Tìm x
x-6/1979+x-4/1981+x-2/1983=x+2/1987+x+4/1989+x+6/1991
Những câu hỏi liên quan
x-6/1979+x-4/1981+x-2/1983=x+2/1987+x+4/1889+x+6/1991
(1991 + 1992 + ... + 1999) - ( 11 + 12 + ... +19) =b. 21 x 23 x 25 x 27 - 11 x 13 x 15 x17 =16358 - 6 x 16 x 46 x 56 = Hãy cho biết chữ số tận cùng của mỗi kết quả sau :1981 + 1982 + 1983 + ... + 1989A.6 B.7 C.9 D.5
Xem chi tiết
Tìm x ; y biết : x - y - 10 và x/y = 7/5
Tính nhanh : 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 +1/128
Tính : 1993 - 1981 + 1989 - 1987 ..... + 5 - 3 + 1
1+1/3+1/6+1/10+.......+2/Xx(X+1)=1và 1989/1991
ta có
\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}\) \(=\)\(1+2\)\(\left(\frac{1}{2}-\frac{1}{3}\right)+2\left(\frac{1}{3}-\frac{1}{4}\right)+...+2\left(\frac{1}{x}-\frac{1}{x+1}\right)\)\(=2-\frac{2}{x+1}\)
Nên ta có
\(2-\frac{2}{x+1}=1+\frac{1989}{1991}\Leftrightarrow\frac{2}{x+1}=\frac{2}{1991}\Leftrightarrow x=1990\)
tìm x biết :
2+2/3+2/6+2/12+....+2/x*(x+1) = 1+1989/1991
2+2/3+2/6+...+2/x*(x+1)=1*1989/1991
2+ 2/3+2/6+2/12+...+2/x(x+1)= 1 1989/ 1991
Tìm x
\(2+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+......+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)
\(2+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)
\(2\left(1+\frac{1}{3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=1\frac{1989}{1991}\)
\(2\left(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=1\frac{1989}{1991}\)
\(2\left(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{x+1}\right)=1\frac{1989}{1991}\)
\(\frac{8}{3}+2-\frac{2}{x+1}=1\frac{1989}{1991}\)
\(\frac{2}{x+1}=\frac{13}{10}\)( số thập phân dài quá nên mk lấy số tròn thôi nha )
\(x+1=2:\frac{13}{10}\)
\(x+1=\frac{20}{13}\)
\(\Leftrightarrow x=\frac{7}{13}\)
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Tìm x biết \(\frac{x-1}{1995}-\frac{x+3}{1991}=\frac{x+7}{1987}-\frac{x+11}{1983}\)
=> \(\frac{x-1}{1995}+1-1-\frac{x+3}{1991}=\frac{x+7}{1987}+1-1-\frac{x+11}{1983}\)
=> \(\left(\frac{x-1}{1995}+1\right)-\left(1+\frac{x+3}{1991}\right)=\left(\frac{x+7}{1987}+1\right)-\left(1+\frac{x+11}{1983}\right)\)
=> \(\frac{x+1994}{1995}-\frac{x+1994}{1991}=\frac{x+1994}{1987}-\frac{x+1994}{1983}\)
=> \(\left(x+1994\right)\left(\frac{1}{1995}-\frac{1}{1991}-\frac{1}{1987}+\frac{1}{1983}\right)=0\)
=>x + 1994 = 0 Vì \(\left(\frac{1}{1995}-\frac{1}{1991}-\frac{1}{1987}+\frac{1}{1983}\right)\ne0\)
=> x = -1994
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