x/(y+z)+y/(z+x)+z/(x+y)=1 chứng minh rằng x*x/(y+z)+y*y/(z+x)+z*z/(x+y)= 0
hộ mk với
0 bình luận về “x/(y+z)+y/(z+x)+z/(x+y)=1 chứng minh rằng x*x/(y+z)+y*y/(z+x)+z*z/(x+y)= 0
hộ mk với”
$\begin{array}{l}\underline{\text{Đáp án:}}\\x.\dfrac{x}{y+z}+y.\dfrac{y}{z+x}+z.\dfrac{z}{x+y}=0\\\underline{\text{Giải thích các bước giải:}}\\x.\dfrac{x}{y+z}+y.\dfrac{y}{z+x}+z.\dfrac{z}{x+y}\\=x.(\dfrac{x}{y+z}+1-1)+y.(\dfrac{y}{z+x}+1-1)+z.(\dfrac{z}{x+y}+1-1)\\=x.\dfrac{x+y+z}{y+z}-x+y.\dfrac{x+y+z}{z+x}-y+z.\dfrac{x+y+z}{x+y}-z\\=(x+y+z)(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y})-(x+y+z)\\=x+y+z-(x+y+z)\\=0(ĐPCM)\\\underline{\text{CHÚC BẠN HỌC TỐT}}\\\end{array}$
$\begin{array}{l}\underline{\text{Đáp án:}}\\x.\dfrac{x}{y+z}+y.\dfrac{y}{z+x}+z.\dfrac{z}{x+y}=0\\\underline{\text{Giải thích các bước giải:}}\\x.\dfrac{x}{y+z}+y.\dfrac{y}{z+x}+z.\dfrac{z}{x+y}\\=x.(\dfrac{x}{y+z}+1-1)+y.(\dfrac{y}{z+x}+1-1)+z.(\dfrac{z}{x+y}+1-1)\\=x.\dfrac{x+y+z}{y+z}-x+y.\dfrac{x+y+z}{z+x}-y+z.\dfrac{x+y+z}{x+y}-z\\=(x+y+z)(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y})-(x+y+z)\\=x+y+z-(x+y+z)\\=0(ĐPCM)\\\underline{\text{CHÚC BẠN HỌC TỐT}}\\\end{array}$
Giải thích các bước giải :
`x/(y+z)+y/(z+x)+z/(x+y)=1`
`<=>(x+y+z)(x/(y+z)+y/(z+x)+z/(x+y))=x+y+z`
`<=>(x(x+y+z))/(y+z)+(y(x+y+z))/(z+x)+(z(x+y+z))/(x+y)=x+y+z`
`<=>(x^2+x(y+z))/(y+z)+(y^2+y(z+x))/(z+x)+(z^2+z(x+y))/(x+y)=x+y+z`
`<=>x^2/(y+z)+(x(y+z))/(y+z)+y^2/(z+x)+(y(z+x))/(z+x)+z^2/(x+y)+(z(x+y))/(x+y)=x+y+z`
`<=>x^2/(y+z)+x+y^2/(z+x)+y+z^2/(x+y)+z=x+y+z`
`<=>x^2/(y+z)+y^2/(z+x)+z^2/(x+y)=x+y+z-x-y-z`
`<=>x^2/(y+z)+y^2/(z+x)+z^2/(x+y)=0`
Vậy ….
~Chúc bạn học tốt !!!~