a+b=11 => b= 11-a

c+a=2 => c=2-a

b+c= 3 nên 11-a +2-a= 3

11+2-2a=3

13-2a =3

13=3+2a

13-3=2a

10=2a => a=5

Vậy a=5

5+b=11 => b=11-5=6

Vậy b=6

c+5=2 => c=2-5= (-3)

Vậy c= -3

ĐÀO CÔNG ĐẠT d đâu ra???

Ta có : a + b + b + c + c + a = 11 + 3 + 2

 => 2a + 2b + 2c = 16

=> 2(a+b+c) = 16

=> a + b + c = 8

+) a + b = 11 

=> 11 + c = 8

=> c = 8 - 11 = -3

+) b + c = 3

=> a + 3 = 8

=> a = 8 - 3

=> a = 5

+) c + a = 2

=> 2 + b = 8

=> b = 8 - 2 = 6

Vậy a = 5,b = 6,c = -3

TL:

a+b+b+ c+c+a= 11+3+2

<=> 2( a+b+c) = 16

<=> a+b+c =8 => c=8-11=-3 ; a=8-3=5 ; b=8-2 =6

Hc tốt

Bài 1 :

\(a,\left(a-b\right)+\left(c-d\right)-\left(a-c\right)=-\left(b+d\right)\)

Ta có : \(VT=\left(a-b\right)+\left(c-d\right)-\left(a-c\right)\)

                 \(=a-b+c-d-a+c\)

                 \(=-\left(b+d\right)=VP\)

\(\Rightarrow\left(a-b\right)+\left(c-d\right)-\left(a-c\right)=-\left(b+d\right)\)

\(b,\left(a-b\right)-\left(c-d\right)+\left(b+c\right)=a+d\)

Ta có : \(VT=\left(a-b\right)-\left(c-d\right)+\left(b+c\right)\)

                 \(=a-b-c+d+b+c\)

                 \(=a+d=VP\)

\(\Rightarrow\left(a-b\right)-\left(c-d\right)+\left(b+c\right)=a+d\)

a.d = b.c ⇒ \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}\) = \(\dfrac{3a}{3c}=\dfrac{2b}{2d}\)

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

\(\dfrac{a}{c}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{2a+5b}{2c+5d}\) (1)

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

\(\dfrac{a}{c}=\dfrac{3a}{3c}=\dfrac{2b}{2d}=\dfrac{3a-2b}{2c-2d}\) (2)

Từ (1) và(2) ta có:

\(\dfrac{2a+5b}{2c+5d}\) =  \(\dfrac{3a-2b}{3c-2d}\)(đpcm)

a.d = b.c ⇒ \(\dfrac{a}{c}=\dfrac{b}{d}\)  ⇒ \(\dfrac{a.b}{c.d}\) = \(\dfrac{a^2}{c^2}\) = \(\dfrac{b^2}{d^2}\)

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

\(\dfrac{a.b}{c.d}=\dfrac{a^2}{c^2}\) = \(\dfrac{b^2}{d^2}\) = \(\dfrac{a^2+b^2}{c^2+d^2}\) (đpcm)

\(a,\left(a-b+c\right)-\left(a+c\right)=a-b+c-a-c=-b\)

\(b,\left(a+b\right)-\left(b-a\right)+c=a+b-b+a+c=2a+c\)

\(c,-\left(a+b-c\right)+\left(a-b-c\right)=-a-b+c+a-b-c=-2b\)

\(d,a\left(b+c\right)-a\left(b+d\right)=ab+ac-ab-ad=ac-ad=a\left(c-d\right)\)

\(e,a\left(b-c\right)+a\left(d+c\right)=ab-ac+ad+ac=ab+ad=a\left(b+d\right)\)

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

= a - b + c - a - c

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

= -b

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

= a + b - b + a + c

= 2a + (b - b) + c

= 2a + c

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

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

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

= -2b

d) a(b + c) - a(b + d)

= ab + ac - ab - ad

= (ab - ab) + (ac - ad)

= ac - ad

= a(c - d)

e) a(b - c) + a(d + c)

= a(b - c + d + c)

= a[b - (c - c) + d]

= d(b + d)

a ) 

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

 a - b + c - a -c = - b

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

0 + 0 - b = -b ( đpcm )

\(b)\)

\(4n-3⋮3n-2\)

\(\Leftrightarrow3\left(4n-3\right)⋮3n-2\)

\(\Leftrightarrow12n-9⋮3n-2\)

\(\Leftrightarrow\left(12n-8\right)-1⋮3n-2\)

\(\Leftrightarrow4\left(3n-2\right)-1⋮3n-2\)

\(\Leftrightarrow1⋮3n-2\)

\(\Leftrightarrow3n-2\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(\Rightarrow3n\in\left\{1;3\right\}\)

Mà: \(3n⋮3\)

\(\Leftrightarrow3n=3\)

\(\Leftrightarrow n=1\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)

  suy ra:   x/5 = 45   =>  x  =  225

               y/7 = 45  =>  y  =  315

               z/9 = 45  =>  z  =  405