \(a=b\Rightarrow0x=2b^2\rightarrow\)phương trình vô nghiệm

Với: \(a=-b\Rightarrow0x=0\rightarrow\) phương trình vô số nghiệm 

Với : \(a=b=0\Rightarrow0x=0\rightarrow\)phương trình vô số nghiệm

b mk thấy nó sai đề sao ý 

c) \(C=\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2\)

\(=\left(x^2+x+4\right)^2+2.4x.\left(x^2+x+4\right)+16x^2-x^2\)

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

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

\(=\left(x^2+5x+4-x\right)\left(x^2+5x+4+x\right)=\left(x^2+4x+4\right)\left(x^2+6x+4\right)\)

b) \(B=\)ghi lại đề nha bn 

Đặt \(x^2+4x-3=t\)  ta có:

\(B=t^2-5xt+6x^2\)

\(B=t^2-2xt-3xt+6x^2\)

\(B=t\left(t-2x\right)-3x\left(t-2x\right)=\left(t-2x\right)\left(t-3x\right)\)

\(B=\left(x^2+4x-3-2x\right)\left(x^2+4x-3-3x\right)\)

\(B=\left(x^2+2x-3\right)\left(x^2+x-3\right)\)

bn làm tương tự câu c) cũng như vậy nha!!!

phần a thừa một số 6 ở cuối cùng