Gọi n,n+1,n-1 là 3 số tự nhiên liên tiếp 

TH1: n=3k 

=> \(n⋮3\)

TH2: n=3k+1

=> n-1=3k+1-1=3k \(⋮\) 3

TH3: n=3k+2

=> n+1=3k+2+1=3k+3 \(⋮\) 3

Vậy,,,,

Có:

\(a\left(a+b+c\right)=-12\)

\(b\left(a+b+c\right)=18\)

\(c\left(a+b+c\right)=30\)

=> \(a\left(a+b+c\right)+\)\(b\left(a+b+c\right)+\)\(c\left(a+b+c\right)=-12+18+30\)

\(\left(a+b+c\right)\left(a+b+c\right)=36\)

\(\left(a+b+c\right)^2=6^2\)

=>\(a+b+c=6\)

Có: \(a\left(a+b+c\right)=-12\)

       => \(6a=-12\)

             a=-2

\(b\left(a+b+c\right)=18\Rightarrow6b=18\Rightarrow b=3\)

\(c\left(a+b+c\right)=30\Rightarrow6c=30\Rightarrow c=5\)

Vậy....

Diên tích hình chữ nhật là: 

\(12.8=96\left(cm^2\right)\)

Diện tích 2 hình tam giác là:

\(2.\left(\dfrac{4.4}{2}\right)=16\left(cm^2\right)\)

Diện tích phần tô màu là:

\(96-16=80\left(cm^2\right)\)

ĐS....

42

ĐK:\(x\ge1\)

\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)

\(\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\)

\(-2\sqrt{x-1}=-2\)

\(\sqrt{x-1}=1\)

\(x-1=1\)

x=2

Để \(A.B=\dfrac{\sqrt{x}}{3}\) thì

\(\dfrac{5}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{\sqrt{x}}{3}\)

\(\Leftrightarrow\dfrac{5}{\sqrt{x}+2}=\dfrac{\sqrt{x}}{3}\)

\(\Leftrightarrow x+2\sqrt{x}=15\)

\(\Leftrightarrow x+2\sqrt{x}-15=0\)

\(\Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)=0\)

\(\Rightarrow\sqrt{x}-3=0\) ( do \(\sqrt{x}+5\ge5>0\) với mọi x)

     \(\sqrt{x}=3\)

       x=9

\(\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\)

\(\Leftrightarrow\sqrt{b}\cdot\sqrt{\dfrac{a}{b}}=\sqrt{a}\)

\(\Leftrightarrow\sqrt{b\cdot\dfrac{a}{b}}=\sqrt{a}\)

\(\Leftrightarrow\sqrt{a}=\sqrt{a}\left(lđ\right)\)