1 .x+5  và 2y+1 là Ư(42) lập bảng tính

2.vd tc chia hết 

\(\left(x-5\right)^6=\left(x-5\right)^8\)

\(\Leftrightarrow\left(x-5\right)^6\left[1-\left(x-5\right)^2\right]=0\)

\(\Leftrightarrow\left(x-5\right)^6\left(x-4\right)\left(6-x\right)=0\Leftrightarrow x=4;x=5;x=6\)

Ta có: 6 = 2.3

       20 = 2² .5

      15 = 3.5

=> BCNN(6; 20; 15) = 2². 3.5 = 60

=> BC(6; 20; 15) = B(60) = {0; 60; 120; 180; 240; 300; ...} 

=> (x + 1) \(\in\){0; 60; 120; 180; 240; 300; ...}

=> x \(\in\){-1; 59; 119; 179; 239; 299; ...}

Do 0 \(\le\)x \(\le\)300

=> x \(\in\){59; 119; 179; 239; 299}

Bài 2:

10^n có tổng các chữ số là 1

5^3 có tổng các chữ số là 8

=>10^n+5^3 có tổng các chữ số là 9

=>10^n+5^3 chia hết cho 9

thiếu đề

Thiếu đề nha bn

hình như tự chế

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}\div2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}\)

\(\Rightarrow\frac{1}{x+1}=\frac{2}{4018}=\frac{1}{2009}\)

\(\Rightarrow x+1=2009\)

\(\Rightarrow x=2008\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

=>\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{4018}\)(nhân cả hai vế với \(\frac{1}{2}\))

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)= \(\frac{2007}{4018}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\frac{1}{x+1}\)=\(\frac{1}{2}-\frac{2007}{4018}\)

\(\frac{1}{x+1}=\frac{1}{2009}\)

x+1=2009

x=2009-1=2008

Vậy x bằng 2008

x=2008

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2005}\)

\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)

\(2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)

\(2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)

\(=>2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4008}{2005}\)

\(2.\left(1-\frac{1}{x+1}\right)=\frac{4008}{2005}\)

=> \(1-\frac{1}{x+1}=\frac{4008}{2005}:2=\frac{2004}{2005}\)

\(\frac{1}{x+1}=1-\frac{2004}{2005}=\frac{1}{2005}\)

=>x+1=2005

=>x=2004

1/3 + 1/6 + 1/10 +...+ 2/x(x+1) = 2014/2015

Đ/A là 2004

chúc đồng chí Chế Minh Hải học tốt

Ta có: 1/3+1/6+1/10+...+2/x*(x+1)

=2/6+2/12+2/20+...+2/x*(x+1)

=2/2*3+2/3*4+2/4*5+...+2/x*(x+1)

=2*(1/2*3+1/3*4+1/4*5+...+1/x*(x+1))

=2*(1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1)

=2*(1/2-1/x+1)=2000/2002

=>1/2-1/x+1=2000/2002:2

=>1/2-1/x+1=500/1001

=>1/x+1=1/2-500/1001

=>1/x+1=1/2002

=>x+1=2002

=>x=2002-1

=>x=2001 thuộc N

Vậy x=2001

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