1 + 1 + 1+ 1 + 1 = 1x 5 = 5

a)\(\frac{x+2}{-5}\ge0\Leftrightarrow x+2\ge0\Leftrightarrow x\ge-2\)

b)\(\frac{x-1}{x-3}>1\Leftrightarrow\frac{x-1}{x-3}-1>0\Leftrightarrow\frac{2}{x-3}>0\Leftrightarrow x=\frac{2}{0}+3\)=> vô nghiệm 

ĐKXD: X khác 3

125874963:45698=2754,496105

Cho \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)

Tính \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)

Giải:

TH1: a + b + c = 0

\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)

\(\Rightarrow M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)

\(=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}\)

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

TH2: a + b + c ≠ 0

áp dụng tính chất của dãy TSBN ta có:

\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}b+c=2a\\c+a=2b\\a+b=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c=3a\\a+b+c=3b\\a+b+c=3c\end{matrix}\right.\)

\(\Rightarrow3a=3b=3c\)

\(\Rightarrow a=b=c\)

\(\Rightarrow M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)

\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)

\(=2.2.2=8\)

vì On là tia phân giác của \(\widehat{AOD}\Rightarrow\widehat{nOD}=\frac{1}{2}\widehat{AOD}\)

vì Op là tia phân giác của \(\widehat{DOB}\Rightarrow\widehat{DOp}=\frac{1}{2}\widehat{DOB}\)

ta có: \(\widehat{AOD}+\widehat{DOB}=180^o\) (2 góc kề bù)

\(\Rightarrow\widehat{nOD}+\widehat{DOp}=\frac{1}{2}\widehat{AOD}+\frac{1}{2}\widehat{DOB}\)

\(=\frac{1}{2}\left(\widehat{AOD}+\widehat{DOB}\right)\)

\(=\frac{1}{2}.180^o=90^o\)

hay \(\widehat{nOp}=90^o\)

\(\Rightarrow On\perp Op\)

\(6\times4=24\)

Nhanh nhứt nà