\(\frac{5^{10}.7^3-25^4.7^4}{\left(5^3.7\right)^3+125^3.14^3}\)

\(=\frac{5^{10}.7^3-\left(5^2\right)^4.7^4}{5^9.7^3+\left(5^3\right)^3.\left(7.2\right)^3}\)

\(=\frac{5^{10}.7^3-5^8.7^4}{5^9.7^3+5^9.7^3.2^3}\)

\(=\frac{5^8.7^3\left(5^2-7\right)}{5^9.7^3\left(1+2^3\right)}\)

\(=\frac{18}{5.9}\)

\(=\frac{2}{5}\)

hok tốt!!

\(8\frac{4}{17}-\left(2\frac{5}{9}+3\frac{4}{17}\right)=\frac{140}{17}-\left(\frac{23}{9}+\frac{55}{17}\right)=\frac{140}{17}-\frac{886}{153}=\frac{22}{9}=2,444444444444\)

a) \(\frac{16\cdot64\cdot8^2}{4^3\cdot2^5\cdot16}=\frac{2^6\cdot2^6}{2^6\cdot2^5}=2\)

Có: (3/5)^2017.(5/3)^2016

=(5/3)^-2017.(5/3)^2016

=(5/3)^-2017+2016

=(5/3)^-1

=3/5.

Bạn phải học số mũ âm thì mới hiểu được nhé: 

Công thức nè: x^-n=1/x^n.

#)Giải :

\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{2}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\times0\)

\(=0\)

\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{2}{6}+\frac{3}{6}\right)\)

=\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).0\)

\(=0\)

-7x=21

x   =21:(-7)

x   =-3

b,=> x-5=9 hoặc -9

Nếu x-5=9 =>x=14

Nếu x-5=-9=>x=-4

c,-206+2015+206+(-2016)

   =    1809  +  (-1810)

   =    -1

d, (-8).(-5).16.(-125)

 = 8.125.(-5).16

 = 1000.  (-80)

  = -80000

22,

1, Đặt √(3-√5) = A

=> √2A=√(6-2√5)

=> √2A=√(5-2√5+1)

=> √2A=|√5 -1|

=> A=\(\dfrac{\sqrt{5}-1}{\text{√2}}\)

=> A= \(\dfrac{\sqrt{10}-\sqrt{2}}{2}\)

2, Đặt √(7+3√5) = B

=> √2B=√(14+6√5)

 => √2B=√(9+2√45+5)

=> √2B=|3+√5|

=> B= \(\dfrac{3+\sqrt{5}}{\sqrt{2}}\)

=> B= \(\dfrac{3\sqrt{2}+\sqrt{10}}{2}\)

3, 

Đặt √(9+√17) - √(9-√17) -\(\sqrt{2}\)=C

=> √2C=√(18+2√17) - √(18-2√17) -\(2\)

=> √2C=√(17+2√17+1) - √(17-2√17+1) -\(2\)

=> √2C=√17+1- √17+1 -\(2\)

=> √2C=0

=> C=0

26,

|3-2x|=2\(\sqrt{5}\)

TH1: 3-2x ≥ 0 ⇔ x≤\(\dfrac{-3}{2}\)

3-2x=2\(\sqrt{5}\)

-2x=2\(\sqrt{5}\) -3

x=\(\dfrac{3-2\sqrt{5}}{2}\) (KTMĐK)

TH2: 3-2x < 0 ⇔ x>\(\dfrac{-3}{2}\)

3-2x=-2\(\sqrt{5}\)

-2x=-2√5 -3

x=\(\dfrac{3+2\sqrt{5}}{2}\) (TMĐK)

Vậy x=\(\dfrac{3+2\sqrt{5}}{2}\)

2, \(\sqrt{x^2}\)=12 ⇔ |x|=12 ⇔ x=12, -12

3, \(\sqrt{x^2-2x+1}\)=7

⇔ |x-1|=7 

TH1: x-1≥0 ⇔ x≥1

x-1=7 ⇔ x=8 (TMĐK)

TH2: x-1<0 ⇔ x<1

x-1=-7 ⇔ x=-6 (TMĐK)

Vậy x=8, -6

4, \(\sqrt{\left(x-1\right)^2}\)=x+3

⇔ |x-1|=x+3

TH1: x-1≥0 ⇔ x≥1

x-1=x+3 ⇔ 0x=4 (KTM)

TH2: x-1<0 ⇔ x<1

x-1=-x-3 ⇔ 2x=-2 ⇔x=-1 (TMĐK)

Vậy x=-1