a) 3a + 4b - 5c - 2a - 3b + 5c

= ( 3a - 2a ) + ( 4b - 3b ) - ( 5c - 5c )

= a + b

b) 7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c

= ( 7a - 3a - 4a ) + ( 3b + 2b + b ) - ( 4c + 2c + 2c ) 

= 6b - 8c

a) 3a + 4b - 5c - 2a - 3b + 5c

= (3a - 2a) + (4b - 3b) - (5c - 5c)

= a + b - 0 = a + b

b) 7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c

= (7a - 3a - 4a) + (3b + 2b + b) - ( 4c + 2c + 2c)

= 0 + 6b - 8c = 6b - 8c

a)

3a + 4b - 5c - 2a - 3b + 5c

=( 3a - 2a ) + ( 4b - 3b ) + ( -5c + 5c )

= a + b

b)

7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c

=( 7a - 3a - 4a ) + ( 3b + 2b + b ) + ( -4c - 2c - 2c )

= 6b + (-8c)

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:

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

\(\Rightarrow2VP\ge2VT\)

\(\RightarrowĐPCM\)

\(\hept{\begin{cases}2a+b+2c=6\\3a+4b-3c=4\end{cases}}\)\(\Rightarrow a+3b-5c=-2\)

\(\Rightarrow3b=-2+5c-a\)\(\Rightarrow3b+2a-4c=-2+5c-a+2a-4c\)

\(\Rightarrow P=-2+a+c\)

Lại có : \(2a+b+2c=6\Rightarrow2\left(a+c\right)\le6\)

\(\Rightarrow a+c\le3\)

\(\Rightarrow P\le-2+3=1\Rightarrow P\le1\)

Dấu " = " sảy ra \(\Leftrightarrow\hept{\begin{cases}b=0\\3a-3c=4\\2a+2c=6\end{cases}}\)\(\Rightarrow\hept{\begin{cases}b=0\\3a-3c=4\\3a+3c=9\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=\frac{13}{6}\\b=0\\c=\frac{5}{6}\end{cases}}\)

Chị chỉ tìm được Max thui 

\(\hept{\begin{cases}2a+b+2c=6\\3a+4b-3c=4\end{cases}}\)

<=> \(\hept{\begin{cases}b+2c=6-2a\\4b-3c=4-3a\end{cases}}\)

<=> \(\hept{\begin{cases}c=\frac{20}{11}-\frac{5a}{11}\\b=\frac{26}{11}-\frac{12}{11}a\end{cases}}\)

P = \(2a+3\left(\frac{26}{11}-\frac{12}{11}a\right)-4\left(\frac{20}{11}-\frac{5a}{11}\right)\)

\(=-\frac{2}{11}+\frac{6}{11}a\ge-\frac{2}{11}\)

Dấu "=" xảy ra <=> a = 0 => c =20/11 và b = 26/11

Vậy min P = -2/11 tại a = 0; b = 26/11 và c= 20/11

Cách tìm max khác:

Ta có: \(\hept{\begin{cases}2a+b+2c=6\\3a+4b-3c=4\end{cases}}\)

<=> \(\hept{\begin{cases}2a+2c=6-b\\3a-3c=4-4b\end{cases}}\) <=> \(\hept{\begin{cases}a+c=3-\frac{b}{2}\\a-c=\frac{4}{3}-\frac{4b}{3}\end{cases}}\)

<=> \(\hept{\begin{cases}a=\frac{13}{6}-\frac{11b}{12}\\c=\frac{5}{6}+\frac{5}{12}b\end{cases}}\)

khi đó P = \(2\left(\frac{13}{6}-\frac{11b}{12}\right)+3b-4\left(\frac{5}{6}+\frac{5}{12}b\right)=1-\frac{1}{2}b\le1\)

Dấu bằng xảy ra khi và chỉ khi b = 0 khi đó a = 13/6 và c = 5/6( thỏa mãn)

Vậy maxP = 1 tại a = 13/6 ;  b = 0 ; c = 5/6.

\(\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}=\frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}\)

\(=\frac{15a-10b+6c-15a+10b-6c}{25+9+4}=0\)

\(\Rightarrow\left\{{}\begin{matrix}3a=2b\\2c=5a\\5b=3c\end{matrix}\right.\Rightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=\frac{a+b+c}{10}\)

\(\Rightarrow\left\{{}\begin{matrix}a=\frac{a+b+c}{5}\\b=\frac{3\left(a+b+c\right)}{10}\\c=\frac{a+b+c}{2}\end{matrix}\right.\)

\(\Rightarrow P=\frac{\frac{33\left(a+b+c\right)}{10}}{\frac{43\left(a+b+c\right)}{10}}=\frac{33}{43}\)

cần 1 lời giải đáp cụ thể

trên face có đấy,lên đó mà tìm

làm ơn trả lời hộ mk với ah mai mk phải nộp bài r

a/

\(\frac{3a-b}{a+b}=\frac{3\left(a+b\right)-4b}{a+b}=3-\frac{4b}{a+b}=\frac{3}{4}.\)

\(\Rightarrow\frac{4b}{a+b}=\frac{9}{4}\Rightarrow9a+9b=16b\Rightarrow9a=7b\Rightarrow\frac{a}{b}=\frac{7}{9}\)

b/

\(\frac{a}{b}=\frac{3}{7}\Rightarrow\frac{a}{3}=\frac{b}{7}=\frac{3a}{9}=\frac{4b}{28}=\frac{3a-4b}{9-28}=\frac{3a-4b}{-19}\)

\(\frac{a}{3}=\frac{b}{7}\Rightarrow\frac{2a}{6}=\frac{3b}{21}\Rightarrow\frac{2a+3b}{6+21}=\frac{2a+3b}{27}\)

\(\Rightarrow\frac{3a-4b}{-19}=\frac{2a+3b}{27}\Rightarrow\frac{3a-4b}{2a+3b}=-\frac{19}{27}\)