\(VP=\frac{x}{y+z+t}+\frac{y}{z+t+x}+\frac{z}{t+x+y}+\frac{t}{x+y+z}+\frac{y+z+t}{x}+\frac{z+t+x}{y}+\frac{t+x+y}{z}+\frac{x+y+z}{t}=\left(\frac{x}{y+z+t}+\frac{y+z+t}{9x}\right)+\left(\frac{y}{z+t+x}+\frac{z+t+x}{9y}\right)+\left(\frac{z}{t+x+y}+\frac{t+x+y}{9z}\right)+\left(\frac{t}{x+y+z}+\frac{x+y+z}{9t}\right)+\frac{8}{9}\left(\frac{y+z+t}{x}+\frac{z+t+x}{y}+\frac{t+x+y}{z}+\frac{x+y+z}{t}\right)\)\(\ge8\sqrt[8]{\frac{x}{y+z+t}.\frac{y}{z+t+x}.\frac{z}{t+x+y}.\frac{t}{x+y+z}.\frac{y+z+t}{9x}.\frac{z+t+x}{9y}.\frac{t+x+y}{9z}.\frac{x+y+z}{9t}}+\frac{8}{9}\left(\frac{y}{x}+\frac{z}{x}+\frac{t}{x}+\frac{z}{y}+\frac{t}{y}+\frac{x}{y}+\frac{t}{z}+\frac{x}{z}+\frac{y}{z}+\frac{x}{t}+\frac{y}{t}+\frac{z}{t}\right)\)\(\ge\frac{8}{3}+\frac{8}{9}.12\sqrt[12]{\frac{y}{x}.\frac{z}{x}.\frac{t}{x}.\frac{z}{y}.\frac{t}{y}.\frac{x}{y}.\frac{t}{z}.\frac{x}{z}.\frac{y}{z}.\frac{x}{t}.\frac{y}{t}.\frac{z}{t}}=\frac{8}{3}+\frac{8}{9}.12=\frac{40}{3}=VT\left(đpcm\right)\)

Đẳng thức xảy ra khi x = y = z = t > 0 

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

\(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)

\(\Rightarrow\frac{x}{y+z+t}+1=\frac{y}{z+t+x}+1=\frac{z}{t+x+y}+1=\frac{t}{x+y+z}+1\)

\(\Rightarrow\frac{x+y+z+t}{y+z+t}=\frac{x+y+z+t}{z+t+x}=\frac{x+y+z+t}{t+x+y}=\frac{x+y+z+t}{x+y+z}\)

- Nếu \(x+y+z+t\ne0\Rightarrow x=y=z=t\)

=> \(P=\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}=1+1+1+1=4\)

- Nếu \(x+y+z+t=0\Rightarrow x+y=-\left(z+t\right);y+z=-\left(t+x\right);z+t=-\left(x+y\right);t+x=-\left(y+z\right)\)

=> \(P=\frac{-\left(z+t\right)}{z+t}+\frac{-\left(t+x\right)}{t+x}+\frac{-\left(x+y\right)}{x+y}=\frac{-\left(y+z\right)}{y+z}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

Vậy P = 4 hoặc P = -4

áp dụng định lí Pain có

\(\frac{\left(x+y+z+t\right)}{3\left(x+y+z+t\right)}=\frac{1}{3}\)

tương tự

theo định lí Pain có

\(E=\frac{2\left(x+y+z+t\right)}{2\left(x+y+z+t\right)}=1\)

P/S : chém bừa ( i love you)

\(\text{Xét 2 khoảng ta có:}\)

  *  \(\text{Nếu x + y + z + t = 0 thì }E=-1+-1+-1+-1=-4\)

  *  \(\text{Nếu }x+y+z+t\ne0\text{ thì }\)

 \(\frac{x}{y+z+t}=\frac{y}{x+z+t}=\frac{z}{x+y+t}=\frac{t}{x+y+z}=\frac{x+y+z+t}{y+z+t+x+z+t+x+y+t+x+y+z}=\frac{1}{3}\left(\text{Dãy tỉ sô băng nhau}\right)\)

  \(\Rightarrow x=\frac{1}{3\left(y+z+t\right)};y=\frac{1}{3\left(x+z+t\right)};z=\frac{1}{3\left(x+y+t\right)};t=\frac{1}{3\left(x+y+z\right)}\)

\(\Rightarrow x=y=z=t\)

Lấy ví dụ là x ta có:

\(E=\frac{2x}{2x}+\frac{2x}{2x}+\frac{2x}{2x}+\frac{2x}{2x}=4\)

Giải thích: ta có: x+y+z+t=0 (*)

*=> x+y = -(z+t) => \(\frac{x+y}{z+t}=-1\)
*=> y+z = -(t+x)=> \(\frac{y+z}{t+x}=-1\)
*=>z+t = -(x+y) =>\(\frac{z+t}{x+y}=-1\)
*=> t+x = -(z+y)=>\(\frac{t+x}{z+y}=-1\)

Vậy......

TA CÓ : ( x / y + z + t ) + 1 = ( y / z +t + x ) + 1 = ( t / x + y + z ) + 1 

Suy ra : x+y+z+t / y+z+t = x+y+z+t / z+t+x = x+y+z+t / t+x+y = x+y+z+t / x+y+z 

do x+y+z+t khác 0 suy ra x=y=z=t suy ra M= 1+1+1+1 =4  tích đúng nha

xét x+y+z+t=0

=>x+y=-(z+t)

y+z=-(t+x)

\(\Rightarrow M=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}=\frac{x+y}{-\left(x+y\right)}+\frac{y+z}{-\left(y+z\right)}+\frac{z+t}{-\left(z+t\right)}+\frac{t+x}{-\left(t+x\right)}\)

\(=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

xét x+y+z+t\(\ne0\)

\(\frac{x}{y+z+t}+\frac{y}{x+z+t}+\frac{z}{x+y+t}+\frac{t}{x+y+z}=\frac{x+y+z+t}{3\left(x+y+z+t\right)}=\frac{1}{3}\)

=>3x=x+y+z

=>4x=x+y+z+t

3y=x+z+t

=>4y=x+y+z+t

3z=x+y+t

=>4z=x+y+z+t

3t=x+y+z+t

=>4t=x+y+z+t

=>4x=4y=4z=4t

=>x=y=z=t

\(\Rightarrow P=\frac{x+y}{z+t}+\frac{y+z}{x+t}+\frac{z+t}{x+y}+\frac{x+t}{y+z}=\frac{x+y}{x+y}+\frac{x+y}{x+y}+\frac{x+y}{x+y}+\frac{x+y}{x+y}\)

=1+1+1+1=4

Vậy P=-4 khi \(x+y+z+t=0\)

P=4 khi \(x+y+z+t\ne0\)

\(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)

\(\Rightarrow\frac{x}{y+z+t}+1=\frac{y}{z+t+x}+1=\frac{z}{t+x+y}+1=\frac{t}{x+y+z}+1\)

\(\frac{x+y+z+t}{y+z+t}=\frac{y+z+t+x}{z+t+x}=\frac{z+t+x+y}{t+x+y}=\frac{t+x+y+z}{x+y+z}\)

Khi đó \(P=1+1+1+1=4\)

Khi đó \(P=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

ms đúng \(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)