Nguyễn Thanh Hằng

Vì các tổ có số nam số nữ như nhau nên :

\(18\) chia hết cho số tổ

\(24\) chia hết cho số tổ

\(\Leftrightarrow\) Số tổ \(\inƯC\left(18;24\right)=\left\{1;2;3;6\right\}\)

\(\Leftrightarrow\) Có thể chia thành \(1;2;3;6\) tổ

\(\Leftrightarrow\) Có 4 cách chia

Đặt :

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+..........+\dfrac{1}{n^2}\)

Ta thấy :

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

..........................

\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)

\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{\left(n-1\right)n}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(\Leftrightarrow A< 1-\dfrac{1}{n}< 1\)

\(\Leftrightarrow A< 1\)

Vậy ......

Bài 1 :

Sửa đề :

Tìm \(n\in Z\) để những phân số sau đồng thời có giá trị nguyên

\(\dfrac{-12n}{n};\dfrac{15}{n-2};\dfrac{8}{n+1}\)

Làm

Ta có :

\(\dfrac{-12n}{n}=-12\)

\(\Leftrightarrow\) Với mọi \(n\) thì \(\dfrac{-12n}{n}\) đều có giá trị nguyên \(\left(1\right)\)

Để \(\dfrac{15}{n-2}\in Z\) \(\Leftrightarrow n-2\inƯ\left(15\right)=\left\{\pm1;\pm15;\pm3;\pm5\right\}\)

\(\Leftrightarrow n\in\left\{-13;\pm3;\pm1;5;7;17\right\}\left(1\right)\)

Để \(\dfrac{8}{n+1}\in Z\Leftrightarrow n+1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(\Leftrightarrow n\in\left\{-9;-5;\pm3;-2;0;1;7\right\}\left(3\right)\)

Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow n\in\left\{\pm3;1;7\right\}\)

Đã đảo!!!! đả đảo!!!!!!!!!!!!!!!!!!!!!!!

2. Để phân số \(A=\dfrac{3n-5}{n+4}\in Z\)

\(\Leftrightarrow\left\{{}\begin{matrix}3n-5⋮n+4\\3n+12⋮n+4\end{matrix}\right.\)

\(\Leftrightarrow17⋮n+4\)

Vì \(n\in Z\Leftrightarrow n+4\in Z;n+4\inƯ\left(17\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}n+4=1\\n+4=17\\n+4=-1\\n+4=-17\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=-3\\n=13\\n=-5\\n=-21\end{matrix}\right.\)

Vậy ....

Ta có :

\(2n-3⋮n+1\)

Mà \(n+1⋮n+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}2n-3⋮n+1\\2n+2⋮n+1\end{matrix}\right.\)

\(\Leftrightarrow5⋮n+1\)

Vì \(n\in Z\Leftrightarrow n+1\in Z;n+1\inƯ\left(5\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}n+1=1\\n+1=5\\n+1=-1\\n+1=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}n=0\\n=4\\n=-2\\n=-6\end{matrix}\right.\)

Vậy .....

ko coog bằng :((

Ta có :

\(x-y=18\)

\(5x=7y\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{5}\)

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

\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x-y}{7-5}=\dfrac{18}{2}=9\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{7}=9\Leftrightarrow x=63\\\dfrac{y}{5}=9\Leftrightarrow y=45\end{matrix}\right.\)

Vậy ...

\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)..............\left(\dfrac{1}{100^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right).............\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)

\(=\dfrac{-1}{2}.\dfrac{3}{2}.\dfrac{-2}{3}.\dfrac{4}{3}.............\dfrac{-99}{100}.\dfrac{101}{100}\)

\(=\dfrac{-\left(1.2.3....99\right)}{2.3......100}.\dfrac{3.4...101}{2.3....100}\)

\(=\dfrac{-1}{100}.\dfrac{101}{2}\)

\(=\dfrac{-101}{200}< \dfrac{-1}{2}\)

\(\Leftrightarrow A< \dfrac{-1}{2}\)

Ta có :

\(x+2⋮x+4\)

Mà \(x+4⋮x+4\)

\(\Leftrightarrow2⋮x+4\)

Vì \(x\in Z\Leftrightarrow x+4\in Z;x+4\inƯ\left(2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=2\\x+4=-1\\x+4=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=-5\\x=-6\end{matrix}\right.\)

Vậy ......

\(P=\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,27-2,2+\dfrac{11}{7}+\dfrac{11}{13}}\)

\(=\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}\)

\(=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}\)

\(=\dfrac{3}{11}\)

Gọi số nhỏ là a \(\left(a< 96\right)\)

Ta có :

\(ƯCLN\left(a,96\right)=16\)

\(\Leftrightarrow a=16k\left(k\in N\right)\)

\(\Leftrightarrow96=16.6\)

Mà \(ƯCLN\left(a,96\right)=96\LeftrightarrowƯCLN\left(k,6\right)=1\)

Do \(a< 96\Leftrightarrow k< 6\)

\(\Leftrightarrow k\in\left\{1;5\right\}\)

\(\Leftrightarrow a\in\left\{16;80\right\}\)

Vậy .....

Biết tìm chỗ đăng qá nghen nhưng xin lỗi tớ mù công nghệ :D

\(\left(\dfrac{-2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left(\dfrac{-2}{3}+\dfrac{3}{7}\right).\dfrac{5}{4}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right).\dfrac{5}{4}\)

\(=\dfrac{5}{4}\left[\left(\dfrac{-2}{3}+\dfrac{3}{7}\right)+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right)\right]\)

\(=\dfrac{5}{4}\left[\left(\dfrac{-2}{3}+\dfrac{-1}{3}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\right]\)

\(=\dfrac{5}{4}\left[\left(-1\right)+1\right]\)

\(=\dfrac{5}{4}.0\)

\(=0\)

\(\dfrac{x-2}{5}=\dfrac{x}{3}\)

\(\Leftrightarrow\left(x-2\right)3=5x\)

\(\Leftrightarrow3x-6=5x\)

\(\Leftrightarrow5x-3x=-6\)

\(\Leftrightarrow2x=-6\)

\(\Leftrightarrow x=-3\)

Vậy .....

b, \(B=1+2+2^2+..........+2^{2017}\)

\(\Leftrightarrow2B=2+2^2+.......+2^{2018}\)

\(\Leftrightarrow2B-B=\left(2+2^2+......+2^{2018}\right)-\left(1+2+......+2^{2017}\right)\)

\(\Leftrightarrow B=2^{2018}-1\)

c, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)

\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)

\(\Leftrightarrow4x+92=3x+120\)

\(\Leftrightarrow4x-3x=120-92\)

\(\Leftrightarrow x=28\)

a, \(\dfrac{x-2}{5}=\dfrac{x}{3}\)

\(\Leftrightarrow3\left(x-2\right)=5x\)

\(\Leftrightarrow3x-6=5x\)

\(\Leftrightarrow5x-3x=6\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

b, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)

\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)

\(\Leftrightarrow4x+92=2x+80\)

\(\Leftrightarrow4x-2x=80-92\)

\(\Leftrightarrow2x=-12\)

\(\Leftrightarrow x=-6\)

c, \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...........+\dfrac{1}{2^{2017}}\)

\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...........+\dfrac{1}{2^{2016}}\)

\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2016}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2017}}\right)\)

\(\Leftrightarrow A=1-\dfrac{1}{2^{2017}}\)

d, \(B=1+2+2^2+........+2^{2017}\)

\(\Leftrightarrow2B=2+2^2+2^3+......+2^{2018}\)

\(\Leftrightarrow2B-B=\left(2+2^2+.....+2^{2018}\right)-\left(1+2+....+2^{2017}\right)\)

\(\Leftrightarrow B=2^{2018}-1\)

\(H=37^8+34^6+53^{41}\)

Ta có :

\(37^8\) có tận cùng là 1 số lẻ

\(34^6\) có tận cùng là 1 số chẵn

\(43^{41}\) có tận cùng là 1 số lẻ

\(\Leftrightarrow\) H có tận cùng là 1 số chẵn

\(\Leftrightarrow H⋮1;H⋮H;H⋮2\)

\(\Leftrightarrow H\) là hợp số

\(\left(5x-1\right)\left(2x-\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)

Vậy ....

\(\left(x-2\right)^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^2=1^2\\\left(x-2\right)^2=\left(-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy ...

\(5^{200}+5^{199}+5^{198}\)

\(=5^{198}\left(5^2+5+1\right)\)

\(=5^{198}.31\) \(⋮31\)

\(\Leftrightarrow5^{200}+5^{199}+5^{198}⋮31\rightarrowđpcm\)

Đặt :

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

\(\Leftrightarrow a=bk;c=dk\)

\(VT=\dfrac{ac}{bd}=\dfrac{bkdk}{bd}=\dfrac{bdk^2}{bd}=k^2\left(1\right)\)

\(VP=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

Ta có :

\(\left(2x-3\right)\left(3y-2\right)=1\)

Vì \(x,y\in N\Leftrightarrow2x-3;3y-2\in N,2x-3;3y-2\inƯ\left(1\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=1\\3y-2=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy \(\left(x,y\right)=\left(2,1\right)\)

Đặt :

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

\(\Leftrightarrow a=bk;c=dk\)

\(VP=\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2\left(bk\right)^2-3bkb+5b^2}{2b^2+3bkb}=\dfrac{2b^2.k^2-2b^2.k+5b^2}{2b^2+3b^2k}=\dfrac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\left(1\right)\)

\(VT=\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}=\dfrac{2\left(dk\right)^2-3dkd+5d^2}{2\left(dk\right)^2+3dkd}=\dfrac{2.d^2.k^2-3d^2.k+5.d^2}{2.d^2.k^2+3d^2k}=\dfrac{d^2\left(2k^2-3k+5\right)}{d^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

Diện tích xung quanh phòng học là :

\(\left(8,5+6,4\right).2.3,5=104,3\left(m^2\right)\)

Diện tích trần nhà là :

\(8,5.6,4=54,4\left(m^2\right)\)

Diện tích các cửa là :

\(104,3.12\%=12,516\left(m^2\right)\)

Diện tích cần quét vôi là :

\(104,3+54,4-12,516=146,184\left(m^2\right)\)

Vậy ...

a, Ta có :

42*7 \(⋮\) 9

\(\Leftrightarrow\) 4 + 2 + * + 7 \(⋮9\) ( \(0\le\)*\(\le9\))

\(\Leftrightarrow13+\) * \(⋮9\)

Vậy .....

\(\Leftrightarrow\) * \(\in\left\{6\right\}\)

b,c tương tự

Gọi số tiền mỗi người nhận được là \(x;y;z\left(x;y;z\in N\right)\)

Theo bài ta có :

\(x+y+z=3280000\)

\(x:y:z=96:120:112=\dfrac{x}{12}=\dfrac{y}{15}=\dfrac{z}{14}\)

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

\(\dfrac{x}{12}=\dfrac{y}{15}=\dfrac{z}{14}=\dfrac{x+y+z}{12+15+14}=\dfrac{3280000}{41}=80000\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{12}=8000\Leftrightarrow x=960000\left(đ\right)\\\dfrac{y}{15}=80000\Leftrightarrow y=1200000\left(đ\right)\\\dfrac{z}{14}=80000\Leftrightarrow z=1120000\left(đ\right)\end{matrix}\right.\)

Vậy .....

Giả sử \(4n+3;2n+3\) chưa nguyên tố cùng nhau

\(\Leftrightarrow4n+3;2n+3\) có ước cũng là số nguyên tố

Gọi \(ƯCLN\left(4n+3;2n+3\right)=d\left(d\in N\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}4n+3⋮d\\2n+3⋮d\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4n+3⋮d\\4n+6⋮d\end{matrix}\right.\)

\(\Leftrightarrow3⋮d\)

Vì \(d\in N;3⋮d\Leftrightarrow d=1;3\)

+) \(d=3\Leftrightarrow2n+3⋮3\)

Mà \(3⋮3\)

\(\Leftrightarrow2n+6⋮3\)

\(\Leftrightarrow2\left(n+3\right)⋮3\)

\(\Leftrightarrow n+3⋮3\)

\(\Leftrightarrow n=3k\left(k\in N\right)\)

Khi \(n=3k\) thì \(4n+3=4\left(3k\right)+3=12k+3⋮3\)

Vậy \(n\in N;n\ne3k\) thì \(4n+3;2n+3\) là 2 số nguyên tố cùng nhau

\(3.8^x-2.2^{3x}-16=0\)

\(\Leftrightarrow3.\left(2^3\right)^x-2.2^{3x}=16\)

\(\Leftrightarrow3.2^{3x}-2^{3x}=16\)

\(\Leftrightarrow2^{3x}\left(3-1\right)=16\)

\(\Leftrightarrow2^{3x}.2=16\)

\(\Leftrightarrow2^{3x+1}=2^4\)

\(\Leftrightarrow3x+1=4\)

\(\Leftrightarrow x=1\)

Vậy ....