2:

-8x^6-12x^4y-6x^2y^2-y^3

=-(8x^6+12x^4y+6x^2y^2+y^3)

=-(2x^2+y)^3

3:

=(1/3)^2-(2x-y)^2

=(1/3-2x+y)(1/3+2x-y)

không biết :))))

a/

Với mọi số thực x;y;z ta luôn có:

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

\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\)

\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2yz\ge3xy+3yz+3zx\)

\(\Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\) (đpcm)

b/

\(M=2\left(\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+zx}+\frac{1}{xy+yz+zx}\right)+\frac{1}{xy+yz+zx}\)

\(M\ge2.\frac{9}{x^2+y^2+z^2+xy+yz+zx+xy+yz+zx}+\frac{1}{\frac{\left(x+y+z\right)^2}{3}}\)

\(M\ge\frac{18}{\left(x+y+z\right)^2}+\frac{3}{\left(x+y+z\right)^2}=\frac{21}{\left(x+y+z\right)^2}=21\)

\(M_{min}=21\) khi \(x=y=z=\frac{1}{3}\)

`B=(x/2+y)^3-6(x/2+y)^2z + 6(x+2y)z^2-8z^3`

`=(x/2+y)^3 - 3. (x/2+y)^2 . 2z + 3. (x/2+y) . (2z)^2 - (2z)^3`

`=(x/2+y-2z)^3`

Sửa đề: Δ\(B=\left(\dfrac{x}{2}+y\right)^3-6\left(\dfrac{x}{2}+y\right)^2z+12\left(x+2y\right)\cdot z^2-8z^3\)

Ta có: \(B=\left(\dfrac{x}{2}+y\right)^3-6\left(\dfrac{x}{2}+y\right)^2z+12\left(x+2y\right)\cdot z^2-8z^3\)

\(=\left(\dfrac{1}{2}x+y\right)^2-3\cdot\left(\dfrac{1}{2}x+y\right)^2\cdot2z+3\cdot\left(\dfrac{1}{2}x+y\right)\cdot\left(2z\right)^2-\left(2z\right)^3\)

\(=\left(\dfrac{1}{2}x+y-2z\right)^3\)

đề dài v~

1.

a) \(f\left(x\right)=5x^2-2x+1\)

\(5f\left(x\right)=25x^2-10x+5\)

\(5f\left(x\right)=\left(25x^2-10x+1\right)+4\)

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

Mà  \(\left(5x-1\right)^2\ge0\)

\(\Rightarrow5f\left(x\right)\ge4\)

\(\Leftrightarrow f\left(x\right)\ge\frac{4}{5}\)

Dấu " = " xảy ra khi :

\(5x-1=0\Leftrightarrow x=\frac{1}{5}\)

Vậy ....

b)  \(P\left(x\right)=3x^2+x+7\)

\(3P\left(x\right)=9x^2+3x+21\)

\(3P\left(x\right)=\left(9x^2+3x+\frac{1}{4}\right)+\frac{83}{4}\)

\(3P\left(x\right)=\left(3x+\frac{1}{2}\right)^2+\frac{83}{4}\)

Mà  \(\left(3x+\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow3P\left(x\right)\ge\frac{83}{4}\)

\(\Leftrightarrow P\left(x\right)\ge\frac{83}{12}\)

Dấu "=" xảy ra khi :

\(3x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{6}\)

Vậy ...

c)  \(Q\left(x\right)=5x^2-3x-3\)

\(5Q\left(x\right)=25x^2-15x-15\)

\(\Leftrightarrow5Q\left(x\right)=\left(25x^2-15x+\frac{9}{4}\right)-\frac{69}{4}\)

\(\Leftrightarrow5Q\left(x\right)=\left(5x-\frac{3}{2}\right)^2-\frac{69}{4}\)

Mà  \(\left(5x-\frac{3}{2}\right)^2\ge0\)

\(\Rightarrow5Q\left(x\right)\ge\frac{-69}{4}\)

\(\Leftrightarrow Q\left(x\right)\ge-\frac{69}{20}\)

Dấu "=" xảy ra khi :

\(5x-\frac{3}{2}=0\Leftrightarrow x=0,3\)

Vậy ...

2.

a)  \(f\left(x\right)=-3x^2+x-2\)

\(-3f\left(x\right)=9x^2-3x+6\)

\(-3f\left(x\right)=\left(9x^2-3x+\frac{1}{4}\right)+\frac{23}{4}\)

\(-3f\left(x\right)=\left(3x-\frac{1}{2}\right)^2+\frac{23}{4}\)

Mà  \(\left(3x-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow-3f\left(x\right)\ge\frac{23}{4}\)

\(\Leftrightarrow f\left(x\right)\le\frac{23}{12}\)

Dấu "=" xảy ra khi :

\(3x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{6}\)

Vậy ...

b)  \(P\left(x\right)=-x^2-7x+1\)

\(-P\left(x\right)=x^2+7x-1\)

\(-P\left(x\right)=\left(x^2+7x+\frac{49}{4}\right)-\frac{53}{4}\)

\(-P\left(x\right)=\left(x+\frac{7}{2}\right)^2-\frac{53}{4}\)

Mà  \(\left(x+\frac{7}{2}\right)^2\ge0\)

\(\Rightarrow-P\left(x\right)\ge-\frac{53}{4}\)

\(\Leftrightarrow P\left(x\right)\le\frac{53}{4}\)

Dấu "=" xảy ra khi :

\(x+\frac{7}{2}=0\Leftrightarrow x=-\frac{7}{2}\)

Vậy ...

c)  \(Q\left(x\right)=-2x^2+x-8\)

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

\(-2Q\left(x\right)=\left(4x^2-2x+\frac{1}{4}\right)+\frac{63}{4}\)

\(-2Q\left(x\right)=\left(2x-\frac{1}{2}\right)^2+\frac{63}{4}\)

Mà :  \(\left(2x-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow-2Q\left(x\right)\ge\frac{63}{4}\)

\(\Leftrightarrow Q\left(x\right)\le-\frac{63}{8}\)

Dấu "=" xảy ra khi :

\(2x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)

Vậy ...

3/

Ta có x + y + z = 0

<=> (x + y + z)² = 0

<=> x² + y² + z² + 2xy + 2xz + 2yz = 0

<=> x² + y² + z² + 2(xy + xz + yz) = 0

Thay xy + yz + xz = 0 vào BT trên

=> x² + y² + z² = 0

Mà \(x^2\ge0;y^2\ge0;z^2\ge0\Rightarrow x^2+y^2+z^2\ge0\)

=> x = y = z (đpcm)

Áp dụng bđt AM-GM ta có

\(P\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2.\left(yz+1\right)^2.\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=A\)

  Ta có   \(A=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

Áp dụng bđt AM-GM ta có

\(A\ge3\sqrt[3]{8\sqrt{\frac{xyz}{xyz}}}=3.2=6\)

\(\Rightarrow P\ge6\)

Dấu "=" xảy ra khi x=y=z=\(\frac{1}{2}\)

Làm tiếp bài ღ๖ۣۜLinh's ๖ۣۜLinh'sღ] ★we are one★ chớ hình như bị ngược dấu ó.Do mình gà nên chỉ biết cô si mù mịt thôi ạ

\(3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

\(=3\sqrt[3]{\left(y+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}\right)\left(z+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}\right)\left(x+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}\right)}\)

\(\ge3\sqrt[3]{5\sqrt[5]{\frac{y}{256x^4}}\cdot5\sqrt[5]{\frac{z}{256y^4}}\cdot5\sqrt[5]{\frac{x}{256z^4}}}\)

\(=3\sqrt[3]{125\sqrt[5]{\frac{xyz}{256^3\left(xyz\right)^4}}}\)

\(=15\sqrt[3]{\sqrt[5]{\frac{1}{256^3\left(xyz\right)^3}}}\)

\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\left(\frac{x+y+z}{3}\right)^9}}\)

\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\frac{1}{2^9}}}=\frac{15}{2}\)

Dấu "=" xảy ra tại \(x=y=z=\frac{1}{2}\)

Không phải ngược đâu nha mọi người,dấu bằng không xảy ra nhé!

Bài này thì AM-GM thôi 

\(P=\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}\)

Sử dụng BĐT AM-GM cho 3 số không âm ta có :

\(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)^2}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}\)

\(=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}\)

\(=3\sqrt[3]{\left(\frac{xy}{x}+\frac{1}{x}\right)\left(\frac{yz}{y}+\frac{1}{y}\right)\left(\frac{zx}{z}+\frac{1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

Tiếp tục sử dụng AM-GM cho 2 số không âm ta được :

\(3\sqrt[3]{\left(2\sqrt[2]{y\frac{1}{x}}\right)\left(2\sqrt[2]{z\frac{1}{y}}\right)\left(2\sqrt[2]{x\frac{1}{z}}\right)}\ge3\sqrt[3]{\left(2\sqrt{\frac{y}{x}}\right)\left(2\sqrt{\frac{z}{y}}\right)\left(2\sqrt{\frac{x}{z}}\right)}\)

\(=3\sqrt[3]{8\left(\sqrt{\frac{y}{x}}.\sqrt{\frac{z}{y}}.\sqrt{\frac{x}{z}}\right)}=3\sqrt[3]{8.\sqrt{\frac{xyz}{xyz}}}=3\sqrt[3]{8}=3.2=6\)

Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\frac{1}{2}\)

Vậy \(Min_P=6\)đạt được khi \(x=y=z=\frac{1}{2}\)