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

 

Bài 2:

a : b : c : d = 2 : 3 : 4 : 5 \(\Rightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{d}{5}\)

Áp dụng tc dãy tỉ:

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{d}{5}=\frac{a+b+c+d}{2+3+4+5}=\frac{-42}{14}=-3\)

Với \(\frac{a}{2}=-3\Rightarrow a=-6\)

Với \(\frac{b}{3}=-6\Rightarrow b=-18\)

Với \(\frac{c}{4}=-6\Rightarrow c=-24\)

Với \(\frac{d}{5}=-6\Rightarrow d=-30\)

\(\dfrac{a}{b}=\dfrac{2}{3}\Rightarrow\dfrac{a}{2}=\dfrac{b}{3};\dfrac{a}{c}=\dfrac{1}{2}\Rightarrow\dfrac{a}{1}=\dfrac{c}{2}\\ \Rightarrow\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^3}{8}=\dfrac{b^3}{27}=\dfrac{c^3}{64}\)

Áp dụng tcdtsnb:

\(\dfrac{a^3}{8}=\dfrac{b^3}{27}=\dfrac{c^3}{64}=\dfrac{a^3+b^3+c^3}{8+27+64}=\dfrac{99}{99}=1\\ \Rightarrow\left\{{}\begin{matrix}a^3=8\\b^3=27\\c^3=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)

\(1,\\ \left(a+1\right)\left(b+2\right)=5\\Vậy:\left(a+1\right);\left(b+2\right)\inƯ\left(5\right)=\left\{1;5\right\}\\ TH1:a+1=1\Rightarrow a=0;b+2=5\Rightarrow b=3\left(Loại,vì:a< b\right)\\ TH2:a+1=5\Rightarrow a=4;b+2=1\Rightarrow b=-1\left(Nhận,vì:a>b\right)\\ \Rightarrow\left(a;b\right)=\left(4;-1\right)\)

\(2,\\ \left(a+1\right).\left(b+3\right)=6\\ \Rightarrow\left(a+1\right);\left(b+3\right)\inƯ\left(6\right)=\left\{1;2;3;6\right\}\\ \Rightarrow TH1:a+1=1\Rightarrow a=0;b+3=6\Rightarrow b=3\left(Loại,vì:a< b\right)\\ TH2:a+1=2\Rightarrow a=1;b+3=3\Rightarrow b=0\left(Nhận,vì:a>b\right)\\ TH3:a+1=3\Rightarrow a=2;b+3=2\Rightarrow b=-1\left(Nhận,vì:a>b\right)\\ TH4:a+1=6\Rightarrow a=5;b+3=1\Rightarrow b=-2\left(Nhận,vì:a>b\right)\\ Vậy:\left(a;b\right)=\left(1;0\right).hoặc\left(a;b\right)=\left(2;-1\right).hoặc\left(a;b\right)=\left(5;-2\right)\)

\(\left(a+1\right)\left(b+2\right)=5\)

\(\Rightarrow\left(a+1\right);\left(b+2\right)\in\left\{1;5\right\}\)

\(\Rightarrow\left(a;b\right)\in\left\{\left(0;3\right)\right\}\)

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 ) 

a a + b + b + c + a + c = 11 + 3 + 2 2a + 2b + 2c = 16 a + b + c = 8 Mà a + b = 11 Suy ra c = - 3 b + c = 3 Vậy b = 6 c + a = 2 a = 5 Vậy a = 5 ; b = 6 ; c = -3 b a + b + c + a + b + d + a + c + d = 4 + 3 + 2 a + 2a + 2b + 2c + 2d = 9 Mà a + b + c + d = 1 Suy ra a + 2 = 9 a = 7 a + c + d = 2 c + d = -5 a + b + d = 3 b + d = -4 a + b + c = 4 b + c = -3 b + c + c + d + d + b = -5 + -4 + -3 2b + 2c + 2d = -12 b + c + d = -6 b + c = -3 d = -3 c + d = -5 c = -2 b + d = -4 b = -1 Vậy a = 7 ; b = -1 ; c = -2 ; d = -3