cho \(\frac{x}{y}=\frac{y}{z}=\frac{z}{t}cmr\) \(\left(\frac{x+y+z}{y+z+t}\right)^3\) =\(\frac{y}{t}\)
Cho x,y,z,t >0
C/m : \(\left(x+y+z+t\right)\left(\frac{1}{x+y+z}+\frac{1}{y+z+t}+\frac{1}{z+t+x}+\frac{1}{t+x+y}\right)\ge\frac{16}{3}\)
bạn dùng BĐT Cauchuy-Swartch cho cs Bt thứ 2 là ra nhé
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 : "$\frac{x+y}{z+t}$=$\frac{y+z}{t+x}$=$\frac{z+t}{x+y}$=$\frac{t+z}{y+z}$"
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 các số nguyên x,y,z,t thỏa mãn
\(\frac{x+y}{y+z}=\frac{y+z}{z+t}=\frac{z+t}{t+x}=\frac{t+x}{x+y}\)
CTR:A =\(\left(\frac{y+z}{x+t}\right)^{2013}+\left(\frac{y+t}{x+y}\right)^{2014}\)có giá trị là số nguyên
Đề sai kìa bạn , xem lại phân số : (y+t/x+y)^2014
vậy bn làm theo cái đúng của bn,mong bn giúp mk
CMR:
\(\left(x+y+z+t\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)\ge16\) \(\forall x,y,z,t>0\)
áp dụng bđt cô si có:
\(x+y+z+t\ge4\sqrt[4]{xyzt}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\ge4\sqrt[4]{\frac{1}{xyzt}}\)
nhân theo vế \(\left(x+y+z+t\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)\ge4\cdot4\sqrt[4]{xyzt\cdot\frac{1}{xyzt}}=16\)
Cho \(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}=0;x\ne y,y\ne z,z\ne x\)
Tính Q=\(\frac{x}{\left(y-z\right)^2}+\frac{y}{\left(z-x\right)^2}+\frac{z}{\left(x-y\right)^2}\)
\(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}=0\\ =\frac{x}{y-z}=-\left(\frac{y}{z-x}+\frac{z}{x-y}\right)\\ =\frac{x}{\left(y-x\right)^2}=-\left(\frac{y}{z-x}+\frac{z}{x-y}\right).\frac{1}{y-x}=\frac{-xy+y^2-z^2+xz}{\left(z-x\right)\left(x-y\right)\left(y-z\right)}\left(1\right)\)
Tự làm với 2 phân thức còn lại, ta có:
\(\frac{y}{\left(z-x\right)^2}=\frac{-x^2+z^2+xy-yz}{\left(z-x\right)\left(x-y\right)\left(y-z\right)}\left(2\right)\)
\(\frac{z}{\left(x-y\right)^2}=\frac{x^2-y^2-xz+yz}{\left(z-x\right)\left(x-y\right)\left(y-z\right)}\left(3\right)\)
Cộng 3 vế lại với nhau ta có: \(Q=\frac{x}{\left(y-x\right)^2}+\frac{y}{\left(z-x\right)^2}+\frac{z}{\left(x-y\right)^2}=0\)
Cho 3 số x, y, z là 3 số khác 0 t/m đ/k \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\). Tính giá trị biểu thức \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)\(\Rightarrow\frac{y+z}{x}-1=\frac{z+x}{y}-1=\frac{x+y}{z}-1\)
\(\Rightarrow\frac{y+z}{x}=\frac{z+x}{y}=\frac{x+y}{z}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{y+z}{x}=\frac{z+x}{y}=\frac{x+y}{z}=\frac{y+z+z+x+x+y}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
Do đó: +) \(\frac{y+z}{x}=2\)\(\Rightarrow y+z=2x\)
+) \(\frac{z+x}{y}=2\)\(\Rightarrow z+x=2y\)
+) \(\frac{x+y}{z}=2\)\(\Rightarrow x+y=2z\)
Ta có: \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{y+x}{y}.\frac{z+y}{z}.\frac{x+z}{x}=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}=2.2.2=8\)
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?
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=\(\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{z+y}\)