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= (1988 x 1996 + 1997 x 11 + 1985) : (1997 x 1996 - 1995 x 1996)
= (1988 x 1996 + (1996 + 1) x 11 + 1985) : (1996 x (1997 - 1995))
= (1988 x 1996 + 1996 x 11 + (11 + 1985)) : (1996 x 2)
= (1996 x (1988 + 11) + 1996 : (1996 x 2)
= (1996 x 1999 + 1996) : (1996 x 2)
= (1996 x 1999 + 1996 x 1) : (1996 x 2)
= (1996 x (1999 + 1)) : (1996 x 2)
= 1996 x 2000 : 1996 x 2
= 2000 : 2
= 1000.
ta có :
\(\frac{1998x1996+1997x11+1985}{1997x1996-1995x1996}\)= \(\frac{1998x1996+1996x11+11+1985}{1996x\left(1997-1995\right)}\)= \(\frac{1996x\left(1998+11+1\right)}{1996x2}\)=\(\frac{2010}{2}=1005\)
vậy giá trị cần tìm là 1005
\(\frac{1998\times1996+1997+1995}{1996\times\left(1997-1995\right)}=\frac{1998\times1996+1996+1996}{1996\times2}=\frac{1996\times1999+1996}{1996\times2}\)
=\(\frac{1996\times2000}{1996.2}=\frac{2000}{2}=1000\)
Ta có: \(1\frac{4}{5}+2\frac{5}{7}+3\frac{4}{5}+4\frac{5}{7}\)
\(=\left(1\frac{4}{5}+3\frac{4}{5}\right)+\left(2\frac{5}{7}+4\frac{5}{7}\right)\)
\(=\left(\frac{9}{5}+\frac{19}{5}\right)+\left(\frac{19}{7}+\frac{33}{7}\right)\)
\(=\frac{28}{5}+\frac{52}{7}=13\frac{1}{35}\)
= ( \(1\frac{4}{5}\)+ \(3\frac{4}{5}\)) + ( \(2\frac{5}{7}\)+ \(4\frac{5}{7}\))
= \(4\frac{4}{5}\) + \(6\frac{5}{7}\)
= \(\frac{24}{5}\) + \(\frac{47}{7}\)
= ...... ( tính nốt nhé )
p=12342 x34566+23453
12342 x(34566+1)+11111
p=12342 x34566+23453
12342 x34566+12342 x1+11111
p=23453
12342+11111
p=23453
23453
p=1
\(B=\)\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)
\(B=\frac{3.1+3.11+3.111+3.1111+3.11111}{4.1+4.11+4.111+4.1111+4.11111}\)
\(B=\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)
\(B=\frac{3}{4}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(A.2=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right).2\)
\(A.2=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
=>\(A.2-A=\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right)\)
\(A=\frac{2}{3}-\frac{1}{192}\)
\(A=\frac{127}{192}\)
\(\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
Đặt \(C=\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
\(C=\frac{1995.1990.1997.1993.997}{1997.1993.1994.1995.995}\)
\(C=\frac{1990.997}{1994.995}\)
\(C=\frac{995.2+997}{997.2+995}=1\)
\(B=\frac{3+33+333+3333+ 33333}{4+44+444+4444+44444}\)
\(\Rightarrow B=\frac{3\left(1+11+111+1111+11111\right)}{4\left(1+11+111+1111+11111\right)}=\frac{3}{4}\)
1988/1996
Hok tốt !!
=\(\frac{1988x1996+1996+1996}{1996x\left(1997-1995\right)}\)=\(\frac{1996x\left(1988+2\right)}{1996x2}\)
=\(\frac{1996x1990}{1996x2}\)=\(\frac{1990}{2}\)=\(995\)
Hello!!!