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a: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2001}{2000}=\dfrac{2001}{2}\)
b: \(=10.1\left(340+120-270\right)=10.1\cdot190=101\cdot19=1919\)
c: \(=24\%+8\%+59\%+9\%=1\)
1=ĐS=1
2=ĐS=2000
3=ĐS=0
4=ĐS=7036
K MÌNH NHA
a) A = 2 + 4 + 6 + 8 + ... + 1000
Ta có : A = 2 + 4 + 6 + 8 + ... + 1000 ( có 500 số )
= (1000 + 2) . 500 : 2 = 250500
c) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
Bài 1:
a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)
\(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{4}\)
\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) x \(\dfrac{4}{3}\)
\(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{3}\)
\(\dfrac{1}{4}x\) = \(\dfrac{1}{3}\) + \(\dfrac{1}{8}\)
\(\dfrac{1}{4}\) \(x\)= \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)
\(\dfrac{1}{4}x=\dfrac{11}{24}\)
\(x=\dfrac{11}{24}:\dfrac{1}{4}\)
\(x=\dfrac{11}{24}\times4\)
\(x=\dfrac{11}{6}\)
b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)
\(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)
\(x\) = \(\dfrac{9}{10}\)
a).( x - 1/2 )+( x - 1/4 ) = 1
x + x - 1/2 - 1/4 = 1
2x = 1/2 + 1/4 + 1
2x = 3/4 + 1
2x = 7/4
x = 7/4 : 2
x = 1/2
mik chỉ làm được như thế thôi xin lỗi bạn nhé
1a)6,28x18,24+18,24x3,72 d)0,9x95+1,8x2+0,9
=18,24x(6,28+3,72) = 0,9x95+0,9x4+0,9
=18,24x10 =0,9(95+4+1)
=182,4 =0,9x100=90
b) 35,7x99+35+0,7 e)0,25x611,7x40
=35,7x99+35,7 =(0,25x40)x611,7
=35,7x(99+1) =10x611,7
=35,7x100 =6117
=357 g)37,2x101-37-0,2
c)17,34x99+18-0,66 = 101x37,2-(37+0,2)
=17,34x99+17,34 =101x37,2-37,2
=17,34x(99+1) =37,2x(101-1)
=17,34x100 =37,2x100
=1734 =3720
Bạn ơi đăng từng bài thôi mk giải cho. Nhiều quá nhìn hoa mắt mặc dù bài rất dễ!!!
A. \(\left(x+1\right)+\left(x+2\right)+......+\left(x+100\right)=5750\)
\(x+1+x+2+....+x+100=5750\)
\(100x+\left(1+2+3+.......+100\right)=5750\)
\(100x+5050=5750\)
\(100x=700\)
\(x=700:100=7\)
B. x+(1+2+......+100) = 2000
x + 5050 = 2000
x = 2000 - 5050
x= -3050
C. ( x-1 )+(x-2)+......+( x - 100 ) = 50
x-1+x-2+.........+x-100 = 50
100x + ( -1-2-........-100 ) = 50
100x + ( - 5050 ) = 50
100x = 50 + 5050
100 x = 5100
x = 5100 : 100
x = 51
A . \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(\Rightarrow x=\frac{700}{100}=7\)
B. \(x+\left(1+2+3+4+5+....+100\right)=2000\)
\(x+\frac{\left(100+1\right).100}{2}=2000\)
\(x+5050=2000\)
\(\Rightarrow x=2000-5050=-3050\)
C. \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+....+\left(x-100\right)=50\)
\(\left(x+x+x+...+x\right)-\left(1+2+3+...+100\right)=50\)
\(100x-5050=50\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)
a: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2001}{2000}=\dfrac{2001}{2}\)
b: \(=101\left(34+13-27\right)=101\cdot20=2020\)
c: \(=24\%+8\%+59\%+9\%=1\)