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

\(ab=\frac{1}{2};bc=\frac{2}{3};ac=\frac{3}{4}\)

Nhân từng vế các đẳng thức trên,ta đc:

\(ab.bc.ac=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}\Rightarrow\left(abc\right)^2=\frac{1}{4}\Rightarrow\hept{\begin{cases}abc=\frac{1}{2}\\abc=-\frac{1}{2}\end{cases}}\)

+)abc=1/2

Có \(ab=\frac{1}{2}\Rightarrow c=abc:ab=\frac{1}{2}:\frac{1}{2}=1\)

có \(bc=\frac{2}{3}\Rightarrow a=abc:bc=\frac{1}{2}:\frac{2}{3}=\frac{3}{4}\)

có \(ac=\frac{3}{4}\Rightarrow b=abc:ac=\frac{1}{2}:\frac{3}{4}=\frac{2}{3}\)

+)abc=-1/2,xét tương tự abc=1/2 : a=-3/4;b=-2/3;c=-1

Vậy (a;b;c) \(\in\){(-3/4;-2/3;-1);(3/4;2/3;1)}

1)Từ đề bài:

`=>a^2+4b+4+b^2+4c+4+c^2+4a+4=0`

`<=>(a+2)^2+(b+2)^2+(c+2)^2=0`

`<=>a=b=c-2`

`ab+bc+ca=abc`

`<=>1/a+1/b+1/c=1`

`<=>(1/a+1/b+1/c)^2=1`

`<=>1/a^2+1/b^2+1/c^2+2/(ab)+2/(bc)+2/(ca)=1`

`<=>1/a^2+1/b^2+1/c^2=1-(2/(ab)+2/(bc)+2/(ca))`

`a+b+c=0`

Chia 2 vế cho `abc`

`=>1/(ab)+1/(bc)+1/(ca)=0`

`=>2/(ab)+2/(bc)+2/(ca)=0`

`=>1/a^2+1/b^2+1/c^2=1-0=1`

Ta có:

\(S=1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{2001!}\)

\(=2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2001!}\)

Ta lại có:

\(\frac{1}{2!}=\frac{1}{1.2}\)

\(\frac{1}{3!}

Bài 1 là có n số hạng các bạn nhé 

mình quên mất

\(M=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+a}\)

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

Vậy M>1

2^x+1.3^y=36^x=(4.9)^x=4^x.9^x=2^2x.3^2x

=>2^x+1=2^2x

=>x+1=2x=>2x-x=1=>x=1

và 3^y=3^2x=>y=2x=>y=2.1=2

Vậy (x;y)=(1;2)

\(a,A=\dfrac{-3\left(2n-3\right)-8}{2n-3}=-3-\dfrac{8}{2n-3}\in Z\\ \Leftrightarrow2n-3\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\\ \Leftrightarrow n\in\left\{1;2\right\}\left(n\in Z\right)\)

\(b,\dfrac{ab}{a+2b}=\dfrac{3}{2}\Leftrightarrow\dfrac{a+2b}{ab}=\dfrac{2}{3}\Leftrightarrow\dfrac{1}{b}+\dfrac{2}{a}=\dfrac{2}{3}\\ \dfrac{bc}{b+2c}=\dfrac{4}{3}\Leftrightarrow\dfrac{b+2c}{bc}=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{c}+\dfrac{2}{b}=\dfrac{3}{4}\\ \dfrac{ca}{c+2a}=3\Leftrightarrow\dfrac{c+2a}{ca}=\dfrac{1}{3}\Leftrightarrow\dfrac{1}{a}+\dfrac{2}{c}=\dfrac{1}{3}\)

Cộng vế theo vế \(\Leftrightarrow\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}=\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{3}=\dfrac{7}{4}\)

\(\Leftrightarrow3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{7}{4}\\ \Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{7}{12}\\ \Leftrightarrow\dfrac{ab+bc+ca}{abc}=\dfrac{7}{12}\\ \Leftrightarrow T=\dfrac{12}{7}\)

bc bằng 25 còn abc =325

mình mới học  lớp 5