Cho \(\frac{x}{y+z+t}\)=\(\frac{y}{z+t+x}\)=\(\frac{z}{t+x+y}\)=\(\frac{t}{x+y+z}\)
CMR : bieu thuc sau co gia tri nguyen P = \(\frac{x+y}{z+t}\)+\(\frac{y+z}{t+x}\)+\(\frac{z+t}{x+y}\)+\(\frac{t+x}{y+z}\)
Cho \(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)
Tinh gia tri cua da thuc\(P=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)
Cho x ; y ; z ; t : CM : \(M=\frac{x}{x+y+z}+\frac{y}{x+y+t}+\frac{z}{y+z+t}+\frac{t}{x+z+t}\)co gia tri khong phai la so tu nhien
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 : "$\frac{x+y}{z+t}$=$\frac{y+z}{t+x}$=$\frac{z+t}{x+y}$=$\frac{t+z}{y+z}$"
cmr
M=\(\frac{x}{x+y+z}+\frac{y}{x+y+t}+\frac{z}{y+z+t}+\frac{t}{x+z+t}\) (x,y,z,t\(_{\in}\) \(ℕ^∗\) ) co gia tri khong phai la so tu nhien
ta có \(\frac{x}{x+y+z}>\frac{x}{x+y+z+t}\)
\(\frac{y}{x+y+t}>\frac{y}{x+y+z+t}\)
\(\frac{z}{y+z+t}>\frac{z}{x+y+z+t}\)
\(\frac{t}{x+z+t}>\frac{t}{x+y+z+t}\)
\(\Rightarrow M\)\(>\frac{x}{x+y+z+t}+\frac{y}{x+y+z+t}+\frac{z}{x+y+z+t}+\frac{t}{x+y+z+t}=1\)
ta lại có \(\frac{x}{x+y+z}< \frac{x+t}{x+y+z+t}\)
\(\frac{y}{x+y+t}< \frac{y+z}{x+y+z+t}\)
\(\frac{z}{y+z+t}< \frac{z+x}{x+y+z+t}\)
\(\frac{t}{x+z+t}< \frac{t+y}{x+y+z+t}\)
\(\Rightarrow M>\frac{x+x+y+y+z+z+t+t}{x+y+z+t}=\frac{2x+2y+2z+2t}{x+y+z+t}=2\)
\(\Rightarrow1< M< 2\)
vậy M không phải là số tự nhiên
\(M>\frac{x}{x+y+z+t}+\frac{y}{x+y+z+t}+\frac{z}{x+y+z+t}+\frac{t}{x+y+z+t}=1\)
\(CM:\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m};m\in\)N*
Biến đổi tương đương.
\(\Rightarrow M< \frac{x+t}{x+y+z+t}+\frac{y+z}{x+y+z+t}+\frac{x+z}{x+y+z+t}+\frac{t+y}{x+y+z+t}=2\)
Vì 1<M<2=> M ko phải stn
Cho \(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)
CMR biểu thức sau có giá trị nguyên
P=\(\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)
ĐK:y+z+t,z+t+x,t+x+z,x+z+y khác 0
x+y+t+z khác 0
\(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}=\frac{x+y+z+t}{3\left(x+y+z+t\right)}\)
mà x+y+z+t khác 0 nên:
\(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}=\frac{1}{3}\Rightarrow x=y=z=t\)
\(\Rightarrow P=4\left(\text{nguyên}\right).\text{Vậy: P nguyên}\)
@shitbo : Cơ sở đâu mà bạn cho rằng: x + y + z + t khác 0? Nếu x + y + z + t = 0 thì P = -1 ok?
Cho :\(\frac{x}{y+z+t}=\frac{y}{x+z+t}=\frac{z}{y+x+t}=\frac{t}{y+x+z}\)
CMR biểu thức sau có GT nguyên:
\(P=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)
cho:
\(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)
CMR biểu thức sau có giá trị nguyên:
\(P=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+y}{y+z}\)
Cho \(\frac{x}{y+z+t}=\frac{y}{x+z+t}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)
CMR biểu thức sau có giá trị nguyên
\(\frac{x+y}{z+t}+\frac{y+z}{x+t}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)
CMR \(\frac{X}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)
Cho biểu thức P=\(\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{z+y}\)