\(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{\left(5.20\right)^4}{\left(25.4\right)^5}=\dfrac{100^4}{100^5}=\dfrac{1}{100}\)

a, =2.0,1-0,5                                  b,=\(\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)\)

    =0,2-0,5                                        =  -10:(\(\dfrac{-5}{7}\))

    =-0,3                                             = 14

           c, \(=\dfrac{5^4.\left(4.5\right)^4}{\left(5^2\right)^4.4^5}=\dfrac{5^4.4^4.5^4}{5^8.4^5}=\dfrac{1}{4^{ }}\)

\(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.5^4.4^4}{5^{10}.4^5}=\frac{1}{5^2.4}=\frac{1}{100}\)

Chúc hok tốt

1. 2008.\(\left(\dfrac{1}{2007}-\dfrac{2009}{1004}\right)-2009\left(\dfrac{1}{2007}-2\right)\)

=\(\left(2008.\dfrac{1}{2007}-2008.\dfrac{2009}{1004}\right)-\left(2009.\dfrac{1}{2007}-2009.2\right)\)

=\(\left(\dfrac{2008}{2007}-2.2009\right)-\left(\dfrac{2009}{2007}-2.2009\right)\)

=\(\left(\dfrac{2008}{2007}-4018\right)-\left(\dfrac{2009}{2007}-4018\right)\)

=\(\dfrac{2008}{2007}-4018-\dfrac{2009}{2007}+4018\)

=\(\left(\dfrac{2008}{2007}-\dfrac{2009}{2007}\right)+\left[\left(-4018\right)+4018\right]\)

=\(\dfrac{1}{2007}.\left(2008-2009\right)+0\)

=\(\dfrac{1}{2007}.\left(-1\right)+0\)

=\(\dfrac{-1}{2007}\)

2.\(\dfrac{5^5.20^3-5^4.20^3+5^7.4^5}{\left(20+5\right)^3+4^5}\)

=\(\dfrac{5^5.\left(2^2.5\right)^3-5^4.\left(2^2.5\right)^3+5^7.\left(2^2\right)^5}{\left[\left(2^2.5\right)+5\right]^3+\left(2^2\right)^5}\)

=\(\dfrac{5^5.2^6.5^3-5^4.2^6.5^3+5^7.2^{10}}{2^6.5^3+5^3+2^{10}}\)

=\(\dfrac{5^9.2^6-5^7.2^6+5^7.2^{10}}{5^3.\left(2^6+1\right)+2^{10}}\)

=\(\dfrac{5^7.2^6\left(5^2-1-2^4\right)}{5^3\left(2^6+1\right)+2^{10}}\)

bí rồi

\(=\frac{5^5\cdot\left(4.5\right)^3-5^4\cdot\left(4.5\right)^3+5^7\cdot4^5}{\left(5^3\right)^3\cdot4^5}=\frac{5^8.4^3-5^7.4^3+5^7.4^5}{5^9.4^5}=\frac{5^7.4^3.\left(5-1+4^2\right)}{5^7.4^3.\left(5^2.4^2\right)}\)

= \(\frac{4+4^2}{5^2.4^2}=\frac{4.5}{5^2.4^2}=\frac{1}{4.5}=\frac{1}{20}\)

\(\frac{5^4.20^4}{25^5.4^5}=\frac{\left(5.20\right)^4}{\left(25.4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)

mình đầu tiên

1/100 nhes http://olm.vn/hoi-dap/question/116928.html

Câu 1: \(\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{5^4\cdot\left(2^2\cdot5\right)^4}{\left(5^2\right)^5\cdot\left(2^2\right)^5}=\dfrac{5^4\cdot2^8\cdot5^4}{5^{10}\cdot2^{10}}=\dfrac{5^8\cdot2^8}{5^{10}\cdot2^{10}}=\dfrac{1}{100}\)

Câu 2 : \(2^{150}\&3^{100}\)

Ta có : \(2^{150}=\left(2^3\right)^{50}=8^{50}\)

\(3^{100}=\left(3^2\right)^{50}=9^{50}\)

Vì \(8^{50}< 9^{50}nên2^{150}< 3^{100}\)

3/ \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)

\(\Leftrightarrow2x-6-3+6x=4+4-4x\)

\(\Leftrightarrow8x-9=8-4x\)

\(\Leftrightarrow8x+4x=8+9\)

\(\Leftrightarrow12x=17\)

\(\Leftrightarrow x=\dfrac{17}{12}\)

Vậy \(x=\dfrac{17}{12}\)

4/ \(\dfrac{x-2}{2}-\dfrac{1+x}{3}=\dfrac{4-3x}{4}-1\)

\(\Leftrightarrow6\left(x-2\right)-4\left(1+x\right)=3\left(4-3x\right)-12\)

\(\Leftrightarrow6x-12-4-4x=12-9x-12\)

\(\Leftrightarrow6x-4-4x=12-9x\)

\(\Leftrightarrow2x-4=12-9x\)

\(\Leftrightarrow2x+9x=12+4\)

\(\Leftrightarrow11x=16\)

\(\Leftrightarrow x=\dfrac{16}{11}\)

Vậy \(x=\dfrac{16}{11}\)