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a)
\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\Leftrightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
x =8.2 =16
y =12.2 =24
z=15.2 =30
b) \(\frac{x}{3}=\frac{y}{-2}=\frac{z}{4}=\frac{4x+y-2z}{4.3+\left(-2\right)-2.4}=-\frac{18}{2}=-9\)
x =-9.3 =-27
y =-9.(-2) = 18
z =-9.4 = -36
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) nha bạn!
ko hỉu thì ib
\(\left(x+y+z\right).\left(\frac{1}{z}+\frac{1}{y}+\frac{1}{x}\right)\ge9\) với x,y,z dương hay jj đó chứ? (cái này t k bt -.-) VD: x=2, y=-2,z=4
=> \(\left(x+y+z\right).\left(\frac{1}{z}+\frac{1}{y}+\frac{1}{x}\right)=\left(2-2+4\right).\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{4}\right)=1\)
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\(\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)
\(\Leftrightarrow\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-\frac{x+y+z}{x+y+z}=0\)
\(\Leftrightarrow\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
vì x+y+z khác 0 => \(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{xy+yz+xz}{xyz}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{\left(xy+yz+xz\right).\left(x+y+z\right)-xyz}{xzy.\left(x+y+z\right)}=0\)
\(\Leftrightarrow\frac{x^2y+xy^2+xyz+zyx+y^2z+yz^2+x^2z+xyz+xz^2-xzy}{xyz.\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x^2y+xyz\right)+\left(xy^2+y^2z\right)+\left(yz^2+xzy\right)+\left(x^2z+xz^2\right)=0\)
\(\Leftrightarrow xy.\left(x+z\right)+y^2.\left(x+z\right)+yz.\left(z+x\right)+xz.\left(x+z\right)=0\)
\(\Leftrightarrow\left(x+z\right).\left(xy+y^2+yz+xz\right)=0\)
\(\Leftrightarrow\left(x+z\right).\left[x.\left(y+z\right)+y.\left(y+z\right)\right]=0\)
\(\Leftrightarrow\left(x+y\right).\left(y+z\right).\left(x+z\right)=0\Leftrightarrow\orbr{\begin{cases}x=-y\\y=-z\end{cases}\text{hoặc }x=-z}\)
\(\Rightarrow P=\left(\frac{1}{x}-\frac{1}{y}\right).\left(\frac{1}{y}+\frac{1}{z}\right).\left(\frac{1}{z}+\frac{1}{x}\right)=0\)
ps: bài này t làm cách l8, ai có cách ez hơn giải vs ak :') morongtammat
Do \(x^2+y^2+z^2=1\Rightarrow x^2< 1\Rightarrow x< 1\)
\(\Rightarrow x^5< x^2\)
Tương tự ta có: \(y< 1\Rightarrow y^6< y^2\); \(z< 1\Rightarrow z^7< z^2\)
\(\Rightarrow x^5+y^6+z^7< x^2+y^2+z^2\)
\(\Rightarrow x^5+y^6+z^7< 1\)