a/ (X+1)/35+1+(x+3)/33+1 =(x+5)/31+(x+7)/29+1+1
=>(x+36)/35+(x+36)/33-(x+36)/31-(x+36)/27=0
=>(X+36)(1/35+1/33-1/31-1/29)=0
=> x+36=0(vì c=vế 2 luôn luôn khác 0)
=>x=-36
b/ CMTT câu a
trừ tung phân số cho 1 ta được x=2004
x-10/1994+X-8/1996+X-6/1998/+ X-4/2000+X-2/2002=X-2002/2+X-2000/4+X-1998/6+X-1996/8+X-1994/10
Bài 1: Tìm x
a) \(\frac{1}{4}+\frac{3}{4}:x=\frac{5}{8}\)
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{x}{x\left(x+1\right)}=\frac{2000}{2002}\)
Bài 2: Chứng tỏ
\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}< \frac{1}{2}\)
Mình cần gấp lắm mấy ban ưi ~~~
Bài 1: Tìm x
a) \(\frac{1}{4}+\frac{3}{4}:x=\frac{5}{8}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)
b) Bài 2: Chứng tỏ
\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}< \frac{1}{2}\)
Mình cần gấp lắm mấy ban ưi ~~~
a)\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\)\(\frac{x+1}{6}\)
b)\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\)\(\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
c)\(2.2^2.2^3.2^4.........2^x=1024\)\(\left(x\in z\right)\)
Tìm x
tìm số tự nhiên x biết rằng
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)
Tìm x
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\frac{x-5}{2000}+\frac{x-4}{2001}+\frac{x-3}{2002}=\frac{x-2}{2003}+\frac{x-1}{2004}+\frac{x}{2005}\)
Bài \(1:\)TÌM \(x:\)
\(a,\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{2013}\)
\(b,\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(c,\frac{x+5}{205}+\frac{x+4}{204}+\frac{x+3}{203}=\frac{x+166}{366}+\frac{x+167}{367}+\frac{x+168}{368}\)
\(d,\) \(x.\)\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}{\frac{2011}{1}+\frac{2011}{2}+\frac{2010}{3}+\frac{2009}{4}+...+\frac{2}{2011}+\frac{1}{2012}}=1\)
tìm số tự nhiên x biết
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)
