a) Vì x và y là 2 đại lượng tỉ lệ nghịch nên: x.y = k ( k\(\ne0\)) (*)

Thay x =2 ; y = 3 vào (*), ta có:

2.3=k => 6=k

Vậy k = 6

b) \(y=\dfrac{6}{x}\)

nhanh len cac ban

a) Ta có: y = k.x (vs k \(\ne\) 0 ) (*)

Thay x = 6 ; y = 4 vào (*) , ta có :

4 = k . 6

=> k = 4/6 = 2/3

Vậy k = 2/3

b) y = 2/3.x

c) khi x=10 thì y = 2/3.10 => y = 20/3

Ta có: /x-y/ \(\ge\) 0 với mọi x,y

/y+9/36/ \(\ge\) 0 với mọi y

=> /x-y/ + /y+9/36/ \(\ge\) 0 vs mọi x,y

Ta có: /x-y/ + /y+9/36/ \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-y\right|=0\\\left|y+\dfrac{9}{36}\right|\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{36}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=-\dfrac{9}{36}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-9}{36}\\y=-\dfrac{9}{36}\end{matrix}\right.\)

\(\dfrac{-3}{5}.y=\dfrac{21}{10}\)

\(y=\dfrac{21}{10}:\dfrac{-3}{5}=\dfrac{21}{10}.\dfrac{-5}{3}\)

\(y=\dfrac{-7}{2}\)

Vậy \(y=\dfrac{-7}{2}\)

Theo đề bài, ta có:\(\left\{{}\begin{matrix}\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\\a+2b-3c=-20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}\\a+2b-3c=-20\end{matrix}\right.\)

Áp dụng tính chất dãy tỉ số bằng nhau , ta có:

\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\)

=> a=2.5=10; 2b=5.6=30 => b=15; 3c=5.12=60 =>c=20

Vậy a=10; b=15; c=20

\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-3+\dfrac{7}{21}+2\)

= \(\dfrac{15}{34}+\dfrac{19}{34}+\dfrac{7}{21}+\dfrac{7}{21}-3+2\)

= \(\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{7}{21}\right)-\left(3-2\right)\)

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

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

= \(\dfrac{2}{3}\)

GOOD LUCK!

Chọn đáp án C

x thuộc {2;3}. Tick mình nha mấy bạn

Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=> a = b.k ; c = d.k

Ta lại có : \(\dfrac{a-b}{a+b}=\dfrac{b.k-b}{b.k+b}=\dfrac{b.\left(k-1\right)}{b.\left(k+1\right)}=\dfrac{k-1}{k+1}\)

\(\dfrac{c-d}{c+d}=\dfrac{d.k-d}{d.k+d}=\dfrac{d.\left(k-1\right)}{d.\left(k+1\right)}=\dfrac{k-1}{k+1}\)

Vì \(\dfrac{a-b}{a+b}=\dfrac{k-1}{k+1}\) ; \(\dfrac{c-d}{c+d}=\dfrac{k-1}{k+1}\) nên \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)

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