a, \(\sqrt{25}-3\sqrt{\dfrac{4}{9}}=5-3.\dfrac{2}{3}=3\)

b, \(\left(2-\dfrac{5}{3}\right):\left(\dfrac{2}{7}+\dfrac{5}{21}-1\right)\)

\(=\dfrac{1}{3}:\dfrac{6+5-21}{21}\)

\(=-\dfrac{1}{3}.\dfrac{21}{10}\)

\(=-\dfrac{7}{10}\)

\(\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)

\(=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}.\left(2^4\right)^2.7^{48}}\)

\(=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)

\(=\frac{7^{48}.5^{30}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{48}}\)

\(=5.\left(1-7.4\right)\)

\(=5.\left(1-28\right)\)

\(=5.\left(-27\right)=-135\)

Đặt : \(P=\frac{48^2\cdot8^5\cdot100^9}{12^2\cdot2^{15}\cdot4^2}\)

\(=\frac{\left(2^4\cdot3\right)^2\cdot\left(2^3\right)^5\cdot\left(2^2\cdot5^2\right)^9}{\left(2^2\cdot3\right)^2\cdot2^{15}\cdot\left(2^2\right)^2}\)

\(=\frac{2^8\cdot3^2\cdot2^{15}\cdot2^{18}\cdot5^{18}}{2^4\cdot3^2\cdot2^{15}\cdot2^4}\)

\(=\frac{2^{41}\cdot3^2\cdot5^{18}}{2^{23}\cdot3^2}=2^{18}\cdot5^{18}=\left(2\cdot5\right)^{18}=10^{18}\)

Vậy : \(P=10^{18}\)

a) \(\frac{1}{2}-\left(2x-\frac{3}{4}\right)=-\frac{5}{8}\)

\(\frac{1}{2}-2x+\frac{3}{4}=-\frac{5}{8}\)

\(\frac{1}{2}+\frac{3}{4}-2x=-\frac{5}{8}\)

\(1,25-2x=-0,625\)

\(2x=1,875\)

\(x=0,9375\)

b) \(-\frac{1}{2}-\frac{3}{2}.\left(x+\frac{5}{3}\right)=-\frac{1}{4}\)

\(-\frac{1}{2}-\frac{3}{2}.x-\frac{5}{2}=-\frac{1}{4}\)

\(-\frac{1}{2}-\frac{5}{2}-\frac{3}{2}.x=-\frac{1}{4}\)

\(-3-\frac{3}{2}.x=-\frac{1}{4}\)

...

đến đây thì b tự tính nha!

 (2^x-8)^3=(4^x+2^x+5)^3-(4^x+13)^3 
(2^x-8)^3=[(4^x+2^x+5)-(4^x+13)]*[(4^x... + (4^x+13)^2] 
(2^x-8)^3=(2^x-8)*[(4^x+2^x+5)^2+(4^x+... + (4^x+13)^2] 
2^x=8=>x=3 
hoặc (2^x-8)^2=(4^x+2^x+5)^2+(4^x+2^x+5)(4^x+... + (4^x+13)^2 
(4^x+2^x+5)^2 - (2^x-8)^2+(4^x+2^x+5)(4^x+13) + (4^x+13)^2=0 
[(4^x+2^x+5)-(2^x-8)]*[(4^x+2^x+5)+(2^... + (4^x+3)*[(4^x+2^x+5)+(4^x+13)]=0 
(4^x+13)*(4^x+2*2^x-3) + (4^x+3)*(2*4^x+2^x+18)=0 
(4^x+13)[(4^x+2*2^x-3) + (2*4^x+2^x+18)]=0 
4^x+13=0 (VN) 
hoặc 3*4^x + 3*2^x +15=0 
đặt t=2^x ( t>0) 
t^2 + t + 5=0 ptvn

\(=\dfrac{5}{21}+\dfrac{16}{21}-\left(\dfrac{19}{23}+\dfrac{4}{23}\right)+\dfrac{1}{2}=\dfrac{1}{2}\)

Đặt A = 1+2+2^2+2^3+....+2^60

2A = 2+2^2+2^3+2^4+.....+2^61

2A-A= ( 2+2^2+2^3+....+2^61)-(1+2+2^2+.....+2^60)

A = 2^61-1

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