Cho x/y = y/z = z/t
CMR x/t = (x+y+z/y+z+t)^3
Những câu hỏi liên quan
cho x/y+z+t=y/z+t+x=z/t+x+y=t/x+y+z cmr bieu thuc sau co gia tri nguyen P=(x+y/z+t)+(y+z/t+x)+(z+t/x+y)=(t+x/y+z)
Cho x/(y+z+t)=y/(z+t+x)=z/(t+x+y)=t/(x+y+z) cmr P=(x+y)/(z+t)+(y+z)/(t+x)+(z+t)/(x+y)+(t+x)/(y+z) là biểu thức có giá trị nguyên
cho dãy tỉ số bằng nhau :$\frac{x}{y+z+t}$=$\frac{y}{z+t+x}$=$\frac{z}{t+x+y}$=$\frac{t}{x+y+z}$ cmr : "$\frac{x+y}{z+t}$=$\frac{y+z}{t+x}$=$\frac{z+t}{x+y}$=$\frac{t+z}{y+z}$"
cho x /(y+z+t)= y /(x+z+t)= z /(x+y+t)=t /(x+y+z)
CMR: P=(x+y)/(z+t) + (y+z)/(t+x) + (z+t)/(x+y) + (t+x)/(y+z)
cho x/(y+z+t)=y/(z+t+x)=z/(t+x+y)=t/(x+y+z)
cmr bieu thuc sau co gia tri nguyen
P=(x+y)/(z+t)+(y+z)/(t+x)+(z+t)/(x+y)+(t+x)/(y+z)
Cho \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)
CMR biểu thức sau có giá trị nguyên
\(A=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+z}{y+z}\)
TH1: \(x+y+z+t\ne0\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)\(\dfrac{x}{y+z+t}=\dfrac{1}{3}\Rightarrow3x=y+z+t\Rightarrow4x=x+y+z+t\\ \dfrac{y}{z+t+x}=\dfrac{1}{3}\Rightarrow3y=x+z+t\Rightarrow4y=x+y+z+t\\ \dfrac{z}{t+x+y}=\dfrac{1}{3}\Rightarrow3z=x+y+t\Rightarrow4z=x+y+z+t\\ \dfrac{t}{x+y+z}=\dfrac{1}{3}\Rightarrow3t=x+y+z\Rightarrow4t=x+y+z+t\)
\(\Rightarrow4x=4y=4z=4t\\ \Rightarrow x=y=z=t\)
\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =1+1+1+1\\ =4\)
TH2: \(x+y+z+t=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\t+x=-\left(y+z\right)\end{matrix}\right.\)
\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =\dfrac{-\left(z+t\right)}{z+t}+\dfrac{-\left(t+x\right)}{t+x}+\dfrac{-\left(x+y\right)}{x+y}+\dfrac{-\left(y+z\right)}{y+z}\\ =-1-1-1-1\\ =-4\)
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cho các số thực x,y,z,t thoả mãn\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)
cmr P = \(\dfrac{x+y}{z+t}=\dfrac{y+z}{t+x}=\dfrac{z+t}{x+y}=\dfrac{t+x}{y+z}\)
có giá trị nguyên
Từ gt của đề bài :
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{y+z+t}\text{=}\dfrac{y}{z+t+x}\text{=}\dfrac{z}{x+y+t}\text{=}\dfrac{t}{x+y+z}\text{=}\dfrac{x+y+z+t}{3.\left(x+y+z+t\right)}\left(\cdot\right)\)
Xét TH : \(x+y+z+t\text{=}0\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z\text{=}-\left(x+t\right)\\z+t\text{=}-\left(x+y\right)\\x+t\text{=}-\left(y+z\right)\end{matrix}\right.\)
Do đó : \(P\text{=}-1+-1+-1+-1\)
\(P\text{=}-4\in Z\)
TH : \(x+y+z+t\ne0\)
\(\Rightarrow\left(\cdot\right)\text{=}\dfrac{1}{3}\)
Do đó : \(\dfrac{x}{y+z+t}\text{=}\dfrac{1}{3}\Rightarrow3x\text{=}y+z+t\)
\(\Rightarrow4x\text{=}x+y+z+t\)
\(CMTT:\left\{{}\begin{matrix}4y\text{=}x+y+z+t\\4z\text{=}x+y+z+t\\4t\text{=}x+y+z+t\end{matrix}\right.\)
Mà : \(\dfrac{x}{y+z+t}\text{=}\dfrac{y}{x+z+t}\text{=}\dfrac{z}{x+y+t}\text{=}\dfrac{t}{x+y+z}\)
\(\Rightarrow4x\text{=}4y\text{=}4z\text{=}4t\)
\(\Rightarrow x\text{=}y\text{=}z\text{=}t\)
Do đó : \(P\text{=}4\in Z\)
\(\Rightarrowđpcm\)
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Kham khảo :
https://olm.vn/cau-hoi/cho-cac-so-thuc-xyzt-thoa-mandfracxyztdfracyztxdfracztxydfractxyz-cmr-p-dfracxyztdfracyztx.8377111224063.
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Cho x/y+z+t= y/z+t+x=z/t+x+y=t/x+y+z
Cmr: x+y/z+t=y+z/t+x=z+t/x+y=t+x/y+z có giá trị nguyên
GIÚP MÌNH VỚI MÌNH CẢM ƠN
Cho M = \(\dfrac{x}{x+y+z}\)+\(\dfrac{y}{y+z+t}\)+\(\dfrac{z}{z+t+x}\)+\(\dfrac{t}{t+x+y}\) với x, y, z, t ϵ N*
CMR: M10 < 1025