Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) .Cmr:
\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\) \(\dfrac{a-c}{a+c}=\dfrac{b-d}{b+d}\)
\(\dfrac{a}{a+c}=\dfrac{b}{b+d}\) \(\dfrac{a}{a-c}=\dfrac{b}{b-d}\)
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\(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)
\(\dfrac{a}{c}\) = \(\dfrac{b}{d}\)
\(\dfrac{a}{c}\) = \(\dfrac{5a}{5c}\) = \(\dfrac{3b}{3d}\) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}\) = \(\dfrac{5a+3b}{5c+3d}\) (1)
\(\dfrac{a}{c}\) = \(\dfrac{5a-3b}{5c-3d}\) (2)
Kết hợp (1) và (2) ta có:
\(\dfrac{5a+3b}{5c+3d}\) = \(\dfrac{5a-3b}{5c-3d}\)
⇒ \(\dfrac{5a+3b}{5a-3b}\) = \(\dfrac{5c+3d}{5c-3d}\) (đpcm)
b; \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)
\(\dfrac{a}{b}\) = \(\dfrac{3a}{3b}\) = \(\dfrac{2c}{2d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}\) = \(\dfrac{3a+2c}{3b+2d}\) (đpcm)
\(\dfrac{a}{a+b+c}>\dfrac{a}{a+b+c+d}\)
Làm tương tự với 3 phân số còn lại và cộng vế với vế
\(\dfrac{a}{a+b+c}< \dfrac{a+d}{a+b+c+d}\)
Làm tương tự với 3 phân số còn lại và cộng vế với vế
\(\dfrac{a}{b}< \dfrac{a+c}{b+c}\)
\(\Leftrightarrow a\left(b+c\right)< b\left(a+c\right)\)
\(\Leftrightarrow ab+ac< ba+bc\)
\(\Leftrightarrow ac< bc\)
\(\Leftrightarrow a< b\)(đúng)
a)Áp dụng
\(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}=2\left(1\right)\)
Lại có:\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a}{a+b+c}+\dfrac{b}{b+c+a}+\dfrac{c}{c+a+b}=1\left(2\right)\)
Từ (1) và (2)=> đpcm
Vì \(\dfrac{a}{b}< 1\Rightarrow a< b\Rightarrow ac< bc\Rightarrow ac+ab< bc+ab\Rightarrow a\left(b+c\right)< b\left(a+c\right)\Rightarrow\dfrac{a\left(b+c\right)}{b\left(b+c\right)}< \dfrac{b\left(a+c\right)}{b\left(b+c\right)}\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+c}\)a) ta có
\(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}\)\(\Leftrightarrow\dfrac{a+b+c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{2\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\\ \Rightarrow\dfrac{b+c+d}{a}=\dfrac{a+c+d}{b}=\dfrac{a+b+d}{c}=\dfrac{a+b+c}{d}=\dfrac{3\left(a+b+c+d\right)}{a+b+c+d}=3\\ \Rightarrow\left\{{}\begin{matrix}b+d+c=3a\\a+c+d=3b\\a+b+d=3c\\a+b+c=3d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c+d=4a\\a+b+c+d=4b\\a+b+c+d=4c\\a+b+c+d=4d\end{matrix}\right.\\ \Rightarrow4a=4b=4c=4d\Rightarrow a=b=c=d\\ \Rightarrow P=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\\ \Rightarrow\left\{{}\begin{matrix}b+c+d=3a\\a+c+d=3b\\a+b+d=3c\\a+b+c=3d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c+d=2a\\a+b+c+d=2b\\a+b+c+d=2c\\a+b+c+d=2d\end{matrix}\right.\\ \Rightarrow2a=2b=2c=2d\\ \Rightarrow a=b=c=d\\ \Rightarrow A=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)
Có \(\dfrac{a}{b}=\dfrac{c}{d}< =>ad=bc\)
Xét \(\dfrac{a+c}{b+d}-\dfrac{a-c}{b-d}\)
= \(\dfrac{\left(a+c\right)\left(b-d\right)-\left(b+d\right)\left(a-c\right)}{\left(b+d\right)\left(b-d\right)}\)
= \(\dfrac{ab-ad+bc-cd-ab+bc-da+cd}{\left(b+d\right)\left(b-d\right)}\)
= 0
<=> \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k\)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)
Do đó: \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
b: \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
nên \(\dfrac{a+c}{a-c}=\dfrac{b+d}{b-d}\)
c: \(\dfrac{a}{a+c}=\dfrac{bk}{bk+dk}=\dfrac{b}{b+d}\)