Theo bài ra ta có : \(a+b=11\Rightarrow a=11-b\)(1) ; \(b+c=3\Rightarrow c=3-b\)(2) 

\(\Leftrightarrow c+a=2\)hay \(11-b+3-b=0\Leftrightarrow14-2b=0\Leftrightarrow b=7\)

Thay lại vào (1) ; (2) ta có : 

\(\Leftrightarrow a=11-b=11-7=4\)

\(\Leftrightarrow c=3-b=3-7=-4\)

Do a ; b ; c \(\in Z\)Vậy a ; b ; c = 4 ; 7 ; -4 ( thỏa mãn điều kiện ) 

2.Giải:

Theo bài ra ta có:

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{d}{5}\) và a + b + c + d = -42

Theo tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{d}{5}=\frac{a+b+c+d}{2+3+4+5}=\frac{-42}{14}=-3\)

+) \(\frac{a}{2}=-3\Rightarrow a=-6\)

+) \(\frac{b}{3}=-3\Rightarrow b=-9\)

+) \(\frac{c}{4}=-3\Rightarrow c=-12\)

+) \(\frac{d}{5}=-3\Rightarrow d=-15\)

Vậy a = -6

        b = -9

        c = -12

        d = -15

Bài 3:

Ta có:\(\frac{a}{2}=\frac{b}{3}\Leftrightarrow\frac{a}{10}=\frac{b}{15}\); \(\frac{b}{5}=\frac{c}{4}\Leftrightarrow\frac{b}{15}=\frac{c}{12}\)

\(\Rightarrow\frac{a}{10}=\frac{b}{15}=\frac{c}{12}\)

Áp dụng tc dãy tỉ:

\(\frac{a}{10}=\frac{b}{15}=\frac{c}{20}=\frac{a+b+c}{10+15+12}=\frac{-49}{37}\)

Với \(\frac{a}{10}=\frac{-49}{37}\Rightarrow a=10\cdot\frac{-49}{37}=\frac{-490}{37}\)

Với \(\frac{b}{15}=\frac{-49}{37}\Rightarrow b=15\cdot\frac{-49}{37}=\frac{-735}{37}\)

Với \(\frac{c}{12}=\frac{-49}{37}\Rightarrow c=12\cdot\frac{-49}{37}=\frac{-588}{37}\)

chia hết cho n+1 nha các bạn

? nghĩa là    sao

Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{b+c+1}{a}=\frac{a+c+2}{b}=\frac{a+b-3}{c}=\frac{b+c+1+a+c+2+a+b-3}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2=\frac{1}{a+b+c}\)

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

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

\(\Rightarrow a+b+c+1=3a\)

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

\(\Rightarrow3a=\frac{3}{2}\)

\(\Rightarrow a=\frac{1}{2}\)

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

\(\Rightarrow a+b+c+2=3b\)

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

\(\Rightarrow\frac{5}{2}=3b\)

\(\Rightarrow b=\frac{5}{6}\)

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

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

\(\Rightarrow\frac{1}{2}-3=3c\)

\(\Rightarrow\frac{-5}{2}=3c\)

\(\Rightarrow c=\frac{-5}{6}\)

Vậy bộ số \(\left(a;b;c\right)\) là \(\left(\frac{1}{2};\frac{5}{6};\frac{-5}{6}\right)\)

\(\frac{b+c+1}{a}=\frac{a+c+2}{b}=\frac{a+b-3}{c}=\frac{b+c+1+a+c+2+a+b-3}{a+b+c}=2\)(T/C...)

\(\Rightarrow\frac{1}{a+b+c}=2\Rightarrow a+b+c=\frac{1}{2}=0,5\)

\(\Rightarrow\frac{b+c+1}{a}=2\Rightarrow\frac{0,5-a+1}{a}=2\Rightarrow1,5-a=2a\Rightarrow a=\frac{1}{2}\)

\(\Rightarrow\frac{a+c+2}{b}=2\Rightarrow\frac{0,5-b+2}{b}=2\Rightarrow2,5-b=2b\Rightarrow b=\frac{5}{6}\)

\(\Rightarrow c=0,5-\frac{1}{2}-\frac{5}{6}=-\frac{5}{6}\)

áp dụng t.c dãy tỉ số bằng nhau ta có:

\(\frac{b+c+1}{a}=\frac{a+c+2}{b}=\frac{a+b-3}{c}=\frac{b+c+1+a+c+2+a+b-3}{a+b+c}=2\)(vì a+b+c khác 0)

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

\(\frac{b+c+1}{a}=2\Rightarrow2a=b+c+1\Rightarrow3a=a+b+c+1\Rightarrow a=\frac{1}{2}\)

\(\frac{a+c+2}{b}=2\Rightarrow2b=a+c+2\Rightarrow3b=a+b+c+2\Rightarrow b=\frac{5}{6}\)

\(\frac{a+b-3}{c}=2\Rightarrow2c=a+b-3\Rightarrow3c=a+b+c-3\Rightarrow c=-\frac{5}{6}\)

Vậy \(a=\frac{1}{2},b=\frac{5}{6},c=-\frac{5}{6}\)

A=2

B=3

C=5

D=7

E=17

B=3

C=5

D=7

E=17

Lời giải:
$b=(a+b+c+d)-(a+c+d)=1-2=-1$

$c=(a+b+c+d)-(a+b+d)=1-3=-2$

$d=(a+b+c+d)-(a+b+c)=1-4=-3$

$a = (a+b+c+d)-(b+c+d)=1-[(-1)+(-2)+(-3)]=1-(-6)=5$