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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 : \(y^2+2xz=y^2+xz+xz\)

                                    \(=y^2+xz-xy-yz\)

                                    \(=y\left(y-x\right)+z\left(x-y\right)\)

                                    \(=\left(x-y\right)\left(z-y\right)\)

                  \(z^2+2xy=\left(x-z\right)\left(y-z\right)\)

\(\Rightarrow M=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(x-y\right)\left(z-y\right)}+\frac{xy}{\left(x-z\right)\left(y-z\right)}\)  \(M=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)

\(M=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{yz\left(y-z\right)-xz\left(x-y+y-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

\(A=\frac{\left(yz-xz\right)\left(y-z\right)+\left(xy-xz\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)

Tính thế làm gì bạn ê

\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\\ \Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\\ \Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\\ \Rightarrow x=y=z\\ \Rightarrow A=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)

đề bài thiếu hả bạn

ơ quên thêm đề bài là: thì trong ba số đó đối nhau

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{\left(x+y+z\right)\cdot\left(xy+yz+zx\right)}{xyz\left(x+y+z\right)}=\frac{xyz}{xyz\left(x+y+z\right)}\)

\(\Leftrightarrow\left(x+y+z\right)\cdot\left(xy+yz+zx\right)=xyz\)

\(\Leftrightarrow x^2y+xy^2+yz^2+y^2z+z^2x+zx^2+3xyz=xyz\)

\(\Leftrightarrow x^2y+xy^2+y^2z+yz^2+z^2x+zx^2+2xyz=0\)

\(\Leftrightarrow\left(x^2y+xy^2\right)+\left(y^2z+xyz\right)+\left(yz^2+z^2x\right)+\left(zx^2+xyz\right)=0\)

\(\Leftrightarrow xy\left(x+y\right)+yz\left(x+y\right)+zx\left(x+y\right)+z^2\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)\left(xy+yz+zx+z^2\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+y=0\\\orbr{\begin{cases}y+z=0\\z+x=0\end{cases}}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-y\\\orbr{\begin{cases}y=-z\\z=-x\end{cases}}\end{cases}}}\)

Vậy tồn tại trong 3 số x, y, z có 2 số đối nhau

Ta có: \(\left(\sqrt{x+y}\right)^2=\left(\sqrt{x-z}+\sqrt{y-z}\right)^2\)

\(\Leftrightarrow\)\(x+y=x+y-2z+2\sqrt{\left(x-z\right)\left(y-z\right)}\)

\(\Leftrightarrow2z=2\sqrt{\left(x-z\right)\left(y-z\right)}\)

Theo giả thiết, ta có: 

theo giả thiết, ta có: \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0\Rightarrow\frac{1}{z}-\frac{1}{x}=\frac{1}{y}\)\(\Rightarrow\frac{x-z}{zx}=\frac{1}{y}\Rightarrow x-z=\frac{zx}{y}\)

Tương tự, ta có: \(y-z=\frac{zy}{x}\)

Do đó: \(2\sqrt{\left(x-z\right)\left(y-z\right)}=2\sqrt{\frac{zx}{y}.\frac{zy}{x}}=2z\) (1)

ta có: \(\left(\sqrt{x+y}\right)^2=\left(\sqrt{x-z}+\sqrt{y-z}\right)^2\)

\(\Leftrightarrow2z=2\sqrt{\left(x-z\right)\left(y-z\right)}\)(2)

Thay (2) vào (1) ta thấy (2) luôn đúng

Suy ra ĐPCM

Vì \(x>0,y>0\Rightarrow\frac{1}{x}>0;\frac{1}{y}>0\)

mà \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0\Rightarrow\frac{1}{z}=\frac{1}{x}+\frac{1}{y}\Rightarrow\frac{1}{z}>0\Rightarrow z>0\)

Ta có: \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0\Leftrightarrow yz+zx-xy=0\)

\(\Leftrightarrow-z^2=-z^2+yz+zx-xy=-\left(x-z\right)\left(y-z\right)\)

\(\Leftrightarrow z^2=\left(x-z\right)\left(y-z\right)>0\)

\(\Rightarrow z=\sqrt{\left(x-z\right)\left(y-z\right)}\left(z>0\right)\)

Lại có: \(x+y=x-z+y-z+2z\)

\(=\left(x-z\right)+\left(y-z\right)+2\sqrt{\left(x-z\right)\left(y-z\right)}=\left(\sqrt{x-z}+\sqrt{y-z}\right)^2\)

Suy ra \(\sqrt{x+y}=\sqrt{x-z}+\sqrt{y-z}\) (ĐPCM)