Chung minh rang neu 2(x+y) = 5(y+z) = 3(z+x) thi \(\frac{x-y}{4}\) \(\frac{y-z}{5}\)
cho x, y,z >0 chung minh rang\(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}< hoac=\frac{3}{ }4\)3/4
c/m: neu 2(x+y)=5(y+z)=3(z+x) thi \(\frac{x-y}{4}=\frac{y-z}{5}\)
giải hẳn ra
2(x+y) = 5(y+z) = 3(z+x)
<=> (x+y)/(1/2) = (y+z)/(1/5) = (z+x)/(1/3) = (x+y-z-x)/(1/2-1/3) = (z+x-y-z)/(1/3-1/5)
=> (y-z)/(1/2-1/3) = (x-y)/(1/3-1/5) => (y-z)/(1/6) = (x-y)/(2/15)
=> 6(y-z) = 15(x-y)/2 <=> 2(y-z) = 5(x-y)/2 <=> (y-z)/5 = (x-y)/4 đpcm
cho x,y,z>0 va xyz=1 chung minh rang neu x+y+z>1/x+1/y+1/z thi trong 3 so co it nhat 1 so lon hon 1
cho \(x,y,z>1\)thoar man :\(xyz=x+y+z\)chung minh rang:
\(\frac{x-2}{z^2}+\frac{y-2}{x^2}+\frac{z-2}{y^2}\ge\sqrt{3}-2\)
Tim x,y va z neu \(x+y+z=\frac{x}{y+z-2}=\frac{y}{z+x-3}=\frac{z}{x+y+5}\)
Theo tính chất dãy tỉ số bằng nhau ta có:
\(x+y+z=\frac{x}{y+z-2}=\frac{y}{z+x-3}=\frac{z}{x+y+5}=\frac{x+y+z}{\left(y+z-2\right)+\left(z+x-3\right)+\left(x+y+5\right)}=\frac{x+y+z}{2.\left(x+y+z\right)}=\frac{1}{2}\)
=> x + y +z = 1/2 => y + z = 1/2 - x
\(\frac{x}{y+z-2}=\frac{1}{2}\Rightarrow y+z-2=2x\) => \(\frac{1}{2}-x-2=2x\) => \(-\frac{3}{2}=3x\Rightarrow-\frac{1}{2}=x\)
tương tự, \(\frac{y}{z+x-3}=\frac{1}{2}\Rightarrow2y=z+x-3\) => \(2y=\frac{1}{2}-y-3\) => 3y = -5/2 => y = -5/6
z = 1/2 - (x+y) = \(\frac{1}{2}-\left(-\frac{1}{2}-\frac{5}{6}\right)=\frac{1}{2}-\left(-\frac{8}{6}\right)=\frac{1}{2}+\frac{8}{6}=\frac{11}{6}\)
Chung minh rang:
\(\frac{x}{x+y}\)+\(\frac{y}{y+z}\)+\(\frac{z}{z+x}\)< 2
Ta co:\(\frac{x}{x+y}\)<1\(\Rightarrow\)\(\frac{x}{x+y}\)<\(\frac{x+y}{x+y+z}\)(1)
\(\frac{y}{y+z}\)<1\(\Rightarrow\)\(\frac{y}{y+z}\)<\(\frac{y+x}{y+z+x}\)(2)
\(\frac{z}{z+x}\)<1\(\Rightarrow\)\(\frac{z}{z+x}\)<\(\frac{z+y}{z+x+y}\)(3)
Tu(1)(2)(3)\(\Rightarrow\)\(\frac{x}{x+y}\)+\(\frac{y}{y+z}\)+\(\frac{z}{z+x}\)< \(\frac{x+z}{x+y+z}\)+ \(\frac{y+x}{y+z+x}\) + \(\frac{z+y}{z+x+y}\)
\(\Rightarrow\)A <\(\frac{2x+2y+2z}{x+y+z}\)
\(\Rightarrow\)A < \(\frac{2\left(x+y+z\right)}{x+y+z}\)
\(\Rightarrow\)A< 2
Bạn định kiểm tra chỉ số thông minh IQ người khác hà mà sao biết bài toán rồi vẫn hỏi?
cm rang neu x+y+z=0 thi( x^2+ y^2+z^2 )=2(x^4+y^4+z^2)
cho cac so x,y,z khac 0 va thoa man \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\) Chung minh rang x2(y+z)+y2(z+x ) +z2(x+z)+3xyz
ai nnha nhat minh tik dung luon
cho 2(x+y) = 5(y+z) = 3(x+z) chứng minh \(\frac{x-y}{4}=\frac{y-z}{5}\)
Ta có: 2.(x + y) = 5.(y + z) = 3.(x + z)
\(\Rightarrow\frac{2.\left(x+y\right)}{30}=\frac{5.\left(y+z\right)}{30}=\frac{3.\left(x+z\right)}{30}\)
\(\Rightarrow\frac{x+y}{15}=\frac{y+z}{6}=\frac{x+z}{10}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{x+y}{15}=\frac{y+z}{6}=\frac{x+z}{10}=\frac{\left(x+z\right)-\left(y+z\right)}{10-6}=\frac{\left(x+y\right)-\left(x+z\right)}{15-10}\)
\(=\frac{x-y}{4}=\frac{y-z}{5}\left(đpcm\right)\)
Vì 5 (y + z) = 3 (z + x) \(\Rightarrow\) \(\frac{z+x}{5}=\frac{y+z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{z+x}{5}=\frac{y+z}{3}=\frac{z+x-y-z}{5-3}=\frac{x-y}{2}\)
Do đó: \(\frac{z+x}{5}=\frac{x-y}{2}\Rightarrow\frac{z+x}{10}=\frac{x-y}{4}\left(1\right)\)
Ta lại có: 2 (x + y) = 3 (z + x)
\(\Rightarrow\) \(\frac{z+x}{2}=\frac{x+y}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{z+x}{2}=\frac{x+y}{3}=\frac{z+x-x-y}{2-3}=y-x\)
Do đó: \(\frac{z+x}{2}=y-z\Rightarrow\frac{z+x}{10}=\frac{y-z}{5}\left(2\right)\)
Từ (1) và (2) suy ra: \(\frac{x-y}{4}=\frac{y-z}{5}\) (đpcm)