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 : \(P=\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2\left(a+3c\right)^3}=\frac{\left(4a+4a\right)^5}{\left(4a+4a\right)^2\left(a+3a\right)^3}=\frac{\left(8a\right)^3}{\left(4a\right)^3}=8\)
\(\frac{2a-b}{a+b}=\frac{2}{3}\)
\(\Leftrightarrow6a-3b=2a+2b\)
\(\Leftrightarrow6a-2a=2b+3b\)
\(\Leftrightarrow4a=5b\)
\(\frac{b-c+a}{2a-b}=\frac{2}{3}\)
\(\Leftrightarrow4a-2b=3b-3c+3a\)
\(\Leftrightarrow4a-3a=3b-3c+2b\)
\(\Leftrightarrow a=5b-3c\)
\(\Leftrightarrow a=4a-3c\)
\(\Leftrightarrow3a=3c\)
\(\Rightarrow a=c\)
\(\Rightarrow P=\frac{\left(4a+4a\right)^5}{\left(4a+4a\right)^2\left(a+3a\right)^3}=\frac{\left(8a\right)^5}{\left(8a\right)^2\left(4a\right)^3}=\frac{\left(8a\right)^3}{\left(4a\right)^3}=\frac{8^3}{4^3}=2^3=8\)
\(\frac{2a-b}{a+b}=\frac{b-c+a}{2a-b}=\frac{2}{3}\)
\(\Rightarrow\frac{2a-b}{a+b}=\frac{b-c+a}{2a-b}=\frac{\left(2a-b\right)+\left(b-c+a\right)}{\left(a+b\right)+\left(2a-b\right)}=\frac{3a-c}{3a}=\frac{2}{3}\)
\(\Rightarrow2\times3a=3\times\left(3a-c\right)\)
\(\Rightarrow6a=9a-3c\)
\(\Rightarrow6a-9a=-3c\)
\(\Rightarrow-3a=-3c\)
\(\Rightarrow\frac{-3a}{-3}=\frac{-3c}{-3}\)
\(\Rightarrow a=c\)
\(\Rightarrow\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2\left(a+3c\right)^3}=\frac{\left(5b+4a\right)^5}{\left(5b+4a\right)^2\left(a+3a\right)^3}=\frac{\left(5b+4a\right)^3}{\left(4a\right)^3}\)
\(\frac{2a-b}{a+b}=\frac{2}{3}\)
\(\Rightarrow3\times\left(2a-b\right)=2\left(a+b\right)\)
\(\Rightarrow6a-3b=2a+2b\)
\(\Rightarrow6a-2a=3b+2b\)
\(\Rightarrow4a=5b\)
\(\Rightarrow b=\frac{4a}{5}\)
\(\Rightarrow\frac{\left(5b+4a\right)^3}{\left(4a\right)^3}=\left(\frac{5\times\frac{4a}{5}+4a}{4a}\right)^3=\left(\frac{4a+4a}{4a}\right)^3\)
\(\Rightarrow\left(\frac{8a}{4a}\right)^3=2^3=8\)
Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:
\(\frac{2a+b+c}{a}=\frac{2b+c+a}{b}=\frac{2c+a+b}{c}=\frac{2a+b+c+2b+c+a+2c+a+b}{a+b+c}=\frac{4\left(a+b+c\right)}{a+b+c}=4\)
\(\Rightarrow\frac{2a+b+c}{a}=4\Rightarrow2a+b+c=4a\Rightarrow b+c=4a-2a=2a\)
\(\frac{2b+c+a}{b}=4\Rightarrow2b+c+a=4b\Rightarrow c+a=4b-2b=2b\)
\(\frac{2c+a+b}{c}=4\Rightarrow2c+a+b=4c\Rightarrow a+b=4c-2c=2c\)
Suy ra \(P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2c.2a.2b}{abc}=\frac{8abc}{abc}=8\)
Vậy P=8
Cho hỏi tớ sai chỗ nào ạ :>?Góp ý giúp nha?