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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 : \(P=\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2\left(a+3c\right)^3}=\frac{\left(4a+4a\right)^5}{\left(4a+4a\right)^2\left(a+3a\right)^3}=\frac{\left(8a\right)^3}{\left(4a\right)^3}=8\)
\(\frac{2a-b}{a+b}=\frac{2}{3}\)
\(\Leftrightarrow6a-3b=2a+2b\)
\(\Leftrightarrow6a-2a=2b+3b\)
\(\Leftrightarrow4a=5b\)
\(\frac{b-c+a}{2a-b}=\frac{2}{3}\)
\(\Leftrightarrow4a-2b=3b-3c+3a\)
\(\Leftrightarrow4a-3a=3b-3c+2b\)
\(\Leftrightarrow a=5b-3c\)
\(\Leftrightarrow a=4a-3c\)
\(\Leftrightarrow3a=3c\)
\(\Rightarrow a=c\)
\(\Rightarrow P=\frac{\left(4a+4a\right)^5}{\left(4a+4a\right)^2\left(a+3a\right)^3}=\frac{\left(8a\right)^5}{\left(8a\right)^2\left(4a\right)^3}=\frac{\left(8a\right)^3}{\left(4a\right)^3}=\frac{8^3}{4^3}=2^3=8\)
link đây tham khảo nhé:
https://hoc24.vn/hoi-dap/question/207558.html
a = \(\frac{4a-5b}{6a+b}\) (1) và \(\frac{a}{b}=\frac23\)
Giải:
\(\frac{a}{b}=\frac23\) suy ra: a = \(\frac{2b}{3}\)
Thay a = 2b/3 vào (1) ta có:
2b/3 = \(\frac{\frac{4.2b}{3}-5b}{\frac{6.2b}{3}+b}\) = \(\frac{b.\left(\frac83-5\right)}{b.\left(4+1\right)}\) = (\(\frac83-\frac{15}{3}):5\) = - \(\frac73\times\frac15\) = - 7/15
b = - 7/15 : (2/3) = - \(\frac{7}{10}\)
a = 2b/3 = (-7/10) x (2/3) = -7/15
Vậy a = - 7/15; b = - 7/10
ba=32 suy ra: a = \(\frac{2 b}{3}\)
Thay a = 2b/3 vào (1) ta có:
2b/3 = \(\frac{\frac{4.2 b}{3} - 5 b}{\frac{6.2 b}{3} + b}\) = \(\frac{b . \left(\right. \frac{8}{3} - 5 \left.\right)}{b . \left(\right. 4 + 1 \left.\right)}\) = (\(\frac{8}{3} - \frac{15}{3} \left.\right) : 5\) = - \(\frac{7}{3} \times \frac{1}{5}\) = - 7/15
b = - 7/15 : (2/3) = - \(\frac{7}{10}\)
a = 2b/3 = (-7/10) x (2/3) = -7/15
Vậy a = - 7/15; b = - 7/10