C

C

c?

Bài 3: 

Bác An có:

24:4/7=42(con)

Bác An đã bán đi:

42x3/7=18(con)

Số số hạng là: \(\dfrac{2x-2}{2}+1=x-1+1=x\left(số\right)\)

Tổng của dãy số 2;4;6;...;2x là;

\(\left(2x+2\right)\cdot\dfrac{x}{2}=x\left(x+1\right)\)

2+4+6+8+...+2x=110

=>x(x+1)=110

=>\(x^2+x-110=0\)

=>\(\left(x+11\right)\left(x-10\right)=0\)

=>\(\left[{}\begin{matrix}x=-11\left(loại\right)\\x=10\left(nhận\right)\end{matrix}\right.\)

a: =5/6(1/8+2/3)=5/6*19/24=95/144

b: =3/4(7/5-1/2)=27/40

c: =35/24*2/3=35/36

\(\Leftrightarrow\left[{}\begin{matrix}3\left(m+6\right)x^2-3\left(m+3\right)x+2m-3>3\\3\left(m+6\right)x^2-3\left(m+3\right)x+2m-3< -3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3\left(m+6\right)x^2-3\left(m+3\right)x+2m-6>0\\3\left(m+6\right)x^2-3\left(m+3\right)x+2m< 0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m+6>0\\\Delta=9\left(m+3\right)^2-12\left(m+6\right)\left(2m-6\right)< 0\end{matrix}\right.\\\left\{{}\begin{matrix}m+6< 0\\9\left(m+3\right)^2-24m\left(m+6\right)< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m>-6\\-15m^2-18m+513< 0\end{matrix}\right.\\\left\{{}\begin{matrix}m< -6\\-15m^2-90m+81< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow...\) (kết quả xấu quá)

-2

-19/3

 k mình nha

\(\dfrac{-11}{4}-\dfrac{6}{8}=\dfrac{-11}{4}-\dfrac{3}{4}=\dfrac{-14}{4}=\dfrac{-7}{2}\)

\(\dfrac{-7}{1}-\dfrac{2}{3}=\dfrac{-21}{3}-\dfrac{2}{3}=\dfrac{-19}{3}\)

=>36/18>x>8/15

=>2>x>8/15

mà x nguyên

nên x=1

ta có: \(2x-1=2\left(x-3\right)+5\)

để \(2x-1⋮x-3\Rightarrow2\left(x-3\right)+5⋮x-3\\ m\text{à }x.nguy\text{ê}n\Rightarrow x-3nguy\text{ê}n\\ \Rightarrow x-3\in\text{Ư}\left(5\right)=\left\{-5;5;1;-1\right\}\)

ta có bảng sau :

\(\Leftrightarrow2.\left(x-3\right)+5⋮x-3\)

\(do2.\left(x-3\right)⋮x-3\)

\(\Leftrightarrow5⋮x-3\)

\(\Leftrightarrow x-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\)