\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{3y}{6}=\dfrac{3z}{9}\)

Áp dụng t/c DTSBN, ta có:

\(\dfrac{x}{5}=\dfrac{3y}{6}=\dfrac{3z}{9}=\dfrac{x-3y+3z}{5-6+9}=\dfrac{24}{8}=3\)

\(\dfrac{x}{5}=3\Rightarrow x=15\)

\(\dfrac{y}{2}=3\Rightarrow y=6\)

\(\dfrac{z}{3}=3\Rightarrow=9\)

Vậy...

C

D

\(\dfrac{b+c+d-a}{a}=\dfrac{c+d+a-b}{b}=\dfrac{d+a+b-c}{c}=\dfrac{a+b+c-d}{d}\)

Áp dụng t/c DTSBN:

\(=\dfrac{b+c+d-a+c+d+a-b+d+a+b-c+a+b+c-d}{a+b+c+d}\)

\(=2\)

\(\dfrac{b+c+d-a}{a}+1=\dfrac{c+d+a-b}{b}+1=\dfrac{d+a+b-c}{c}+1=\dfrac{a+b+c-d}{d}+1\)

\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=3\\ \Rightarrow\left\{{}\begin{matrix}3a=b+c+d\\3b=c+d+a\\3c=d+a+b\\3d=a+b+c\end{matrix}\right.\Rightarrow a=b=c=d\)

\(\Rightarrow A=1+1^2+1^3+1^4=4\)

\(\dfrac{x-1}{x+2}=\dfrac{x+2-3}{x+2}=1-\dfrac{3}{x+2}\)

Do \(1\in Z\Rightarrow\)\(\dfrac{3}{x+2}\in Z\)

\(\Rightarrow x+2\inƯ\left(3\right)=\left\{\pm3;\pm1\right\}\)

Có 4 giá trị của x+2 tương ứng với có 4 giá trị x

\(5+x=2^3-1\\ 5+x=8-1\\ 5+x=7\\ x=2\)

\(\dfrac{x^2+9}{x\left(x+3\right)}+\dfrac{6}{x+3}\\ =\dfrac{x^2+6x+9}{x\left(x+3\right)}\\ =\dfrac{\left(x+3\right)^2}{x\left(x+3\right)}\\=\dfrac{x+3}{x}\)

\(\dfrac{3}{2}-x=-0,5\\ x=2\)

A

\(f\left(-2\right)=2.\left(-2\right)+3=-4+3=-1\)