c: Ta có: \(\dfrac{2}{5}\cdot\left[\left(\dfrac{3}{5}\right)^2:\left(-\dfrac{1}{5}\right)^2-7\right]\cdot\left(1000\right)^0\cdot\left|-\dfrac{11}{15}\right|\)

\(=\dfrac{2}{5}\cdot\left(\dfrac{9}{25}:\dfrac{1}{25}-7\right)\cdot1\cdot\dfrac{11}{15}\)

\(=\dfrac{2}{5}\cdot\dfrac{11}{15}\cdot2\)

\(=\dfrac{44}{75}\)

\(n\left(n+1\right)=6\)

Có \(6=1.6=2.3=3.2=6.1\)

Mà n(n+1) là tích của 2 số tự nhiên liên tiếp, n < n+1

\(\Rightarrow\left\{{}\begin{matrix}n=2\\n+1=3\end{matrix}\right.\Rightarrow n=2\)

Vậy n = 2 là giá trị cần tìm.

Từ đề bài suy ra $n;n+1$ là cặp ước của 6

Mà $n;n+1$ là 2 số nguyên liên tiếp

$6=2.3=(-3).(-2)$

$n+1>n$ 

Nên có 2 trường hợp $n+1=3;n=2$ và $n+1=-2;n=-3$

Vậy $n∈{-3;2}$

Ta có: \(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)\left(\dfrac{1}{2}-\dfrac{3}{7}x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\\dfrac{3}{7}x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{7}{6}\end{matrix}\right.\)

2. Victoria used to have long hair.

3.Victoria didn't use to have many friend.

4.Victoria used to wear glasses.

5.Victoris didn't used to go joging every morning

6. Victoria didn't used to live in the countyside

Victoria used to have long hair.

Victoria didn't use to have many friend.

Victoria used to wear glasses.

Victoria didn't use to go joging every morning

Victoria didn't use to live in the countyside

\(\dfrac{x-1}{5}=\dfrac{y-2}{3}=\dfrac{z-2}{2}=\dfrac{2y-4}{6}=\dfrac{x-1+2y-4-z+2}{5+6-2}=\dfrac{6-5}{9}=\dfrac{1}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}9x-9=5\\9y-18=3\\9z-18=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{9}\\y=\dfrac{7}{3}\\z=\dfrac{20}{9}\end{matrix}\right.\)

mình đang cần gấp m.n giúp mình nha :33

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Chọn B

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