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Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=-\frac{1}{z^3}\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+3\cdot\frac{1}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{z^3}=0\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=-3\cdot\frac{1}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)=-3\cdot\frac{1}{xy}\cdot\left(-\frac{1}{z}\right)=\frac{3}{xyz}\)
Khi đó có : \(P=\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}=xyz.\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz\cdot\frac{3}{xyz}=3\)
GT \(\Leftrightarrow xy+yz+zx=0\). Khi đó: \(\left(xy\right)^3+\left(yz\right)^3+\left(zx\right)^3=3.xy.yz.zx=3x^2y^2z^2\).
Do đó: \(P=\frac{\left(xy\right)^3+\left(yz\right)^3+\left(zx\right)^3}{x^2y^2z^2}=3\)
\(\left(\frac{x}{z+y}+\frac{y}{z+x}+\frac{z}{x+y}\right)\left(x+y+z\right)=1\\ \)
Nhân phân phối ra
\(\left(\frac{x^2}{z+y}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\left(x+y\right).\frac{z}{x+y}+\left(x+z\right).\frac{y}{x+z}+\left(z+y\right).\frac{x}{z+y}=1\)
\(\left(\frac{x^2}{z+y}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)=0\)
Chờ các bạn lâu quá nên mình giải luôn: (x+y+z)\(\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)\)) = \(\frac{x^2}{y+z}+\frac{xy}{x+z}+\frac{xz}{x+y}+\frac{xy}{y+z}+\frac{y^2}{x+z}+\frac{yz}{x+y}+\frac{xz}{y+z}+\frac{yz}{x+z}+\frac{z^2}{x+y}=1\)
\(\left(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\right)+\left(x+y+z\right)=1\)
Do đó: \(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=0\)
a) \(A=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}+\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{2\left(y-z\right)\left(z-x\right)+2\left(x-y\right)\left(z-x\right)+2\left(x-y\right)\left(y-z\right)+\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left[\left(x-y\right)+\left(y-z\right)+\left(z-x\right)\right]^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y+y-z+z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)
Áp dụng: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
b)Ta có: \(\frac{x^2}{y+z}+x=\frac{x^2+x\left(y+z\right)}{y+z}=\frac{x^2+xy+xz}{y+z}=\frac{x\left(x+y+z\right)}{y+z}\)
Tương tự: \(\frac{y^2}{x+z}+y=\frac{y^2+xy+zy}{x+z}=\frac{y\left(x+y+z\right)}{x+z}\)
\(\frac{z^2}{x+y}+z=\frac{z^2+xz+zy}{x+y}=\frac{z\left(x+y+z\right)}{x+y}\)
Suy ra: \(A+\left(x+y+z\right)\)
\(=\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{z+x}+\frac{z\left(x+y+z\right)}{x+y}+\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}+1\right)\)
\(=2.\left(x+y+z\right)\)
Nên \(A=2.\left(x+y+z\right)-\left(x+y+z\right)=x+y+z\)
Mình có sai chỗ nào không nhỉ?
\(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1\)
\(\Rightarrow\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{x+z}+\frac{z\left(x+y+z\right)}{x+y}=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+x+\frac{y^2}{x+z}+y+\frac{z^2}{x+y}+z=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)
\(\Rightarrow M=2019+0=2019\)
\(\Rightarrow\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right).\left(x+y+z\right)=x+y+z\)
\(\Rightarrow\frac{x^2+x\left(z+x\right)}{y+z}+\frac{y^2+y\left(x+z\right)}{x+z}+\frac{z^2+z\left(x+y\right)}{x+y}=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+x+\frac{y^2}{x+z}+y+\frac{z^2}{x+y}+z=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=0\)
Bạn có thể sử dụng BĐT thức Cô-si và xét trường hợp dấu bằng xảy ra nhé bạn !
Câu hỏi của Trần Ngọc Tú - Toán lớp 8 - Học toán với OnlineMath
\(\Rightarrow\left(\frac{x}{x+y}+\frac{y}{z+x}+\frac{z}{x+y}\right)\cdot\left(x+y+z\right)=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{xy}{y+z}+\frac{xz}{y+z}+\frac{y^2}{z+x}+\frac{xy}{z+x}+\frac{yz}{z+x}+\frac{z^2}{x+y}+\frac{xz}{x+y}+\frac{yz}{x+y}=x+y+z\)
Rồi bạn cộng 2 phân thức 2,3 5,6 8,9 lại thì được
\(\Rightarrow\frac{x^2}{y+z}+x+\frac{y^2}{z+x}+y+\frac{z^2}{x+y}+z=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)
Lời giải:
Từ \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\)
\(\Rightarrow \left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)(x+y+z)=x+y+z\)
\(\Leftrightarrow \frac{x^2}{y+z}+\frac{x}{y+z}(y+z)+\frac{y^2}{z+x}+\frac{y}{z+x}(z+x)+\frac{z^2}{x+y}+\frac{z}{x+y}(x+y)=x+y+z\)
\(\Leftrightarrow \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}+(x+y+z)=x+y+z\)
\(\Leftrightarrow \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)
Vậy $M=0$