\(\dfrac{5^{102}\cdot9^{1009}}{3^{2018}\cdot25^{50}}\)

\(=\dfrac{5^{102}\cdot\left(3^2\right)^{1009}}{3^{2018}\cdot\left(5^2\right)^{50}}\)

\(=\dfrac{5^{102}\cdot3^{2018}}{3^{2018}\cdot5^{100}}\)

\(=\dfrac{5^2\cdot1}{1\cdot1}\)

\(=25\)

\(=\dfrac{5^{102}.\left(3^2\right)^{1009}}{3^{2018}.\left(5^2\right)^{50}}\)

\(=\dfrac{5^{102}.3^{2018}}{3^{2018}.5^{100}}=5^2=25\)

\(\frac{3x-7}{5}=\frac{2x-1}{3}\)

\(\Leftrightarrow9x-21=10x-5\)

\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)

\(\frac{4x-7}{12}-x=\frac{3x}{8}\)

\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow-56-64x=36x\)

\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)

\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)

Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0

Vậy x = 2019

\(\frac{5x-8}{3}=\frac{1-3x}{2}\)

\(\Leftrightarrow10x-16=3-9x\)

\(\Leftrightarrow19x=19\Leftrightarrow x=1\)

\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)

\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)

\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)

\(\Rightarrow-16x-272=120x+312\)

\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)

\(\dfrac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)

\(=\dfrac{5^{102}.\left(3^2\right)^{1009}}{3^{2018}.\left(5^2\right)^{50}}\)

\(=\dfrac{5.1}{1.1}=5\)

\(\dfrac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)=\(\dfrac{5^{102}.3^{2018}}{3^{2018}.5^{100}}\) =5\(^2\) =25

nik lộn

2^x+1. 2^{2009 =} 2²⁰¹⁰

10 - 2x = 25 - 3x

a) x=7

b)x=2/3

c)x=-13

d)x=3

e)x=-1

c)

\(4\left(3x-4\right)-2=18\)

<=> \(12x-16-2=18\)
<=> \(12x=36\)

<=> \(x=3\)

Vậy x=3

d)

\(\left(3x-10\right):10=50\)

<=> \(3x-10=500\)

<=> \(3x=510\)

<=> x= \(170\)

Vậy x= 170

f)

\(x-\left[42+\left(-25\right)\right]=-8\)

<=> \(x-17=-8\)

<=> x= \(9\)

Vậy x=9

h)

\(x+5=20-\left(12-7\right)\)

<=> \(x+5=15\)

<=> \(x=10\)

Vậy x= 10

k)

\(\left|x-5\right|=7-\left(-3\right)\)

<=> \(\left|x-5\right|=10\)

* Với \(x>=5\) ; ta được:

\(x-5=10\)

<=> x= 15 (thoả mãn điều kiện )

*Với \(x< 5\) ; ta được:

\(-\left(x-5\right)=10\)

<=> \(-x+5=10\)

<=> \(-x=5\)

<=> \(x=-5\) (thoả mãn điều kiện)

Vậy x=15 ; x= -5

i)

\(\left|x-5\right|=\left|7\right|\)

<=> \(\left|x-5\right|=7\)

*Với \(x>=5\) ; ta được:

\(x-5=7\)

<=> \(x=12\) (thoả mãn)

*Với \(x< 5\) ; ta được:

\(-\left(x-5\right)=7\)

<=> \(-x=2\)

<=> \(x=-2\) (thoả mãn)

Vậy x= 12; x= -2

m)

\(2^{x+1}.2^{2009}=2^{2010}\)

<=> \(2^{x+1+2009}=2^{2010}\)

<=> \(2^{x+2010}=2^{2010}\)

=> \(x+2010=2010\)

=> \(x=0\)

Vậy x=0

n)

\(10-2x=25-3x\)

<=>\(x=15\)

Vậy x=15

Bài 4 là:

\(\dfrac{1}{2.5}\) + \(\dfrac{1}{5.8}\) +....+\(\dfrac{1}{92.95}\) + \(\dfrac{1}{95.98}\)

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