1) ab=2 (I); bc=3 (II); ca=54 (III)

Lấy (I).(II).(III) ⇒ a² . b² . c² = 324 ⇒ abc = ±18

(II) ⇒ a= ±6 ; (I) ⇒ b= ±1/3 ; (II) ⇒ c= ±9

2) ab=5/3 (I); bc=4/5 (II); ca=3/4 (III)

Lấy (I).(II).(III) ⇒ a² . b² . c² = 1 ⇒ abc = ±1

(II) ⇒ a= ±5/4 ; (I) ⇒ b= ±4/3 ; (II) ⇒ c= ±3/5

3) a(a+b+c)= -12 (I)

    b(a+b+c)= 18 (II)

    c(a+b+c)= 30 (III)

Lấy (I)+(II)+(III) ⇒ (a+b+c)² = 36 ⇒ a+b+c = ±6

TH1 : a=6 ⇒ a= -12/6 = -2 ; b= 18/6 = 3 ; c= 30/6 = 5

TH2 : a=-6 ⇒ a= -12/-6 = 2 ; b= 18/-6 = -3 ; c= 30/-6 = -5

\(\dfrac{a}{1}=\dfrac{b}{2}=\dfrac{c}{3}=\dfrac{4a-3b+2c}{4-6+6}=\dfrac{36}{4}=9\\ \Rightarrow\left\{{}\begin{matrix}a=9\\b=18\\c=27\end{matrix}\right.\\ \dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{16}=\dfrac{x-y+z}{10-15+16}=\dfrac{-49}{11}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{490}{11}\\y=-\dfrac{735}{11}\\z=-\dfrac{784}{11}\end{matrix}\right.\)

a: \(\left(abc\right)^2=\dfrac{3}{5}\cdot\dfrac{4}{5}\cdot\dfrac{3}{4}=\dfrac{9}{25}\)

Trường hợp 1: \(abc=\dfrac{3}{5}\)

\(\Leftrightarrow\left\{{}\begin{matrix}c=1\\b=\dfrac{3}{5}:\dfrac{3}{4}=\dfrac{4}{5}\\a=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{4}\end{matrix}\right.\)

Trường hợp 2: \(abc=\dfrac{-3}{5}\)

\(\Leftrightarrow\left\{{}\begin{matrix}c=-1\\b=\dfrac{3}{5}:\dfrac{-3}{4}=\dfrac{-4}{5}\\a=\dfrac{3}{5}:\dfrac{-4}{5}=\dfrac{-3}{4}\end{matrix}\right.\)

a. ab=3/5;bc=4/5;ca=3/4

=>(abc)^2=9/25

=>abc=3/5

=> c=1;a=3/4;b=4/5

b. a(a+b+c)=-12; b(a+b+c)=18; c(a+b+c)=30

=>(a+b+c)^2=36

=>a+b+c=6

=> a=-2;b=3;c=5

ko khó nhưng mà bn đăng từng câu 1 hộ mk mk giải giúp cho

gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)

=> Thay vào thì     \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)

\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)

Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào

=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)

=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)

=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\) 

Đặt: \(\sqrt{a}=x;\sqrt{b}=y;\sqrt{c}=z\)

=>     \(P=\frac{xy}{z^2+3xy}+\frac{yz}{x^2+3yz}+\frac{zx}{y^2+3zx}\)

=>     \(3P=\frac{3xy}{z^2+3xy}+\frac{3yz}{x^2+3yz}+\frac{3zx}{y^2+3zx}=1-\frac{z^2}{z^2+3xy}+1-\frac{x^2}{x^2+3yz}+1-\frac{y^2}{y^2+3zx}\)

Ta sẽ CM: \(3P\le\frac{9}{4}\)<=> Cần CM: \(\frac{x^2}{x^2+3yz}+\frac{y^2}{y^2+3zx}+\frac{z^2}{z^2+3xy}\ge\frac{3}{4}\)

Có:    \(VT\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\)

Ta sẽ CM: \(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\ge\frac{3}{4}\)

<=> \(4\left(x+y+z\right)^2\ge3\left(x^2+y^2+z^2\right)+9\left(xy+yz+zx\right)\)

<=> \(4\left(x^2+y^2+z^2\right)+8\left(xy+yz+zx\right)\ge3\left(x^2+y^2+z^2\right)+9\left(xy+yz+zx\right)\)

<=> \(x^2+y^2+z^2\ge xy+yz+zx\)

Mà đây lại là 1 BĐT luôn đúng => \(3P\le\frac{9}{4}\)=> \(P\le\frac{3}{4}\)

Vậy P max \(=\frac{3}{4}\)<=> \(a=b=c\)

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 \(\frac{d}{5}=-6\Rightarrow d=-30\)