áp dụng tính chất của dãy TSBN, ta có:

\(\frac{x}{y+z-2}=\frac{y}{z+x-3}=\frac{z}{x+y+5}=x+y+z\frac{x+y+z}{\left(y+z-2\right)+\left(z+x-3\right)+\left(x+y+5\right)}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+z=\frac{1}{2}\\y+z-2=2x\\z+x-3=2y\\x+y+5=2z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y+z=\frac{1}{2}\\x+y+z-2=3x\\x+y+z-3=3y\\x+y+z+5=3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{2}-2=3x\\\frac{1}{2}-3=3y\\\frac{1}{2}+5=3z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\frac{-3}{2}=3x\\\frac{-5}{2}=3y\\\frac{11}{2}=3z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{-1}{2}\\y=\frac{-5}{6}\\z=\frac{11}{6}\end{matrix}\right.\)

Vậy \(x=\frac{-1}{2};y=\frac{-5}{6};z=\frac{11}{6}\)

ta có: \(\frac{a}{b}=\frac{a+c}{b+d}\) \(\Rightarrow a\left(b+d\right)=b\left(a+c\right)\)

\(\Leftrightarrow ab+ad=ba+bc\)

\(\Leftrightarrow ad=bc\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)

Vậy \(\frac{a}{b}=\frac{a+c}{b+d}\) \(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)

gọi x là đường thẳng đi qua B song song vs AC

vì x // AC \(\Rightarrow\widehat{BDA}=\widehat{DAC}\) ( 2 góc so le trong )

mà \(\widehat{DAC}=\widehat{BAD}\) ( AD là pg của \(\widehat{BAC}\))

\(\Rightarrow\widehat{BDA}=\widehat{BAD}\)

\(\Rightarrow\Delta BAD\) cân tại B

a) Xét \(\Delta AOCvà\Delta BOC,có:\)

AO = BO ( GT )

\(\widehat{AOC}=\widehat{BOC}\) ( OC là pg của \(\widehat{xOy}\))

OC: cạnh chung

\(\Rightarrow\Delta AOC=\Delta BOC\left(c.g.g\right)\)

\(\Rightarrow CA=CB\) ( 2 cạnh tương ứng )

b) Vì \(\Delta AOC=\Delta BOC\left(cmt\right)\)

\(\Rightarrow\widehat{ACO}=\widehat{BCO}\) ( 2 góc tương ứng )

mà \(\widehat{ACO}+\widehat{BCO}=180^O\)

\(\Rightarrow\widehat{ACO}=\widehat{BCO}=90^o\)

Vậy \(OC\perp AB\)

mk chỉ nghĩ j lm v thôi ko chắc đúng dou nha!!!

\(\frac{1}{c}+\frac{1}{a-b}=\frac{1}{a}-\frac{1}{b-c}\)

\(\Rightarrow\frac{a-b+c}{c\left(a-b\right)}=\frac{b-c-a}{a\left(b-c\right)}\)

Áp dụng tính chất của dãy TSBN, ta có:

\(\frac{a-b+c}{c\left(a-b\right)}=\frac{b-c-a}{a\left(b-c\right)}=\frac{\left(a-b+c\right)+\left(b-c-a\right)}{c\left(a-b\right)+a\left(b-c\right)}=0\)

\(\Rightarrow b-c-a=0\)

\(\Rightarrow b=a+c\)