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a)(3/2-0,5)/x=7/2+1/4
(3/2-1/2)/x=14/4+1/4
1/x=15/4
x=1:15/4
x=4/15
b)(x*0,25+2010)*2011=(53+2010)*(2012-1)
(x*0,25+2010)*2011=2063*2011
=>0,25x+2010=2063
0,25x=2063-2010
0,25x=53
x=53/0,25
x=212
a) (x + 1) + (x + 2) + (x + 3) + ... + (x + 9) = 54
(x + x + x + ... + x) + (1 + 2 + 3 + ... + 9) = 54
9x + 45 = 54
9x = 54 - 45
9x = 9
x = 1
b) (x + 1) + (x + 2) + (x + 3) + ... + (x + 10) = 2010
(x + x + x + ... + x) + (1 + 2 + 3 + ... + 10) = 2010
10x + 55 = 2010
10x = 2010 - 55
10x = 1955
x = 391/2.
1)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+....+\left(x+9\right)=54\)
\(\Rightarrow x+1+x+2+x+3+.....+x+9=54\)
\(\Rightarrow\left(x+x+x+....+x\right)+\left(1+2+3+....+9\right)=54\)
\(\Rightarrow9x+45=54\)\(\Rightarrow9x=9\Rightarrow x=1\)
2)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+....+\left(x+10\right)=2010\)
\(\Rightarrow x+1+x+2+x+3+.....+x+10=2010\)
\(\Rightarrow\left(x+x+x+....+x\right)+\left(1+2+3+....+10\right)=2010\)
\(\Rightarrow10x+55=2010\Rightarrow10x=1995\Rightarrow x=195,5\)
a) x + 3.x + 5.x + ... + 2009. x = 2010.1005
\(x\times\left(1+3+5+...+2009\right)=2010\times1005\)
\(x\times\left[\left(1+2009\right)\times1005:2\right]=2010\times1005\)
\(x\times2010\times1005\times\frac{1}{2}=2010\times1005\)
\(\Rightarrow x\times\frac{1}{2}=2010\times1005:\left(2010\times1005\right)\)
\(x\times\frac{1}{2}=1\)
x = 2
b) x + (x+1) + (x+2) +...+ (x+30) = 620
x. 31 + ( 1+2+...+30) = 620
x.31 + [ ( 30+1).30:2) = 620
x.31 + 465 = 620
x.31 = 620 - 465
x.31 = 155
x = 155 : 31
x = 5
a) x+3.x+5.x+.....+2009.x = 2010.1005
=> x.(1+3+5+....+2009) = 2010.1005
=> x.1010025 = 2020050
=> x = 2
Vậy x = 2
b) x+(x+1)+(x+2)+....+(x+30) = 620
=> (x+x+x+...+x)+(1+2+3+...+30) = 620
=> 31x + 465 = 620
=> 31x = 155
=> x = 5
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{9}{10}=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot....\cdot9}{2\cdot3\cdot4\cdot....\cdot10}=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1}{10}=\frac{x}{2010}\)
\(\Leftrightarrow x=\frac{2010}{10}=201\)
Ta có : \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)
=> \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{9}{10}=\frac{x}{2010}\)
\(\Rightarrow\frac{1.2.3......9}{2.3.4.....10}=\frac{x}{2010}\)
\(\Rightarrow\frac{1}{10}=\frac{x}{2010}\)
\(\Rightarrow x=\frac{2010}{10}=201\)
\(=\frac{1}{10}\)
\(\frac{2}{5}\cdot\frac{1}{2}-\frac{2}{5}\cdot\frac{1}{3}-\frac{2}{5}\cdot\frac{1}{6}\)
\(=\frac{2}{5}\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\frac{2}{5}\cdot\frac{0}{6}=\frac{2}{5}\cdot0=0\)
tìm x
a) \(0,5+\left(x-\frac{15}{2}\right):\frac{1}{2}=\frac{9}{2}\)
\(\Rightarrow x=\frac{19}{2}=8,5\)
b) \(2012\cdot x-2010\cdot x=2014\)
\(\Leftrightarrow\left(2012-1010\right)\cdot x=2014\)
\(\Leftrightarrow2x=2014\)
\(\Rightarrow x=\frac{2014}{2}=1007\)
hok tốt .
Tính nhanh:
2/5 x 1/2 - 2/5 x 1/3 - 2/5 x 1/6
= 2/5 x ( 1/2 - 1/3 - 1/6 )
= 2/5 x ( 3/6 - 2/6 - 1/6 )
= 2/5 x 0/6
=0
a)
x + (x + 1) + (x + 2) + ... + (x + 2010) = 2029099
=> (x + x + x + ... + x) + (1 + 2 + ... + 2010) = 2029099
có 2011 số x có 2010 số hạng
=> 2011x + (1 + 2010) . 2010 : 2 = 2029099
=> 2011x + 2011 . 2010 : 2 = 2029099
=> 2011x + 2021055 = 2029099
=> 2011x = 2029099 - 2021055
=> 2011x = 8044
=> x = 8044 : 2011
=> x = 4
Vậy x = 4
b, 2+4+6+8+....+2x=210
2.1 + 2.2 + 2.3 + 2.4 + ...........+ 2x = 210
2 . ( 1+ 2 +3+.....+x ) = 210
1+2+3+.......+x = 105 ( vế trái có x số hạng )
(1+x).x : 2 = 105
( 1+x).x = 105.2
(1+x) . x = 210
(1+x).x=15.14
=> x = 14
Vậy x= 14
Ta có : \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{9}{10}=\frac{x}{2010}\)
=> \(\frac{1.2.3.....9}{2.3.4....10}=\frac{x}{2010}\)
=> \(\frac{1}{10}=\frac{x}{2010}\)
=> x = 2010/10
=> x = 201
(2% x X -1) +2 = 0,2 : 1/10
(0,02 x X -1) + 2 =0.2 :0.1=2
(0.02 x X -1) = 2-2=0
0.02x X = 0+ 1 =1
1 : 0.02 = 50.
Thử lại :(2% x 50 - 1) + 2 =0.2 : 1/10 ( cả 2 biểu thức đều bằng 2)
b)ta coi biểu thức đầu(1 x2 x3 x........x2010) là A. Ta có :
A x (x -2010)
vì bất cứ số nào nhân với 0 cũng bằng 0 nên biểu thức chứa x phải có kết quả là 0.
x = 0 +2010 =2010