3
Tìm x,y,z
\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)và x-y+z=-21
DN
Những câu hỏi liên quan
thực hiện phép tính
a,x^3+left[frac{xleft(2y^3-x^3right)}{x^3+y^3}right]^3-left[frac{yleft(2x^3-y^3right)}{x^3+y^3}right]^3
b,frac{frac{xleft(x+yright)}{x-y}+frac{xleft(x+zright)}{x-z}}{1+frac{left(y-zright)^2}{left(x-yright)left(x-zright)}}+frac{frac{yleft(y+zright)}{y-z}+frac{yleft(y+xright)}{y-x}}{1+frac{left(z-xright)^2}{left(y-zright)left(y-xright)}}+frac{frac{zleft(z+xright)}{z-x}+frac{zleft(z+yright)}{z-y}}{1+frac{left(x-yright)^2}{left(z-xright)left(z-yright)}}
c,left[frac{y+z-2x}{fra...
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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 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 khác 0 và x+y+z=0
chứng minh rằng
\(\frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{x^2+z^2}{x+z}=\frac{x^3}{yz}+\frac{y^3}{xz}+\frac{z^3}{xy}\)
ta có:(vế phải)2le3left(frac{x^3}{y+z}+frac{y^3}{z+x}+frac{z^3}{x+y}right)cần chứng minh:(vế trái)2/3gefrac{x^3}{y+z}+frac{y^3}{z+x}+frac{z^3}{x+y}Leftrightarrowfrac{x}{y+z}left(frac{x^3+frac{1}{3}}{y+z}-x^2right)+...ge0Leftrightarrowfrac{x^2}{y+z}left(x-yright)left(x-zright)+frac{y^2}{z+x}left(y-xright)left(y-zright)+frac{z^2}{x+y}left(z-xright)left(z-yright)ge0bđt luôn đúng vì là bđt schur mở rộng
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ta có:(vế phải)2\(\le3\left(\frac{x^3}{y+z}+\frac{y^3}{z+x}+\frac{z^3}{x+y}\right)\)
cần chứng minh:
(vế trái)2/3\(\ge\frac{x^3}{y+z}+\frac{y^3}{z+x}+\frac{z^3}{x+y}\)
\(\Leftrightarrow\frac{x}{y+z}\left(\frac{x^3+\frac{1}{3}}{y+z}-x^2\right)+...\ge0\)
\(\Leftrightarrow\frac{x^2}{y+z}\left(x-y\right)\left(x-z\right)+\frac{y^2}{z+x}\left(y-x\right)\left(y-z\right)+\frac{z^2}{x+y}\left(z-x\right)\left(z-y\right)\ge0\)
bđt luôn đúng vì là bđt schur mở rộng
đặt Afrac{sqrt{yz}}{x+3sqrt{yz}}+frac{sqrt{zx}}{y+3sqrt{zx}}+frac{sqrt{xy}}{z+3sqrt{xy}}Rightarrow1-3Afrac{x}{x+3sqrt{yz}}+frac{y}{y+3sqrt{zx}}+frac{z}{z+3sqrt{xy}}gefrac{x}{x+frac{3}{2}left(y+zright)}+frac{y}{y+frac{3}{2}left(z+xright)}+frac{z}{z+frac{3}{2}left(x+yright)}frac{2x}{2x+3left(y+zright)}+frac{2y}{2y+3left(z+xright)}+frac{2z}{2z+3left(x+yright)}frac{2x^2}{2x^2+3xy+3xz}+frac{2y^2}{2y^2+3yz+3xy}+frac{2z^2}{2z^2+3zx+3yz}gefrac{2left(x+y+zright)^2}{2left(x^2+y^2+z^2right)+6left(xy+yz+zxr...
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đặt \(A=\frac{\sqrt{yz}}{x+3\sqrt{yz}}+\frac{\sqrt{zx}}{y+3\sqrt{zx}}+\frac{\sqrt{xy}}{z+3\sqrt{xy}}\)
\(\Rightarrow1-3A=\frac{x}{x+3\sqrt{yz}}+\frac{y}{y+3\sqrt{zx}}+\frac{z}{z+3\sqrt{xy}}\)
\(\ge\frac{x}{x+\frac{3}{2}\left(y+z\right)}+\frac{y}{y+\frac{3}{2}\left(z+x\right)}+\frac{z}{z+\frac{3}{2}\left(x+y\right)}\)
\(=\frac{2x}{2x+3\left(y+z\right)}+\frac{2y}{2y+3\left(z+x\right)}+\frac{2z}{2z+3\left(x+y\right)}\)
\(=\frac{2x^2}{2x^2+3xy+3xz}+\frac{2y^2}{2y^2+3yz+3xy}+\frac{2z^2}{2z^2+3zx+3yz}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x^2+y^2+z^2\right)+6\left(xy+yz+zx\right)}=\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+2\left(xy+yz+zx\right)}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+\frac{2}{3}\left(x+y+z\right)^2}=\frac{2\left(x+y+z\right)^2}{\frac{8}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)
\(\Rightarrow1-3A\ge\frac{3}{4}\Rightarrow A\le\frac{3}{4}\left(Q.E.D\right)\)
cho x,y,z là các số thực dương chứng minh rằng :
\(\frac{x^3}{y^2}+\frac{y^3}{z^2}+\frac{z^3}{x^2}\ge\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\)
\(\Sigma\frac{x^3}{y^2}=\Sigma\frac{x}{y^2}\left(x-y\right)^2+\frac{\Sigma z\left(x^3-yz^2\right)^2}{xyz\left(x+y+z\right)}+\Sigma\frac{x^2}{y}\ge\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\)
\(VT-VP=\Sigma\frac{\left(x+y\right)\left(x-y\right)^2}{y^2}\ge0\)
$\frac{x+y}{z}$ + $\frac{y+z}{x}$ + $\frac{z+x}{y}$- $\frac{x^3+y^3+z^3}{xyz}$ =2
thực hiện phép tính
a, \(\frac{x^2-yz}{1+\frac{y+x}{x}}+\frac{y^2-xz}{1+\frac{z+x}{y}}+\frac{z^2-xy}{1+\frac{x+y}{z}}\)
b, \(\left(1+\frac{y^2+z^2-x^2}{2yz}\right).\frac{1+\frac{x}{y+z}}{1-\frac{x}{y+z}}.\frac{y^2+z^2-\left(y-z\right)^2}{x+y+z}\)
c,\(\frac{2}{3}\left[\frac{1}{1+\frac{\left(2x+1\right)^2}{3}}+\frac{1}{1+\frac{\left(2x-1\right)^2}{3}}\right]\)
Tìm x , y , z nếu :
a)\(\frac{x+y+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
b)\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)và 2x+3y-z=50
b) \(\frac{x-1}{2}=\frac{2x-2}{4}\)
\(\frac{y-2}{3}=\frac{3y-6}{9}\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{2x+3y-z+3-2-6}{9}=\frac{50+3-2-6}{9}=\frac{45}{9}=5\)=>x-1=5.2=10
=>x=11
y-2=5.3=15
=>y=17
z-3=5.4=20
=>z=23
Vậy (x;y;z)=(11;17;23)
Áp dụng t/c của dãy tỉ số bằng nhau:
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
\(=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+x-3\right)}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)(vì x+y+z khác 0).Do đó x+y+z = 0.5
Thay kq này vào bài ta được:
\(\frac{0,5-x+1}{x}=\frac{0,5-y+2}{y}=\frac{0,5-z-3}{z}=2\)
Tức là : \(\frac{1,5-x}{x}=\frac{2,5-y}{y}=\frac{-2,5-z}{z}=2\)
Vậy \(x=\frac{1}{2};y=\frac{5}{6};z=\frac{-5}{6}\)
a) \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
b) \(\frac{x}{y+z+1}=\frac{y}{x+z+2}=\frac{z}{x+y-2}=x+y+z\)
Tìm x ;y;z
a) \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)
\(\Rightarrow x+y+z=\frac{1}{2}\)(do 1/(x+y+z)=2)
\(\Rightarrow y+z=\frac{1}{2}-x;z+x=\frac{1}{2}-y;x+y=\frac{1}{2}-z\)
Thay vào lần lượt ta có:
\(\frac{\frac{1}{2}-x+1}{x}=2\)\(\Rightarrow x=\frac{1}{2}\)
\(\frac{\frac{1}{2}-y+2}{y}=2\)\(\Rightarrow y=\frac{5}{6}\)
\(\frac{\frac{1}{2}-z-3}{z}=2\)\(\Rightarrow z=-\frac{5}{6}\)
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