ai giúp mình giải bài này giùm :23.(-5)2.(-125)./-4/.(-2)3
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5/23 : ( 3/26 + 7/9) - 5/23 : (23/26 + 2/9)
=\(\frac{5}{23}\)\(\div[(\frac{3}{26}+\frac{7}{9})-(\frac{23}{26}+\frac{2}{9})]\)
=\(\frac{5}{23}\div\)\([(\frac{3}{26}+\frac{23}{26})-(\frac{7}{9}+\frac{2}{9})]\)
=\(\frac{5}{23}\div[1-1]\)
=\(\frac{5}{23}\div0\)
=0
5/23 : ( 3/26 + 7/9 ) - 5/23 : ( 23/26 + 2/9 )
=5/23 : { (23/26 + 3/26) - ( 7/9 + 2/9 )
=5/23 : { 1-1 }
=5/23 : 0
=0
a/ -137.45+ (-45).(-37)
= -6165 + 1665
= - 4500
b/(-18+98):5-(-4)^2
=80 : 5 - (-16)
= 16 - (-16)
= 16 + 16
= 32
c/ (578-950)-(-950+578-23)
= 578 - 950 - 950 - 578 + 23
= { 578 - 578 } - {950 - 950} + 23
= 0 - 0 + 23
= 0 + 23
= 23
d/9-x : (-2) = -15
9 - x = -15 . (-2)
9 - x = 30
x = 9 - 30
x = -21
3/2 . x + ( 5/3 - 3/2) : 2/3 = 5/3
3/2.x + 1/6 : 2/3 = 5/3
3/2.x + 1/4 = 5/3
3/2.x = 5/3 - 1/4
3/2.x=17/12
x= 17/12 : 3/2
x= 17/18
Vậy...
Bài 2:
4/5x7 + 4/7x9 + 4/9x11 +...+4/17x19
= 2(2/5.7 + 2/7.9 + 2/9.11+...+ 2/17/19)
= 2( 1/5 - 1/7 + 1/7 -1/9 + 1/9 -1/11 +...+ 1/17 - 1/19)
= 2( 1/5- 1/19)
= 2 . 14/95
= 28/95
Trả lời:
Bài 1
\(\frac{3}{2}\times x+\left(\frac{5}{3}-\frac{3}{2}\right)\div\frac{2}{3}=\frac{5}{3}\)
\(\Leftrightarrow\frac{3}{2}\times x+\frac{1}{6}\div\frac{2}{3}=\frac{5}{3}\)
\(\Leftrightarrow\frac{3}{2}\times x+\frac{1}{6}\times\frac{3}{2}=\frac{5}{3}\)
\(\Leftrightarrow\frac{3}{2}\times x+\frac{1}{4}=\frac{5}{3}\)
\(\Leftrightarrow\frac{1}{6}\times x=\frac{17}{12}\)
\(\Leftrightarrow x=\frac{17}{2}\)
Vậy \(x=\frac{17}{2}\)
2x - 3 = 54 : 52
2x - 3 = 54 - 2 = 52 = 25
2x = 25 + 3
2x = 28
x = 28 : 2
x = 14
5x + 2 = 53 . 125
5x + 2 = 53 . 53 = 53 + 3 = 56
=> x = 6 - 2
x = 4
Ta có: \(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{2011^2}\)
\(=\left(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}\right)+\left(\frac{1}{8^2}+\frac{1}{9^2}+\frac{1}{10^2}+...+\frac{1}{2011^2}\right)\)
\(>\frac{1}{3^2}+\left(\frac{1}{7^2}+\frac{1}{7^2}+\frac{1}{7^2}+...+\frac{1}{7^2}\right)\)(2007 phân số \(\frac{1}{7^2}\))
\(=\frac{1}{3^2}+\left(\frac{1.2007}{7^2}\right)=\frac{1}{3^2}+\frac{2007}{7^2}>\frac{125}{503}^{\left(đpcm\right)}\)
Đặt S= 1/4^2+1/5^2=1/6^2+...+1/2011^2
Ta có: 1/3.4>1/4^2
1/4.5>1/5^2
.........
1/2010.2011>1/2011^2
Suy ra: S>1/3.4+1/4.5+1/5.6+...+1/2010.2011
S>1/3 -1/4+1/4-1/5+...+1/2010-1/2011
S>1/3-1/2011
S>2008/6033>125/503
từ đó suy ra S.125/503
k cho mình nha
1) 4*x-0.25=512
4*x = 512+0,25
4*x= 512,25
x=512.25:4=128,0625
nhìu quá ko làm nữa