\(P=\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\) ; \(Q=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
Chứng minh rằng: Nếu \(P=1\) thì \(Q=0\)
thực hiện phép tính
a,\(x^3+\left[\frac{x\left(2y^3-x^3\right)}{x^3+y^3}\right]^3-\left[\frac{y\left(2x^3-y^3\right)}{x^3+y^3}\right]^3\)
b,\(\frac{\frac{x\left(x+y\right)}{x-y}+\frac{x\left(x+z\right)}{x-z}}{1+\frac{\left(y-z\right)^2}{\left(x-y\right)\left(x-z\right)}}+\frac{\frac{y\left(y+z\right)}{y-z}+\frac{y\left(y+x\right)}{y-x}}{1+\frac{\left(z-x\right)^2}{\left(y-z\right)\left(y-x\right)}}+\frac{\frac{z\left(z+x\right)}{z-x}+\frac{z\left(z+y\right)}{z-y}}{1+\frac{\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}}\)
c,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
\(Cho:\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{x+y+t}=\frac{t}{x+y+z}.TínhF=\frac{x+y}{z+t}+\frac{y+z}{x+t}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)
từ đó =>3x=y+z+t
=>4x=x+y+z+t
tương tự=>4y=x+y+z+t
4z=x+y+z+t
4t=x+y+z+t
=>x=y=z=t=>F=4
mà bài này lớp 7 chứ,có phải lớp 9 đâu
dãy đó ra bằng 1/3. nhưng sao suy ra đc x=y=z=t vậy?
cho các số dương x,y,z,t . Chứng minh: \(\frac{40}{3}\le\frac{x}{y+z+t}+\frac{y}{z+t+x}+\frac{z}{t+x+y}+\frac{t}{x+y+z}+\frac{y+z+t}{x}+\frac{z+t+x}{y}+\frac{t+x+y}{z}+\frac{x+y+z}{t}\)
\(VP=\frac{x}{y+z+t}+\frac{y}{z+t+x}+\frac{z}{t+x+y}+\frac{t}{x+y+z}+\frac{y+z+t}{x}+\frac{z+t+x}{y}+\frac{t+x+y}{z}+\frac{x+y+z}{t}=\left(\frac{x}{y+z+t}+\frac{y+z+t}{9x}\right)+\left(\frac{y}{z+t+x}+\frac{z+t+x}{9y}\right)+\left(\frac{z}{t+x+y}+\frac{t+x+y}{9z}\right)+\left(\frac{t}{x+y+z}+\frac{x+y+z}{9t}\right)+\frac{8}{9}\left(\frac{y+z+t}{x}+\frac{z+t+x}{y}+\frac{t+x+y}{z}+\frac{x+y+z}{t}\right)\)\(\ge8\sqrt[8]{\frac{x}{y+z+t}.\frac{y}{z+t+x}.\frac{z}{t+x+y}.\frac{t}{x+y+z}.\frac{y+z+t}{9x}.\frac{z+t+x}{9y}.\frac{t+x+y}{9z}.\frac{x+y+z}{9t}}+\frac{8}{9}\left(\frac{y}{x}+\frac{z}{x}+\frac{t}{x}+\frac{z}{y}+\frac{t}{y}+\frac{x}{y}+\frac{t}{z}+\frac{x}{z}+\frac{y}{z}+\frac{x}{t}+\frac{y}{t}+\frac{z}{t}\right)\)\(\ge\frac{8}{3}+\frac{8}{9}.12\sqrt[12]{\frac{y}{x}.\frac{z}{x}.\frac{t}{x}.\frac{z}{y}.\frac{t}{y}.\frac{x}{y}.\frac{t}{z}.\frac{x}{z}.\frac{y}{z}.\frac{x}{t}.\frac{y}{t}.\frac{z}{t}}=\frac{8}{3}+\frac{8}{9}.12=\frac{40}{3}=VT\left(đpcm\right)\)
Đẳng thức xảy ra khi x = y = z = t > 0
Biết:\(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{x+t+y}=\frac{t}{x+y+z}\)
Tìm giá tị của \(P=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)
Ta có: \(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{x+t+y}=\frac{t}{x+y+z}\)
Thêm 1 vào mỗi phân số ta được:
\(\frac{x}{y+z+t}+1=\frac{y}{z+t+x}+1=\frac{z}{x+t+y}+1=\frac{t}{x+y+z}+1\)
\(\Rightarrow\frac{x+y+z+t}{y+z+t}=\frac{x+y+z+t}{z+t+x}=\frac{x+y+z+t}{x+t+y}=\frac{x+y+z+t}{x+y+z}\)
- Nếu x + y + z + t \(\ne\) 0 thì x = y = z = t
\(\Rightarrow P=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}=\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}=1+1+1+1=4\)
- Nếu x + y + z + t = 0 thì x + y = -(z + t)
y + z = -(t + x)
z + t = -(x + y)
t + x = -(y + z)
\(\Rightarrow P=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}=\frac{-\left(z+t\right)}{z+t}+\frac{-\left(t+x\right)}{t+x}+\frac{-\left(x+y\right)}{x+y}+\frac{-\left(y+z\right)}{y+z}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
Cho x+y+z=7. Biết \frac{x}{y+z} +\frac{y}{x+z} +\frac{z}{x+y} = 3. Tính \frac{x^{2}}{y+z} +\frac{y^{2}}{x+z} +\frac{z^{2}}{x+y}
Cho x,y,z,t thỏa mãn \(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{y}{t+x+y}=\frac{t}{x+y+z}\)
Tính \(P=\frac{x+y}{z+t}+\frac{y+z}{x+t}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)
\(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)
\(\Rightarrow\frac{x}{y+z+t}+1=\frac{y}{z+t+x}+1=\frac{z}{t+x+y}+1=\frac{t}{x+y+z}+1\)
\(\frac{x+y+z+t}{y+z+t}=\frac{y+z+t+x}{z+t+x}=\frac{z+t+x+y}{t+x+y}=\frac{t+x+y+z}{x+y+z}\)
Xét \(x+y+z+t\ne0\Rightarrow x=y=z=t\)Khi đó \(P=1+1+1+1=4\)
Xét \(x+y+z+t=0\Rightarrow\begin{cases}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\t+x=-\left(y+z\right)\end{cases}\)Khi đó \(P=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
ms đúng \(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)
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 : "$\frac{x+y}{z+t}$=$\frac{y+z}{t+x}$=$\frac{z+t}{x+y}$=$\frac{t+z}{y+z}$"
Đề:
Cho các số thực x, y, z thoả mãn x + y + z = 1 và \(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1\)
\(\left(x\ne-y;y\ne-z;z\ne-x\right)\)
Giá trị của biểu thức \(P=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\) là . . .
Giải:
x + y + z = 1
=> x = 1 - (y + z)
y = 1 - (x + z)
z = 1 - (x + y)
Thay x = 1 - (y + z); y = 1 - (x + z) và z = 1 - (x + y) vào P, ta có:
\(P=\frac{x\left[1-\left(y+z\right)\right]}{y+z}+\frac{y\left[1-\left(x+z\right)\right]}{x+z}+\frac{z\left[1-\left(x+y\right)\right]}{x+y}\)
\(=\frac{x-x\left(y+z\right)}{y+z}+\frac{y-y\left(x+z\right)}{x+z}+\frac{z-z\left(x+y\right)}{x+y}\)
\(=\frac{x}{y+z}-\frac{x\left(y+z\right)}{y+z}+\frac{y}{x+z}-\frac{y\left(x+z\right)}{x+z}+\frac{z}{x+y}-\frac{z\left(x+y\right)}{x+y}\)
\(=\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)-\left(x+y+z\right)\)
\(=1-1\)
\(=0\)
ĐS: 0
Trịnh Trân Trân <3
Hay quớ ak! Mơn m nhìu nha ný! <3 <3 <3 (not thả thính =))))
Tìm x,y,z thuộc Z biết:
\(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{y}{x}+\frac{z}{y}+\frac{x}{z}=x+y+z=3\)
\(Cho:P=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{z+y}\)
Tính P biết: \(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)