Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lớp 7 gì mà dễ ẹc :))
\(\frac{2a-b}{a+b}=\frac{2}{3}\)
\(\Leftrightarrow6a-3b=2a+2b\)
\(\Rightarrow4a=5b\)
\(\frac{b-c+a}{2a-b}=\frac{2}{3}\)
\(\Leftrightarrow4a-2b=3b-3c+3a\)
\(\Leftrightarrow a=5b-3c\)
\(\Leftrightarrow a-5b=-3c\)
\(\Leftrightarrow a-4a=-3c\)
\(\Leftrightarrow-3a=-3c\)
\(\Rightarrow a=c\)
Ta có : \(P=\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2\left(a+3c\right)^3}=\frac{\left(4a+4a\right)^5}{\left(4a+4a\right)^2\left(a+3a\right)^3}=\frac{\left(8a\right)^3}{\left(4a\right)^3}=8\)
\(\frac{5a+5b-c}{c}=\frac{5b+5c-a}{a}=\frac{5c+5a-b}{b}\)
\(\Leftrightarrow\)\(\frac{5a+5b-c}{c}+1=\frac{5b+5c-a}{a}+1=\frac{5c+5a-b}{b}+1\)
\(\Leftrightarrow\)\(\frac{5a+5b}{c}=\frac{5b+5c}{a}=\frac{5c+5a}{b}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{5a+5b}{c}=\frac{5b+5c}{a}=\frac{5c+5a}{b}=\frac{5a+5b+5b+5c+5c+5a}{a+b+c}=\frac{10\left(a+b+c\right)}{a+b+c}=10\)
Do đó :
\(\frac{5a+5b}{c}=10\)\(\Leftrightarrow\)\(5a+5b=10c\)\(\Leftrightarrow\)\(a+b=2c\) \(\left(1\right)\)
\(\frac{5b+5c}{a}=10\)\(\Leftrightarrow\)\(5b+5c=10a\)\(\Leftrightarrow\)\(b+c=2a\) \(\left(2\right)\)
\(\frac{5c+5a}{b}=10\)\(\Leftrightarrow\)\(5c+5a=10b\)\(\Leftrightarrow\)\(c+a=2b\) \(\left(3\right)\)
Thay (1), (2) và (3) vào \(P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{16120abc}\) ta được :
\(P=\frac{2c.2a.2b}{16120abc}=\frac{8abc}{16120abc}=\frac{1}{2015}\)
Vậy \(P=\frac{1}{2015}\)
Chúc bạn học tốt ~
\(\frac{2a-b}{a+b}=\frac{b-c+a}{2a-b}=\frac{2}{3}\)
\(\Rightarrow\frac{2a-b}{a+b}=\frac{b-c+a}{2a-b}=\frac{\left(2a-b\right)+\left(b-c+a\right)}{\left(a+b\right)+\left(2a-b\right)}=\frac{3a-c}{3a}=\frac{2}{3}\)
\(\Rightarrow2\times3a=3\times\left(3a-c\right)\)
\(\Rightarrow6a=9a-3c\)
\(\Rightarrow6a-9a=-3c\)
\(\Rightarrow-3a=-3c\)
\(\Rightarrow\frac{-3a}{-3}=\frac{-3c}{-3}\)
\(\Rightarrow a=c\)
\(\Rightarrow\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2\left(a+3c\right)^3}=\frac{\left(5b+4a\right)^5}{\left(5b+4a\right)^2\left(a+3a\right)^3}=\frac{\left(5b+4a\right)^3}{\left(4a\right)^3}\)
\(\frac{2a-b}{a+b}=\frac{2}{3}\)
\(\Rightarrow3\times\left(2a-b\right)=2\left(a+b\right)\)
\(\Rightarrow6a-3b=2a+2b\)
\(\Rightarrow6a-2a=3b+2b\)
\(\Rightarrow4a=5b\)
\(\Rightarrow b=\frac{4a}{5}\)
\(\Rightarrow\frac{\left(5b+4a\right)^3}{\left(4a\right)^3}=\left(\frac{5\times\frac{4a}{5}+4a}{4a}\right)^3=\left(\frac{4a+4a}{4a}\right)^3\)
\(\Rightarrow\left(\frac{8a}{4a}\right)^3=2^3=8\)
a-5b=2(b-c)
<=>a=3b+2c
P=\(\frac{a-5c}{b-c}\) <=> \(\frac{3b+2c-5c}{b-c}\)
<=>\(\frac{3b-3c}{b-c}\) <=>\(\frac{3\left(b-c\right)}{b-3}\)
=>P=3
\(\frac{a-5b}{c-b}=2\Leftrightarrow a-5b=2c-2b\)
\(\Leftrightarrow a=2c+3b\)
\(\Rightarrow P=\frac{a-5c}{b-c}=\frac{2c+3b-5c}{b-c}=\frac{3b-3c}{b-c}=3\)
Vậy P = 3