Ta có:A=\(\dfrac{n-2}{n+3}=\dfrac{\left(n+3\right)-5}{n+3}=1-\dfrac{5}{n+3}\)

Để A∈Z=>\(\dfrac{5}{n+3}\)∈Z

=>5⋮ n+3

=>n+3∈Ư₍₅₎=\(\left\{\pm1;\pm5\right\}\)

=>n∈\(\left\{-2;-4;2;-8\right\}\)

Ta có:B=\(\dfrac{2n-1}{n+1}=\dfrac{2\left(n+1\right)-3}{n+1}=2-\dfrac{3}{n+1}\)

Để B∈Z=>\(\dfrac{3}{n+1}\)∈Z=>3⋮n+1

=>n+1∈Ư(3)=\(\left\{\pm1;\pm3\right\}\)

=>n∈\(\left\{0;-2;2;-4\right\}\)

ta có :C=\(\dfrac{2n+3}{n+2}=\dfrac{2.\left(n+2\right)-1}{n+2}=2-\dfrac{1}{n+2}\)

Để C∈Z=>\(\dfrac{1}{n+2}\)∈Z=>1⋮n+2

=>n+2∈Ư₍₁₎=\(\pm\)1

=>n=-1;-3

ta có:x.y.y.z.x.z=\(\dfrac{1}{2}.\dfrac{3}{5}.\dfrac{27}{10}=\dfrac{81}{100}\)

=>(x.y.z)²= \(\left(\dfrac{9}{10}\right)^2=\left(\dfrac{-9}{10}\right)^2\)

Nếu x.y.z=\(\dfrac{9}{10}\)

=>\(\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{1}{3}\\z=\dfrac{9}{5}\end{matrix}\right.\)

Nếu x.y.z=\(\dfrac{-9}{10}\)

=>\(\left\{{}\begin{matrix}x=\dfrac{-3}{2}\\y=\dfrac{-1}{3}\\z=\dfrac{-9}{5}\end{matrix}\right.\)

 chò đấy hay tuyệt vời ông mặt trời

10
Hỏi gì lạ vậy bạn ?

ta có :\(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)

\(=>\dfrac{a^{2013}}{c^{2013}}=\dfrac{b^{2013}}{d^{2013}}=\dfrac{\left(a+b\right)^{2013}}{\left(c+b\right)^{2013}}\left(1\right)\)

Mặt khác:\(\dfrac{a}{c}=\dfrac{b}{d}=>\dfrac{a^{2013}}{c^{2013}}=\dfrac{b^{2013}}{d^{2013}}=\dfrac{2.a^{2013}}{2.c^{2013}}=\dfrac{3.b^{2013}}{3.d^{2013}}=\dfrac{2.a^{2013}-3.b^{2013}}{2.c^{2013}-3.d^{2013}}\left(2\right)\)Từ (1),(2)=>\(\dfrac{\left(a+b\right)^{2013}}{\left(c+d\right)^{2013}}=\dfrac{2.a^{2013}-3.b^{2013}}{2.c^{2013}-3.d^{2013}}\left(đpcm\right)\)

Giả sử x TLN với y theo hệ số tỉ lệ là a₁

=> x .y =a₁(a₁≠0)=>y=\(\dfrac{a_1}{x}\)(2)

Giả sử y TLN với z theo hệ số tỉ lệ là a₂

=>y.z=a₂(a₂≠0)(2)

Từ (1),(2)=>\(\dfrac{a_1}{x}.z=a_2=>\dfrac{z}{x}=\dfrac{a_1}{a_2}\)

mà a₁≠0 và a₂≠0=>\(\dfrac{a_1}{a_2}\)≠0

do đó z TLT với y

b)Bạn tự làm nhé!

số bé là:19,36

số lớn là:86,8

chtt tick nha mấy bạn xin đó

12 : x = 6 

x = 12 : 6

x = 2