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:

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

\(\Rightarrow2VP\ge2VT\)

\(\RightarrowĐPCM\)

A = 2a-3b+5c-5b+4a-2c-3c+4b+7a = 13a-4b

k mk nha

A=-(-2a+3b-5c)+(-5b+4a-2c)-(3c-4b-7a)

   =2a-3b+5c-5b+4a-2c-3c+4b+7a

   =13a-4b.

1 , a - ( a - b - c ) - ( b - c -a ) - ( c - b -a )

= a - a + b + c - b + c + a - c + b + a

= (a-a+a) + (b-b+b) + (c-c+c)

= a+b+c

2 , - ( a + b + c ) - ( b - c -a ) + ( 1 - a - b ) - ( c - 3b )

= -a - b - c - b + c + a + 1 - a - b - c + 3b

= (a+a-a) - (b+b+b) + (c-c+c) + 3b

= a - 3b + c + 3b

= a+c + (3b - 3b)

= a+c + 0

= a+c

3 , ( b - c - 6 ) - ( 7 - a + b ) + c

= b - c - 6 - 7 + a - b + c

= (b-b) + (c-c) - (6+7) + a

= 0 + 0 - 13 + a

= -13 + a

4 , - ( 3b - 2a - c ) - ( a - b - c ) - ( a - 2b -+ 2c )

= -3b + 2a + c - a + b + c - a  + 2b - 2c

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

= -3b + 3b + 2c - 2a + 2a - 2c

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

= 0 + 0 + 0

= 0

chỉ bt lm đến đây thoy

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1)    a - ( a - b - c ) - ( b - c - a ) - ( c - b - a )

= a - a + b + c - b + c + a - c + b + a

= 2a + b + c

2)  - ( a + b + c ) - ( b - c - a ) + ( 1 - a - b ) - ( c - 3b )

= -a - b - c - b + c + a + 1 - a - b - c + 3b

= 1 - a - c

1,a-(a-b-c)-(b-c-a)-(c-b-a)

=a-a+b+c-b+c+a-c+b+a

=2a+b+c

2,-(a+b+c)-(b-c-a)+(1-a-b)-(c-3b)

=-a-b-c-b+c+a+1-a-b-c+3b

=1-a-c

3,(b-c-6)-(7-a+b)+c

=b-c-6-7+a-b+c

=a-13

4,-(3b-2a-c)-(a-b-c)-(a-2b+2c)

=-3b+2a+c-a+b+c-a+2b-2c

=0

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

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

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

=10a-10b+0

=10.(a-b)

6,

2a-{a-b[a-b-(a+b+c)+2b]-c-b}

=2a-{a-b[a-b-a-b-c+2b]-c-b}

=2a-a-bc+c+b

=a-bc+c+b

=(a+b)-b(c-1)

a - ( a - b - c ) - ( b - c - a ) - ( c - b - a)

= a - a + b + c - b + c + a - c + b + a

= ( a -a + a ) + ( b - b + b ) + ( c + c - c)              ( vì mình ko có ngoặc vuông nên chỉ thế này thôi)

= a + b + c

Bạn tự làm hết nha

1)=>a-a+b+b-b+c+a-c+b+a=2a+2b+c=2(a+b)+c

2)=>-a-b-c-b+c+a+1-a-b-c+3b=-a

3)=>b-c-6-7+a-b+c=-13+a

4)-3b+2a+c-a+b+c-a+2b-2c=0

5)=>4a-3b+2c-4b+3c+2a-4c+3a-2b+a-b-c=-2a-10b-2c

 2a - { a - b [ a - b - ( a + b + c ) + 2b ] - c - b }

=2a-{a-b[a-b-a-b-c+2b]-c-b}

=2a-{a-bc-c-b}

=2a-a-bc-c-b

=a-bc-b-c

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

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

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

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

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

2, - ( a + b + c ) - ( b - c -a ) + ( 1 - a - b ) - ( c - 3b )

= -a - b -c - b + c + a + 1 - a - b - c + 3b

= (a-a) - (b+b+b) + (c-c) + (-a) + (-c) + 3b

= 0 - 3b + 0 + (-a) + (-c) + 3b

= (3b-3b) + (-a) + (-c)

= 0 + (-a) + (-c)

= (-a) + (-c)

3, ( b - c - 6 ) - ( 7 - a + b ) + c

= b - c - 6 - 7 + a - b + c

= (b-b) + (c-c) - (6+7) + a

= 0 + 0 + 13 + a

= 13 + a

6, 2a - { a - b [ a - b - ( a + b + c ) + 2b ] - c - b }

= 2a - { a - b [ a - b - a - b - c  + 2b ] - c - b }

= 2a - { a - b [ ( a - a ) - (b+b) - c + 2b ] - c - b }

= 2a - { a - b [ 0 - 0 - 2b - c + 2b ] - c - b }

= 2a - { a- b [ (2b - 2b) - c ] - c - b }

= 2a - { a - b [ 0 - c ] - c - b }

= 2a - { a - b.(-c) - c - b}

= 2a - a - b.(-c) - c - b

= 1a - (-b).c - c - b

= a - (-b).c - c.1 - b

= a - [(-b) - 1].c - b

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