\(\frac{a}{b}=\frac{9}{2}-\frac{1}{2}.\frac{4}{9}=\frac{77}{18}\)

\(\Rightarrow a=77,b=18\)

\(a-2b=77-2.18=41\)

a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)

Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)

b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)

\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)

\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)

\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)

\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)

Do đó:  +)  \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)

+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)

+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)

1. \(\frac{x}{y}=\frac{7}{17}\)

3. Có 6 cặp

4. 0 có cặp nào hết

Câu 2 mình không biết nha. Thông cảm

Dùng bđt AM - GM cho 7 số; 2 số và 3 số không âm, ta được:

\(a^3c^2+a^3c^2+a^3c^2+b^3a^2+b^3a^2+1+1\ge7a\)(1)

\(b^3a^2+b^3a^2+b^3a^2+c^3b^2+c^3b^2+1+1\ge7b\)(2)

\(c^3b^2+c^3b^2+c^3b^2+a^3c^2+a^3c^2+1+1\ge7c\)(3)

\(\frac{a+b+c}{2}+\frac{9}{2\left(a+b+c\right)}\ge3\)

\(a+b+c\ge3\)

Từ (1); (2); (3) suy ra \(a^3c^2+b^3a^2+c^3b^2\ge\frac{7\left(a+b+c\right)}{5}-\frac{6}{5}\)

\(P=\text{Σ}_{cyc}\frac{a}{b^2}+\frac{9}{2\left(a+b+c\right)}=\text{Σ}_{cyc}a^3c^2+\frac{9}{2\left(a+b+c\right)}\)

\(\ge\frac{7\left(a+b+c\right)}{5}+\frac{9}{2\left(a+b+c\right)}-\frac{6}{5}\)

\(=\frac{a+b+c}{2}+\frac{9}{2\left(a+b+c\right)}+\frac{9\left(a+b+c\right)}{10}-\frac{6}{5}\)

\(\ge3+\frac{9}{10}.3-\frac{6}{5}=\frac{9}{2}\)

Đẳng thức xảy ra khi a = b = c = 1

Ta có: P= \(2a+3b+\dfrac{1}{a}+\dfrac{4}{b}\) = \(\text{​​}\text{​​}(\dfrac{1}{a}+a)+\left(\dfrac{4}{b}+b\right)+\left(a+2b\right)\)

Ta thấy: \(\text{​​}\text{​​}(\dfrac{1}{a}+a)\ge2\sqrt{\dfrac{1}{a}\cdot a}=2\)

             \(\text{​​}\text{​​}\left(\dfrac{4}{b}+b\right)\ge2\sqrt{\dfrac{4}{b}\cdot b}=4\)

Do đó: P \(\ge2+4+8=14\)

Vậy: P(min)=14  khi:  \(\left\{{}\begin{matrix}\dfrac{1}{a}=a\\\dfrac{4}{b}=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right..\)

sorry, nhầm đề

làm lại:

Ta có: P= \(2a+3b+\dfrac{4}{a}+\dfrac{9}{b}\) = \(\text{​​}\text{​​}(\dfrac{4}{a}+a)+\left(\dfrac{9}{b}+b\right)+\left(a+2b\right)\)

Ta thấy: \(\text{​​}\text{​​}(\dfrac{4}{a}+a)\ge2\sqrt{\dfrac{4}{a}\cdot a}=4\)

             \(\text{​​}\text{​​}\left(\dfrac{9}{b}+b\right)\ge2\sqrt{\dfrac{9}{b}\cdot b}=6\)

Do đó: P \(\ge4+6+8=18\)

Vậy: P(min)=18  khi:  \(\left\{{}\begin{matrix}\dfrac{4}{a}=a\\\dfrac{9}{b}=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\cdot\)

Bài làm

\(P=2a+3b+\frac{4}{a}+\frac{9}{b}=a+a+2b+b+\frac{4}{a}+\frac{9}{b}\)

\(=\left(a+2b\right)+\left(a+\frac{4}{a}\right)+\left(b+\frac{9}{b}\right)\)

\(\ge8+2\sqrt{a\times\frac{4}{a}}+2\sqrt{b\times\frac{9}{b}}\)( Cauchy )

\(=8+4+6=18\)

Đẳng thức xảy ra khi a = 2 ; b = 3

=> MinP = 18 <=> a = 2 ; b = 3

\(P=2a+3b+\frac{4}{a}+\frac{9}{b}\)

\(\Leftrightarrow P=\left(a+\frac{4}{a}\right)+\left(b+\frac{9}{b}\right)+a+2b\)

Áp dụng BĐT AM-GM ta có:

\(P\ge2.\sqrt{a.\frac{4}{a}}+2.\sqrt{b.\frac{9}{b}}+a+2b=2.2+2.3+a+2b\ge4+6+8=18\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a=\frac{4}{a}\\b=\frac{9}{b}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}\)

Vậy \(P_{min}=18\)\(\Leftrightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}\)

\(P=a+2b+a+\frac{4}{a}+b+\frac{9}{b}\)

Áp dụng cô si ta có:

\(a+\frac{4}{a}\ge4\)

\(b+\frac{9}{b}\ge6\)

\(\Rightarrow P\ge4+6+8\Rightarrow P\ge18\)

dấu "=" xảy ra khi a=2,b=3

Vì \(\frac{1}{4}=\frac{1x4}{5x4}=\frac{4}{20}\)và \(\frac{2}{5}=\frac{2x4}{5x4}=\frac{8}{20}\)

Vì 4 < 5,6,7 < 8

=> Vậy phân số đó là : \(\frac{5}{20},\frac{6}{20},\frac{7}{20}\)

Nhưng vì phân số đó phải tối giản nên phân số cần tìm là : \(\frac{7}{20}\)

\(\frac{1}{4}< \frac{a}{b}< \frac{2}{5}\)

\(\Leftrightarrow\frac{5}{20}< \frac{a}{b}< \frac{8}{20}\)

\(\Rightarrow\frac{a}{b}=\frac{6}{20};\frac{7}{20}\)

\(\Rightarrow\frac{a}{b}=\frac{3}{10};\frac{7}{20}\)