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Hướng dẫn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
Thay vào:\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-zx=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)
Tương tự thay vào mà quy đồng
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Rightarrow xy+yz+xz=0\)
\(\Rightarrow\left\{{}\begin{matrix}xy=-yz--xz\\yz=-xy-xz\\xz=-xy-xz\end{matrix}\right.\)
\(\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
CMTT:
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\\\dfrac{yz}{x^2+2yz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)
A=\(\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}+\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
\(A=\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}\left(1\right)\)
mà \(xy+yz+xz=0\)
Từ \(\Rightarrow\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}=0\)
Vậy A=0
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)nhân lần lượt với x; y; z, ta có:
\(1+\frac{x}{y}+\frac{x}{z}=0\)(1)
\(1+\frac{y}{z}+\frac{y}{x}=0\)(2)
\(1+\frac{z}{x}+\frac{z}{y}=0\)(3)
Từ: (1); (2) và (3) => \(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x}{z}+\frac{y}{x}+\frac{z}{y}=-3\)(*)
Mặt khác: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)quy đồng ta có:
\(\frac{\left(xy+yz+zx\right)}{xyz}=0\)hay xy + yz + zx = 0
Hay: \(\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right).\left(xy+yz+zx\right)=0\)
Khai triển, ta có:
\(\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}+\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{z}{x}+\frac{y}{x}+\frac{z}{y}=0\)
Vậy: \(\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}=-\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right)=3\)
Bạn tham khảo tại đây:
Câu hỏi của trieu dang - Toán lớp 8 - Học toán với OnlineMath
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{\left(yz+xz+xy\right)}{xyz}=0\)
\(\Rightarrow yz+zx+xy=0\)
Ta có : \(x^2+2yz=x^2+yz+yz\)
\(=x^2+yz-zx-xy\)
\(=x\left(x-z\right)-y\left(x-z\right)\)
\(=\left(x-y\right)\left(x-z\right)\)
Tương tự : \(y^2+2xz=y^2+xz+xz\)
\(=y^2+xz-xy-yz\)
\(=y\left(y-x\right)+z\left(x-y\right)\)
\(=\left(x-y\right)\left(z-y\right)\)
\(z^2+2xy=\left(x-z\right)\left(y-z\right)\)
\(\Rightarrow M=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(x-y\right)\left(z-y\right)}+\frac{xy}{\left(x-z\right)\left(y-z\right)}\) \(M=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)
\(M=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{yz\left(y-z\right)-xz\left(x-y+y-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(A=\frac{\left(yz-xz\right)\left(y-z\right)+\left(xy-xz\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
1/x + 1/y +1/z = 0
<=> xy+yz+zx = 0
<=> yz=-xy-zx
<=> yz/x^2+2yz = yz/x^2+yz-xy-zx = yz/(x-y).(x-z)
Tương tự : xz/y^2+2xz = xz/(y-x).(y-z) ; xy/z^2+2xy = xy/(z-x).(z-y)
=> A = yz/(x-y).(x-z) + xz/(y-x).(y-z) + xy/(z-x).(z-y)
= -yz.(y-z)-zx.(z-x)-xy.(x-y)/(x-y).(y-z).(z-x)
= z^2y-y^2z+x^2z-xz^2+y^2x-x^2y/(x-y).(y-z).(z-x)
= (x-y).(y-z).(z-x)/(x-y).(y-z).(z-x)
= 1
Tk mk nha
https://olm.vn/hoi-dap/question/255332.html
Bạn tham khảo ở đây nhé!! Cách của mình cũng giống của bạn này
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\Rightarrow\hept{\begin{cases}xy=-yz-xz\\yz=-xy-xz\\xz=-yz-xy\end{cases}}\)
\(x^2+yz+yz=x^2-xy-xz+yz=x.\left(x-y\right)-z.\left(x-y\right)=\left(x-y\right).\left(x-z\right)\)
tương tự bn phân tích rồi quy đồng về mẫu chung :))
Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0< =>\frac{xy}{xyz}+\frac{yz}{xyz}+\frac{zx}{xyz}=0< =>xy+yz+zx=0\)
Khi đó : \(x^2+2yz=x^2+2yz-xy-yz-zx=x^2-xy+yz-zx=\left(x-z\right)\left(x-y\right)\)
Bằng phép chứng minh tương tự ta được : \(y^2+2xz=\left(y-x\right)\left(y-z\right);z^2+2xy=\left(z-x\right)\left(z-y\right)\)
Đặt \(A=\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}=\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(< =>-A=\frac{x^2}{\left(x-y\right)\left(z-x\right)}+\frac{y^2}{\left(x-y\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(y-z\right)}\)
\(=\frac{x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=...\)đến đây nhân tung rồi ghép cặp sẽ ra kq = 1 thì phải
làm luôn đỡ lòng vòng :(
\(=\frac{x^2\left(y-z\right)+y^2z-y^2x+z^2x-z^2y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{x^2\left(y-z\right)+zy\left(y-z\right)-x\left(y^2-z^2\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{x^2\left(y-z\right)+zy\left(y-z\right)-x\left(y-z\right)\left(y+z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(y-z\right)\left(x^2+zy-xy-xz\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left(y-z\right)\left[x\left(x-y\right)-z\left(x-y\right)\right]}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(y-z\right)\left(x-y\right)\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=-1\)
\(< =>-A=-1< =>A=1\)