\(F=5\frac{6}{4453}.\frac{1}{1997}-\frac{2}{1997}.2\frac{3}{4453}\)
Tính nhanh
F = \(5\frac{6}{4453}x\frac{1}{1997}-\frac{2}{1997}x2\frac{3}{4453}\)
M = \(\frac{1}{4587}x7\frac{1}{3897}-3\frac{4586}{4587}x\frac{2}{3897}-\frac{7}{4587}-\frac{3}{4587x3897}\)
x là dấu nhân
\(5\frac{6}{4453}\)x \(\frac{1}{1997}\)- \(\frac{2}{1997}\)x \(2\frac{3}{4453}\)
Đặt \(a=\frac{1}{4453};b=\frac{1}{1997}\)ta có :
\(5\frac{6}{4453}\cdot\frac{1}{1997}-\frac{2}{1997}\cdot2\frac{3}{4453}\)
\(=\left(5+6a\right)\cdot b-2b\left(2+3a\right)\)
\(=5b+6ab-4b-6ab\)
\(=b=\frac{1}{1997}\)
\(F=5\dfrac{6}{4453}.\dfrac{1}{1997}-\dfrac{2}{1997}.2\dfrac{3}{4453}\)
rút gọn bằng cách thay số bằng chữ
Đặt a=4453, b=1997
Ta có: \(F=5\dfrac{6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot2\dfrac{3}{a}\)
\(=\dfrac{5a+6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\dfrac{2a+3}{a}\)+
\(=\dfrac{5a+6-4a-6}{ab}\)
\(=\dfrac{1}{b}\)
\(=\dfrac{1}{1997}\)
\(5\dfrac{6}{4453}.\dfrac{1}{1997}-\dfrac{2}{1997}.2\dfrac{3}{4453}\)
Đặt 4453=a; 1997=b
\(A=\left(5+\dfrac{6}{a}\right)\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\left(2+\dfrac{3}{a}\right)\)
\(=\dfrac{5a+6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\dfrac{2a+3}{a}\)
\(=\dfrac{5a+6-4a-6}{ab}=\dfrac{a}{ab}=\dfrac{1}{b}=\dfrac{1}{1997}\)
Tính :
1 . F = 5\(\dfrac{6}{4453}\). \(\dfrac{1}{1997}\)- \(\dfrac{2}{1997}.2\dfrac{3}{4453}\)
2 . G = 4\(\dfrac{7}{5741}\).\(\dfrac{1}{3759}\) - \(\dfrac{4}{3759}\). 1\(\dfrac{2}{5741}\)+ \(\dfrac{1}{3759}\)+ \(\dfrac{1}{3759.5741}\)
\(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)
Ta có : \(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}\)\(=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)
\(\Rightarrow\left(\frac{x-1991}{9}-1\right)+\left(\frac{x-1993}{7}-1\right)+\left(\frac{x-1995}{5}-1\right)+\left(\frac{x-1997}{3}-1\right)+\left(\frac{x-1999}{1}-1\right)\)
\(=\left(\frac{x-9}{1991}-1\right)+\left(\frac{x-7}{1993}-1\right)+\left(\frac{x-5}{1995}-1\right)+\left(\frac{x-3}{1997}-1\right)+\left(\frac{x-1}{1999}\right)\)
\(\Rightarrow\frac{x-2000}{9}+\frac{x-2000}{7}+\frac{x-2000}{5}+\frac{x-2000}{3}\)
\(=\frac{x-2000}{1991}+\frac{x-2000}{1993}+\frac{x-2000}{1995}+\frac{x-2000}{1997}+\frac{x-2000}{1999}\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)=0\)
\(\Rightarrow\left(x-2000\right)\left[\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\right]=0\)
Vì \(\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\ne0\)
=> x - 2000 = 0
=> x = 2000
Tìm x :
a) \(\frac{x+1}{2000}+\frac{x+2}{1999}+\frac{x+ 3}{1998}+\frac{x+4}{1997}=-4\)
\(b.\frac{x+1}{1999}+\frac{x+2}{2000}+\frac{x+3}{2001}=\frac{x+4}{2002}+\frac{x+5}{2003}+\frac{x+6}{2004}\)
\(a.\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\left(\frac{x+4}{1997}+1\right)=0\)
\(=\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)
\(=\left(x+2001\right).\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)
\(=>x+2001=0\)
\(x=-2001\)
\(b.\left(\frac{x+1}{1999}-1\right)+\left(\frac{x+2}{2000}-1\right)+\left(\frac{x+3}{2001}-1\right)=\left(\frac{x+4}{2002}-1\right)+\left(\frac{x+5}{2003}-1\right)\)\(+\left(\frac{x+6}{2004}-1\right)\)
\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}=\frac{x+1998}{2002}+\frac{x+1998}{2003}+\frac{x+1998}{2004}\)
\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}-\frac{x+1998}{2002}-\frac{x+1998}{2003}-\frac{x+1998}{2004}=0\)
\(\left(x+1998\right).\left(\frac{1}{1999}+\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)
\(=>x+1998=0\)
\(x=-1998\)
dễ quá!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\(\frac{x+1}{2000}+\frac{x+2}{1999}+\frac{x+3}{1998}+\frac{x+4}{1997}=-4\)
\(\Leftrightarrow\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\) \(\left(\frac{x+4}{1997}+1\right)=0\)
\(\Leftrightarrow\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)
\(\Leftrightarrow\left(x+2001\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)
Mà : \(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\ne0\)
\(\Rightarrow x+2001=0\)
\(\Leftrightarrow x=-2001\)
tính nhanh:
B=\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)
A=\(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
\(B=\)\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)
\(B=\frac{3.1+3.11+3.111+3.1111+3.11111}{4.1+4.11+4.111+4.1111+4.11111}\)
\(B=\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)
\(B=\frac{3}{4}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(A.2=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right).2\)
\(A.2=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
=>\(A.2-A=\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right)\)
\(A=\frac{2}{3}-\frac{1}{192}\)
\(A=\frac{127}{192}\)
\(\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
Đặt \(C=\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
\(C=\frac{1995.1990.1997.1993.997}{1997.1993.1994.1995.995}\)
\(C=\frac{1990.997}{1994.995}\)
\(C=\frac{995.2+997}{997.2+995}=1\)
\(B=\frac{3+33+333+3333+ 33333}{4+44+444+4444+44444}\)
\(\Rightarrow B=\frac{3\left(1+11+111+1111+11111\right)}{4\left(1+11+111+1111+11111\right)}=\frac{3}{4}\)
B=\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)
B= \(\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)
B=\(\frac{3}{4}\)
Sau mình làm tiếp vội quá! k mình nha
Tìm n thuộc N thõa : \(1+\frac{1}{3}+\frac{1}{6}+\frac{2}{n\left(n+1\right):2}=1\frac{1997}{1998}\)