\(2,x^4+3x^2+2x+2=\left(x^4+2x^2+1\right)+\left(x^2+2x+1\right)=\left(x^2+1\right)^2+\left(x+1\right)^2>0\left(đpcm\right)\)

\(b,x^2+y^2+z^2+xy+yz+zx\ge0\)

\(\Leftrightarrow2\left(x^2+y^2+z^2+xy+yz+zx\right)\ge0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+2xz+z^2\right)+\left(y^2+2yz+z^2\right)\ge0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x+z\right)^2+\left(y+z\right)^2\ge0\)

Đúng với mọi x , y ,z

c,\(x^2+y^2+xy+x+y+1\ge0\)

\(\Leftrightarrow2\left(x^2+y^2+xy+y+x+1\right)\ge0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+2x+1\right)+\left(y^2+2y+1\right)\ge0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x+1\right)^2+\left(y+1\right)^2\ge0\)

Đúng với mọi x , y

\(A=x^2+3xy+6x+5y^2+7y-2\)

\(=\left[x^2+2x\left(3+\dfrac{3}{2}y\right)+\left(3+\dfrac{3}{2}y\right)^2\right]+5y^2+7y-2-\left(3+\dfrac{3}{2}y\right)^2\)\(=\left(x+3+\dfrac{3}{2}y\right)^2+5y^2+7y-2-9-9y-\dfrac{9}{4}y^2\)\(=\left(x+3+\dfrac{3}{2}y\right)^2+\dfrac{11}{4}y^2-2y-11\)

\(=\left(x+3+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\left(y^2-\dfrac{8}{11}y+\dfrac{16}{121}\right)-\dfrac{125}{11}\)\(=\left(x+3+\dfrac{3}{2}y\right)^2+\dfrac{11}{4}\left(x-\dfrac{4}{11}\right)^2-\dfrac{125}{11}\ge\dfrac{-125}{11}\)Vậy \(Min_A=\dfrac{-125}{11}\) khi \(\left[{}\begin{matrix}x+3+\dfrac{3}{2}y=0\\x-\dfrac{4}{11}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{74}{33}\\x=\dfrac{4}{11}\end{matrix}\right.\)

Biết số nhọ nhưng vẫn làm tiếp:)

\(1,\left(2-x\right)^2-9=0\)

\(\Leftrightarrow\left(2-x-9\right)\left(2-x+9\right)=0\)

\(\Leftrightarrow\left(-7-x\right)\left(11-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\11-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=11\end{matrix}\right.\)

\(b,\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)=15-9\left(x+1\right)^2\)\(\Leftrightarrow x^3-9x^2+27x-27-x^3-27=15-9x^2-18x-9\)\(\Leftrightarrow x^3-9x^2+27x-x^3+9x^2+18x=15+27+27\)\(\Leftrightarrow45x=69\Rightarrow x=\dfrac{23}{15}\)

\(216x^6-125x^9=\left(6x^2\right)^3-\left(5x^3\right)^3=\left(6x^2-5x^3\right)\left(36x^5+60x^2x^3+25x^6\right)=\left(6x^2-5x^3\right)\left(36x^5+60x^5+25x^6\right)=\left(6x^2-5x^3\right)\left(96x^5+25x^6\right)\)\(x^3-\dfrac{8}{27x^3}=\left(x-\dfrac{2}{3x}\right)\left(x^2+\dfrac{4}{3}x+\dfrac{4}{9x^2}\right)\)

\(x+\dfrac{1}{x}=a\Rightarrow\left(x+\dfrac{1}{x}\right)^3=a^3\)

\(\Rightarrow x^3+\dfrac{1}{x^3}+3x\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=a^3\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}=a^3-3a\)

\(x^2+\dfrac{1}{x^2}=\left(x+\dfrac{1}{x}\right)^2-2x\dfrac{1}{x}=a^2-2\)

\(x^4+\dfrac{1}{x^4}=\left(x^2\right)^2+\left(\dfrac{1}{x^2}\right)^2=\left(x^2+\dfrac{1}{x^2}\right)^2-2x^2\dfrac{1}{x^2}=\left(a^2-2\right)^2-2\)\(\Rightarrow\left(x^3+\dfrac{1}{x^3}\right)\left(x^4+\dfrac{1}{x^4}\right)=\left(a^3-3a\right)\left(a^2-2\right)^2\)\(\Leftrightarrow x^7+\dfrac{x^3}{x^4}+\dfrac{x^4}{x^3}+\dfrac{1}{x^7}=\left(a^3-3a\right)\left(a^2-2\right)^2\)\(\Rightarrow x^7+\dfrac{1}{x^7}=\left(a^3-3a\right)\left(a^2-2\right)^2-a\)

Thuộc kết quả mấy cái bài như thế này mất:(

Tham khao cau hoi tuong tu nhe 

Dug quen tick cho minh

Từ giả thiết: \(x+y+z=0\Rightarrow x+y=-z\)

\(\Leftrightarrow\left(x+y\right)^3=\left(-c\right)^3\)

\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=-c^3\)

\(\Leftrightarrow x^3+y^3+z^3=-3xy\left(x+y\right)\)

\(\Leftrightarrow x^3+y^3+z^3=3xyz\) (1)

Nhận cả 2 vế của (1) với \(x^2+y^2+z^2\) ta được:

\(3xyz\left(x^2+y^2+z^2\right)=\left(x^2+y^2+z^2\right)\left(x^3+y^3+z^3\right)=x^5+x^3\left(y^2+z^2\right)+y^5+y^3\left(x^2+z^2\right)+z^5+z^3\left(x^2+y^2\right)\left(2\right)\)Do x + y + z =0 \(\Rightarrow y+z=-x\Rightarrow\left(y+z\right)^2=x^2\Leftrightarrow y^2+z^2=x^2-2yz\)Tương tự ta có:

\(x^2+y^2=z^2-2xy;x^2+z^2=y^2-2xz\)

Thay vào (2) ta được:

\(3xyz\left(x^2+y^2+z^2\right)=x^5+y^5+z^5+x^3\left(x^2-2yz\right)+y^3\left(y^2-2xz\right)+z^3\left(z^2-2xy\right)\)\(=2\left(x^5+y^5+z^5\right)-2xyz\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\left(đpcm\right)\)

Ta có: \(x+y+z=0\Rightarrow x+y=-z\)

\(x+y+z=0\Rightarrow\left(x+y+z\right)^3=0\)

\(\Leftrightarrow\left(x+y\right)^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2+z^3=0\)\(\Leftrightarrow\left(x+y\right)^3+3\left(x+y\right)z\left(x+y+z\right)+c^3=0\)

\(\Leftrightarrow\left(x+y\right)^3+c^3=0\) ( vì x + y+z =0)

\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3+z^3=0\)

\(\Leftrightarrow x^3+y^3+z^3+3xy\left(x+y\right)=0\)

\(\Leftrightarrow x^2+y^2+z^2+3xy\left(-z\right)=0\)

\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow x^3+y^3+z^3=3xyz\) ( đpcm)

\(x+\dfrac{1}{x}=a\Rightarrow\left(x+\dfrac{1}{x}\right)^3=a^3\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3x.\dfrac{1}{x}\left(x+y\right)=a^3\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3a=a^3\)

\(\Rightarrow x^3+\dfrac{1}{x^3}=a^3-3a\)

Mà: \(x^6+\dfrac{1}{x^6}=\left(x^3\right)^2+\left(\dfrac{1}{x^3}\right)^2=\left(x^3+\dfrac{1}{x^3}\right)^2-2.x^3.\dfrac{1}{x^3}\)

\(=\left(a^3-3a\right)^2-a\)

1,

\(x^2-2ax+a^2=\left(x-a\right)^2\)

\(x^2-ax=x\left(x-a\right)\)

Vậy MSC: \(\left(x-a\right)^2x\)

2,

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

\(x^2-x=x\left(x-1\right)\)

\(x^2+x+1\)

vậy MSC là: \(x\left(x-1\right)\left(x^2+x+1\right)\)