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

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

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

Ta CM BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

Ta có: \(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}},a+b\ge2\sqrt{ab}\)( co si với a,b>0)

Suy ra \(\left(\frac{1}{a}+\frac{1}{b}\right)\left(a+b\right)\ge4\RightarrowĐPCM\)\(\Rightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\left(1\right)\)

a/Áp dụng (1) có

\(\frac{1}{a+b+2c}\le\frac{1}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\left(2\right)\).Tương tự ta cũng có:

\(\frac{1}{b+c+2a}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\left(3\right),\frac{1}{c+a+2b}\le\frac{1}{4}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)\left(4\right)\)

Cộng (2),(3) và (4) có \(VT\le\frac{1}{4}.\left(6+6\right)=3\left(ĐPCM\right)\)

b/Áp dụng (1) có:

\(\frac{1}{3a+3b+2c}=\frac{1}{\left(a+b+2c\right)+2\left(a+b\right)}\le\frac{1}{4}\left(\frac{1}{a+b+2c}+\frac{1}{2\left(a+b\right)}\right)\left(5\right)\)

Tương tự có: \(\frac{1}{3a+2b+3c}\le\frac{1}{4}\left(\frac{1}{a+c+2b}+\frac{1}{2\left(a+c\right)}\right)\left(6\right)\)

\(\frac{1}{2a+3b+3c}\le\frac{1}{4}\left(\frac{1}{2a+b+c}+\frac{1}{2\left(b+c\right)}\right)\left(7\right)\)

Cộng (5),(6) và (7) có:

\(VT\le\frac{1}{4}\left(\frac{1}{a+b+2c}+\frac{1}{a+c+2b}+\frac{1}{2a+b+c}+\frac{1}{2}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)\right)\le\frac{1}{4}.9=\frac{3}{2}\)

Để câu trả lời của bạn nhanh chóng được duyệt và hiển thị, hãy gửi câu trả lời đầy đủ và nên:

Để câu trả lời của bạn nhanh chóng được duyệt và hiển thị, hãy gửi câu trả lời đầy đủ và nên:

Ta có: BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)( CM bằng BĐT Shwars nha).Áp dụng ta có:

\(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5a}+\frac{1}{3a+2b+4c}\ge\frac{9}{9a+6b+12c}=\frac{3}{3a+2b+4c}\left(1\right)\)

\(\frac{1}{b+3c+5a}+\frac{1}{c+3a+5b}+\frac{1}{3b+2c+4a}\ge\frac{9}{9b+6c+12a}=\frac{3}{3b+2c+4a}\left(2\right)\)

\(\frac{1}{c+3a+5b}+\frac{1}{a+3b+5c}+\frac{1}{3c+2a+4b}\ge\frac{9}{9c+6a+12b}=\frac{3}{3c+2a+4b}\left(3\right)\)

Cộng (1),(2) và (3) có:

\(2\left(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5c}+\frac{1}{c+3a+5b}\right)+\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\ge3\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\)

\(\Rightarrow2VP\ge2VT\)

\(\RightarrowĐPCM\)

Theo tính chất dãy tỉ số bằng nhau ta có:

\(\frac{2a+b}{c}=\frac{2b+c}{a}=\frac{2c+a}{b}=\frac{2a+b+2b+c+2c+a}{a+b+c}=\frac{3\left(a+b+c\right)}{a+b+c}=3\)

\(\Rightarrow\frac{2a+b}{c}=\frac{3}{3}=1=\frac{a}{2b+c}=\frac{3b}{2c+a}\)

Vậy \(\frac{2a+b}{c}=\frac{a}{2b+c}=\frac{3b}{2c+a}=1\)

Câu hỏi của ✨♔♕ Saiko ♕♔✨ - Toán lớp 6 - Học toán với OnlineMath

Áp dụng bđt Cauchy-schwarz dạng engel ta có:

1. \(\frac{a^2}{a+2b}+\frac{b^2}{b+2c}+\frac{c^2}{c+2a}\ge\frac{\left(a+b+c\right)^2}{\left(a+2b\right)+\left(b+2c\right)+\left(c+2a\right)}=\frac{a+b+c}{3}\)

Dấu "=" \(\Leftrightarrow\frac{a}{a+2b}=\frac{b}{b+2c}=\frac{c}{c+2a}\Leftrightarrow a=b=c\)

2. \(\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{\left(2a+3b\right)+\left(2b+3c\right)+\left(2c+3a\right)}=\frac{a+b+c}{5}\)

Dấu "=" \(\Leftrightarrow a=b=c\)