a: Khi x=6 thì \(A=\dfrac{4}{6-3}=\dfrac{4}{3}\)

b: \(B=\dfrac{4x}{x^2-9}-\dfrac{x-3}{x+3}\)(ĐKXĐ: \(x\notin\left\{3;-3\right\}\))

\(=\dfrac{4x}{\left(x-3\right)\left(x+3\right)}-\dfrac{x-3}{x+3}\)

\(=\dfrac{4x-\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{4x-x^2+6x-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{-x^2+10x-9}{\left(x+3\right)\left(x-3\right)}\)

Ta có: \(27=3^3;13=13\)

\(\Rightarrow BC\left(27;13\right)=3^3.13=351\)

Mà  \(Ư\left(351\right)=\left\{1;3;9;13;27;...351\right\}\)

Vậy a nhỏ nhất để chia hết cho 9 là 9

Vậy a= 9

Ủng hộ tớ nha ^_^

27 nhân 13= 351 chac 100%

a: ĐKXĐ: x∉{0;3;-3}

\(\frac{9}{x^3-9x}+\frac{1}{x+3}\)

\(=\frac{9}{x\left(x^2-9\right)}+\frac{1}{x+3}\)

\(=\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\)

\(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\)

\(=\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\)

\(=\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}=\frac{3x-9-x^2}{3x\cdot\left(x+3\right)}=\frac{-\left(x^2-3x+9\right)}{3x\left(x+3\right)}\)

Ta có: \(A=\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)

\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}:\frac{-\left(x^2-3x+9\right)}{3x\left(x+3\right)}\)

\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\frac{3x\left(x+3\right)}{-\left(x^2-3x+9\right)}=\frac{-3}{x-3}\)

b: A=2

=>\(-\frac{3}{x-3}=2\)

=>\(x-3=-\frac32\)

=>\(x=-\frac32+3=\frac32\) (nhận)

c: A nguyên

=>-3⋮x-3

=>x-3∈{1;-1;3;-3}

=>x∈{4;2;6;0}

Kết hợp ĐKXĐ, ta được: x∈{4;2;6}

Để 59a chia hết cho 2 \(\Rightarrow\)a\(\varepsilon\){0,2,4,6,8}

b,Để 59a chia hết cho 5\(\Rightarrow a\varepsilon\) {0,5}

c,Để 59a chia hết cho 3 \(\Rightarrow5+9+a⋮3\)

\(\Rightarrow a\varepsilon\){1,4,7}

d,Để 59a chia hết cho 9\(\Rightarrow5+9+1⋮9\)

\(\Rightarrow a\varepsilon\)=4

BC=BH+CH=25(cm)

\(AC=\sqrt{CH\cdot BC}=20\left(cm\right)\)

ta có:

\(AC^2=CH.BC\)

\(\Leftrightarrow AC=\sqrt{16.\left(16+9\right)}=\sqrt{400}=20cm\)