a) Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\)

=> ad = bc

Ta có : (a + 2c)(b + d)

= a(b + d) + 2c(b + d)

= ab + ad + 2cb + 2cd (1)

Ta có : (a + c)(b + 2d)

= a(b + 2d) + c(b + 2b)

= ab + a2d + cb + c2b

= ab + c2d + ad + c2b (Vì ad = cd) (2)

Từ (1),(2) => (a + 2c)(b + d) = (a + c)(b + 2d) (ĐPCM)

Sửa đề bài : P = \(\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\)

Ta có : \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)

=> \(\dfrac{y+z+t}{x}=\dfrac{z+t+x}{y}=\dfrac{t+x+y}{z}=\dfrac{x+y+z}{t}\)

=> \(\dfrac{y+z+t}{x}+1=\dfrac{z+t+x}{y}+1=\dfrac{t+x+y}{z}+1=\dfrac{x+y+z}{t}+1\)=> \(\dfrac{y+z+t+x}{x}=\dfrac{z+t+x+y}{y}=\dfrac{t+x+y+z}{z}=\dfrac{x+y+z+t}{t}\)TH1: x + y + z + t # 0

=> x = y = z = t

Ta có : P = \(\dfrac{x+y}{z+t}=\dfrac{y+z}{t+x}=\dfrac{z+t}{x+y}=\dfrac{t+x}{y+z}\)

P = \(\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}\)

P = 1 + 1 + 1 + 1 = 4

TH2 : x + y + z + t = 0

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

y + z = -(t + x)

z + t = -(x + y)

t + x = -(y + z)

Ta có : P = \(\dfrac{x+y}{z+t}=\dfrac{y+z}{t+x}=\dfrac{z+t}{x+y}=\dfrac{t+x}{y+z}\)

P = \(\dfrac{-\left(z+t\right)}{z+t}=\dfrac{-\left(t+x\right)}{t+x}=\dfrac{-\left(x+y\right)}{x+y}=\dfrac{-\left(y+z\right)}{y+z}\)

P = (-1) + (-1) + (-1) + (-1)

P = -4

Vậy ...

Từ \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)

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

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

Vì \(x+y+z+t\ne0\) nên ta đi xét \(x+y+z+t=0\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(t+x\right)\\z+t=-\left(x+y\right)\\t+x=-\left(y+z\right)\end{matrix}\right.\). Khi đó

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

Tuy không hoàn toàn giống nhưng bạn tham khảo rồi chứng minh tương tự nhé !

https://hoc24.vn/hoi-dap/question/459079.html

* Nếu x = y = z = t; vẫn thỏa gt: \(\dfrac{x}{y+z+t}\) = \(\dfrac{y}{x+z+t}\) = \(\dfrac{z}{y+x+t}\) = \(\dfrac{t}{y+z+x}\) = \(\dfrac{1}{3}\)
=> P = \(\dfrac{2x}{2x}+\dfrac{2x}{2x}+\dfrac{2x}{2x}+\dfrac{2x}{2x}=4\)
* Nếu có ít nhất 2 số khác nhau, giả sử x # y. tính chất tỉ lệ thức:
\(\dfrac{x}{y+z+t}\) \(=\dfrac{y}{x+z+t}=\dfrac{x-y}{y+z+t-x-z-t}=\dfrac{x-y}{y-x}=-1\)
\(\rightarrow x=-y+z+t\rightarrow x+y+z+t=0\)
=>
{ x+y = -(z+t) ---- { (x+y)/(z+t) = -1
{ y+z = -(t+x) => { (y+z)/(t+x) = -1
{ z+t = -(x+y) ---- { (z+t)/(x+y) = -1
{ t+x = -(z+y) ---- { (t+x)/(z+y) = -1
=> P = -1 -1 -1 -1 = -4
Vậy P có giá trị nguyên

Ta có:\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)

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

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

*Xét: \(x+y+z+t\ne0\Rightarrow z=y=z=t,\)khi đó:\(P=1+1+1+1=4\)

* Xét \(x+y+z+t=0\Rightarrow x+y=-\left(z+t\right);y+z=-\left(t+x\right);z+t=-\left(x+y\right);t+z=\left(-y+z\right)\)Khi đó: \(P=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

Vậy P luôn luôn có giá trị nguyên

\(M=\dfrac{x}{x+y+z}=\dfrac{y}{x+y+t}=\dfrac{z}{y+z+t}=\dfrac{z}{x+z+t}\)\(\dfrac{x}{x+y+z}< 1\Rightarrow\dfrac{x+t}{x+y+z+t}>\dfrac{x}{x+y+z}\)

\(Tương\)\(tự\):\(\Rightarrow M< \dfrac{2\left(x+y+z+t\right)}{x+y+z+t}\)

\(Ta\) \(có\):\(2>M>1\)

\(\Rightarrow M\notin N\)\(sao\)

\(A=\dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}\)

Giả sử \(A\in N\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+y+z}\in N\\\dfrac{y}{x+y+t}\in N\\\dfrac{z}{y+z+t}\in N\\\dfrac{t}{x+z+t}\in N\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x⋮x+y+z\\y⋮x+y+t\\z⋮y+z+t\\t⋮x+z+t\end{matrix}\right.\)

Vì \(x;y;z;t\in N\circledast\) nên:

\(\left\{{}\begin{matrix}x\ge x+y+z\\y\ge x+y+t\\z\ge y+z+t\\t\ge x+z+t\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-x\ge x+y+z-x\\y-y\ge x+y+t-y\\z-z\ge y+z+t-z\\t-t\ge x+z+t-t\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+z\le0\\x+t\le0\\y+t\le0\\x+z\le0\end{matrix}\right.\)

Vì \(x;y;z;t\in N\circledast\) nên những điều trên không thể xảy ra

\(\Rightarrow\) điều giả sử sai,\(A\notin N\left(đpcm\right)\)