x+2^3=667-1000
Đặt tính rồi tính.
2 667 + 3 825 8 274 - 4 516 324 x 14 74 165 : 5
x-1/2000+x-3/1998+x-5/1996+x/667=6
Tìm x:
1/1.4 + 1/4.7 + 1/7.10 + ... + 1/x.(x+3)= 667/2002
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{667}{2002}\)
\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{667}{2002}\)
\(=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{667}{2002}\)
\(=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{x+3}\right)=\frac{667}{2002}\)
\(\frac{1}{1}-\frac{1}{x+3}=\frac{667}{2002}:\frac{1}{3}\)
\(\frac{1}{1}-\frac{1}{x+3}=\frac{2001}{2002}\)
\(\frac{1}{x+3}=1-\frac{2001}{2002}\)
\(\frac{1}{x+3}=\frac{1}{2002}\)
\(\frac{1}{x}=\frac{1}{2002-3}\)
\(\frac{1}{x}=\frac{1}{1999}\)
Vậy x = 1999
Tìm x:
1/1.4 + 1/4.7 + 1/7.10 + ... + 1/x.(x+3)= 667/2002
Chứng tỏ rằng giá trị của biểu thức Q là một số nguyên
Q=(667-1/7-2/8-3/9-...-667/673):(1/21+1/24+1/27+...+1/2019)
. Tìm x biết :
a) 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/x.(x+1) = 667/668
b) 1/1.4 + 1/4.7 + 1/7.10 + ... + 1/x.(x+3) = 667/2002
c) 1/3 + 1/6 + 1/10 + 1/15 + ... + 1/x.(2x+1) = 1999/2001
lẹ giùm mình T.T mình sắp đi học ròii nên ai nhanh mình tick
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{667}{668}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{667}{668}\)
\(1-\frac{1}{x+1}=\frac{667}{668}\)
\(\frac{1}{x+1}=1-\frac{667}{668}\)
\(\frac{1}{x+1}=\frac{1}{668}\)
\(\Rightarrow x+1=668\)
x = 667
a) 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/x.(x+1) = 667/668
=>1/1-1/2+1/2-1/3+1/3-1/4+.......+1/x-1/x+1=667/668
=>1/1-1/x+1=667/668
=>1/x+1=1/1-667/668
=>1/x+1=1/668
=>x=667
b) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{667}{2002}\)
\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{667}{2002}\)
\(\frac{1}{3}.\left(1-\frac{1}{x+3}\right)=\frac{667}{2002}\)
\(1-\frac{1}{x+3}=\frac{2001}{2002}\)
\(\frac{1}{x+3}=\frac{1}{2002}\)
=> x + 3 = 2002
x = 1999
3 bn lm nhanh nhất mik sẽ tick cho :D
Tìm X
x-1/2000 + x-3/1998 + x-5/1996 + x/667 = 6
\(\frac{x-1}{2000}+\frac{x-3}{1998}+\frac{x-5}{1996}+\frac{x}{667}=6\)
\(\Rightarrow\frac{x-1}{2000}+\frac{x-3}{1998}+\frac{x-5}{1996}+\frac{x}{667}-6=0\)
\(\Rightarrow\left(\frac{x-1}{2000}-1\right)+\left(\frac{x-3}{1998}+1\right)+\left(\frac{x-5}{1996}-1\right)+\left(\frac{x}{667}-3\right)=0\)
\(\Rightarrow\frac{x-1-2000}{2000}+\frac{x-3-1998}{1998}+\frac{x-5-1996}{1996}+\frac{x-3.667}{667}=0\)
\(\Rightarrow\frac{x-2001}{2000}+\frac{x-2001}{1998}+\frac{x-2001}{1996}+\frac{x-2001}{667}=0\)
\(\Rightarrow\left(x-2001\right)\left(\frac{1}{2000}+\frac{1}{1998}+\frac{1}{1996}+\frac{1}{667}\right)=0\)
Ta có: \(\frac{1}{2000}+\frac{1}{1998}+\frac{1}{1996}+\frac{1}{667}\ne0\)
\(\Rightarrow x-2001=0\Rightarrow x=2001\)
667 - [ 44 - x ] =67
667 - 44 - x = 67
623 - x = 67
x = 623 - 67
x = 556
\(667-\left(44-x\right)=67\)
\(=>x=-556\)
667 - [ 44 - x ] =67
[ 44 - x ] = 667 - 67
44 - x = 600
x = 44 - 600
x = - 556
XX3 +3X1+X = 667