4) 

a) x/5 = y/3

=> 3x = 5y

=> x/y = 5/3

=> x= 16 :(5+3) . 5 = 10 ; y = 16 - 10 =6

=> (x;y) thuộc {(10;6)}

b) Ta có : \(\dfrac{2a}{3}=\dfrac{3b}{4}=\dfrac{4c}{5}\)

\(\Leftrightarrow\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}=\dfrac{a+b+c}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Khi đó \(a=12.\dfrac{3}{2}=18;b=12.\dfrac{4}{3}=16;c=12.\dfrac{5}{4}=15\)

Vậy (a,b,c) = (18,16,15) 

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

i-------------7jhmnjbn,j,mn.kmlk.jk,hkghnmgvbvcbvcbcvbcvbcbbccbcbcb

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

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)

\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)

Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)

\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)

3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)

4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)

\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)

Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)