Câu 3:
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{-2}=\dfrac{x-y}{3-\left(-2\right)}=\dfrac{10}{5}=2\)
Do đó:x=6; y=-4
b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}=\dfrac{c-a}{5-2}=\dfrac{6}{3}=2\)
Do đó: a=4; b=6; c=10
Câu 5:
TH1: a+b+c=0
\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)
\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(-a\right)\left(-b\right)\left(-c\right)}{abc}=\dfrac{-\left(abc\right)}{abc}=-1\)
TH1: a+b+c≠0
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}=\dfrac{a+b-c+a-b+c+\left(-a\right)+b+c}{c+b+a}=\dfrac{a+b+c}{a+b+c}=1\)
\(\dfrac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\\ \dfrac{a-b+c}{b}=1\Rightarrow a-b+c=b\Rightarrow a+c=2b\\ \dfrac{-a+b+c}{a}=1\Rightarrow-a+b+c=a\Rightarrow b+c=2a\)
\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{2a.2b.2c}{abc}=\dfrac{8abc}{abc}=8\)
Câu 5:
\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}\)
=>\(\dfrac{a+b}{c}-1=\dfrac{a+c}{b}-1=\dfrac{b+c}{a}-1\)=>\(\dfrac{a+b}{c}=\dfrac{a+c}{b}=\dfrac{b+c}{a}\)
=>\(\dfrac{a+b}{c}=\dfrac{a+c}{b}=\dfrac{b+c}{a}=\dfrac{a+b+a+c+b+c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
=>\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=2.2.2=8\)
Câu 4:
a, xy//uv, xy//zl\(\Rightarrow\) uv//zl
b, Ta có xy//uv \(\Rightarrow\widehat{xAO}=\widehat{AOv}=30^o\) (2 góc so le trong)
uv//zl\(\Rightarrow\widehat{zBO}=\widehat{BOv}=45^o\) (2 góc so le trong)
Ta có: \(\widehat{AOB}=\widehat{AOv}+\widehat{BOv}=30^o+45^o=75^o\)
