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ta có {nhân phân phối ra dẽ hơn} là ghép nhân tử
\(\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\left(x+y+z\right)=\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}....\right)+\left(x+y+z\right)\)
Chia hai vế cho (x+y+z khác 0) chú ý => dpcm
quái lại câu 1 đâu
(a+b+c)=abc tất nhiên theo đầu đk a,b,c khác không
chia hai vế cho abc/2
2/bc+2/ac+2/ab=2 (*)
đăt: 1/a=x; 1/b=y; 1/c=z
ta có
x+y+z=k (**)
x^2+y^2+z^2=k(***)
lấy (*)+(***),<=>(x+y+z)^2=2+k
=> k^2=2+k
=> k^2-k=2
k^2-k+1/4=1/4+2=9/4
\(\orbr{\begin{cases}k=\frac{1}{2}+\frac{3}{2}=\frac{5}{2}\\k=\frac{1}{2}-\frac{3}{2}=-\frac{1}{2}\end{cases}}\)
Mình chưa test lại đâu bạn tự test nhé
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)=> (x+y+z)\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)=0
=> \(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}+3=0\)
=> \(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}=-3\)
\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}=\frac{x+y+z}{z}-1+\frac{x+y+z}{y}-1+\frac{x+y+z}{x}-1\)
\(=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3=0-3=-3\)
\(x^3+y^3+z^3=3xyz\)
\(x^3+y^3+z^3-3xyz=0\)
\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)
\(x^2+y^2+z^2-xy-xz-yz=0\left(x+y+z\ne0\right)\)
\(2\times\left(x^2+y^2+z^2-xy-xz-yz\right)=0\times2\)
\(2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)
\(x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)
\(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)
\(\left[\begin{array}{nghiempt}x-y=0\\x-z=0\\y-z=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=y\\x=z\\y=z\end{array}\right.\)
x = y = z
\(P=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{x}{z}\right)\)
\(=\left(1+\frac{x}{x}\right)\left(1+\frac{y}{y}\right)\left(1+\frac{z}{z}\right)\)
\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)
\(=2^3\)
\(=8\)
trả lời:
ta có:\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\end{cases}}\)
\(Q=\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\)
\(=\frac{x}{z}+\frac{y}{z}+\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}\)
\(=\left(\frac{x}{z}+\frac{x}{y}\right)+\left(\frac{y}{z}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{z}{y}\right)\)
\(=x\left(\frac{1}{z}+\frac{1}{y}\right)+y\left(\frac{1}{z}+\frac{1}{x}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=x\left(-\frac{1}{x}\right)+y\left(-\frac{1}{y}\right)+z\left(-\frac{1}{z}\right)\)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(=-3\)
~hok tốt~
Cách ngắn hơn ạ: \(Q=\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\)
\(=\frac{x+y}{z}+1+\frac{y+z}{x}+1+\frac{z+x}{y}+1-3\)
\(=\frac{x+y+z}{z}+\frac{x+y+z}{x}+\frac{x+y+z}{y}-3\)
\(=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3\)
\(=-3\)