125,5 : 5 + x = 37,86
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Ta có: \(x+18,26:2=112,87\)
=>x+9,13=112,87
=>x=112,87-9,13=103,74

Ta có: \(x+3,25\cdot8,2=38,89\)
=>x+26,65=38,89
=>x=38,89-26,65=12,24

Ta có: x(x+3)=6
=>\(x^2+3x-6=0\)
=>\(x^2+3x+\frac94-\frac{33}{4}=0\)
=>\(\left(x+\frac32\right)^2-\frac{33}{4}=0\)
=>\(\left(x+\frac32\right)^2=\frac{33}{4}\)
=>\(x+\frac32=\pm\frac{\sqrt{33}}{2}\)
=>\(x=\pm\frac{\sqrt{33}}{2}-\frac32\)

\(-\frac{313}{370}=\frac{-370+57}{370}=-1+\frac{57}{370}\)
\(-\frac{314}{371}=\frac{-371+57}{371}=-1+\frac{57}{371}\)
mà \(\frac{57}{370}>\frac{57}{371}\left(370<371\right)\)
nên \(-\frac{313}{370}>-\frac{314}{371}\)
−370313=370−370+57=−1+37057
\(- \frac{314}{371} = \frac{- 371 + 57}{371} = - 1 + \frac{57}{371}\)
mà \(\frac{57}{370} > \frac{57}{371} \left(\right. 370 < 371 \left.\right)\)
nên \(- \frac{313}{370} > - \frac{314}{371}\)

Ta có: \(\frac{x}{20}+\frac{x}{30}+\frac{x}{42}=3\)
=>\(x\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=3\)
=>\(x\left(\frac14-\frac15+\frac15-\frac16+\frac16-\frac17\right)=3\)
=>\(x\left(\frac14-\frac17\right)=3\)
=>\(x\cdot\frac{3}{21}=3\)
=>\(x=3:\frac{3}{21}=21\)

Đặt \(A=\frac{2}{1\times2}+\frac{2}{2\times3}+\cdots+\frac{2}{18\times19}+\frac{2}{19\times20}\)
Ta có:
\(A=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{18\times19}+\frac{1}{19\times20}\right)\)
\(A=2\times\left(\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\cdots+\frac{19-18}{18\times19}+\frac{20-19}{19\times20}\right)\)
\(A=2\times\left(\frac{2}{1\times2}-\frac{1}{1\times2}+\frac{3}{2\times3}-\frac{2}{2\times3}+\cdots+\frac{20}{19\times20}-\frac{19}{19\times20}\right)\)
\(A=2\times\left(1-\frac12+\frac12-\frac13+\ldots+\frac{1}{19}-\frac{1}{20}\right)\)
\(A=2\times\left(1-\frac{1}{20}\right)\)
\(A=2\times\frac{19}{20}\)
\(A=\frac{19}{10}\)
Vậy \(A=\frac{19}{10}\)
\(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\cdots+\frac{2}{19\cdot20}\)
\(=2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{19\cdot20}\right)\)
\(=2\left(1-\frac12+\frac12-\frac13+\cdots+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\left(1-\frac{1}{20}\right)=2\cdot\frac{19}{20}=\frac{19}{10}\)

\(2^{100}\) và \(1024^9\)
\(1024^9=\left(2^{10}\right)^9=2^{90}<2^{100}\)
Vậy \(2^{100}>1024^9\)

\(8^{10}-8^9-8^8\) ⋮ 55
A = \(8^{10}-8^9-8\)
A = \(8^8\).(8\(^2\) - 8 - 1)
A = 8\(^8\).(64 - 8 -1)
A = \(8^8\).(56 - 1)
A = \(8^8\).55 ⋮ 55 (đpcm)

Ta có: \(125,5:5+x=37,86\)
=>25,1+x=37,86
=>x=37,86-25,1=12,76
25,1 + x = 37,86
x = 37,86 - 25,1
x = 12,76
Vậy x=12,76