a) 80

b) 1293440/27

c) 1077936384

có vài chỗ ko thấy

a.

\(\frac{45^{10}\times5^{20}}{75^{15}}=\frac{\left(3^2\times5\right)^{10}\times5^{20}}{\left(3\times5^2\right)^{15}}=\frac{3^{20}\times5^{10}\times5^{20}}{3^{15}\times5^{30}}=3^5=243\)

b.

\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,8\right)^5}{\left(0,4\right)^5}\times\frac{1}{\left(0,4\right)}=\left(\frac{0,8}{0,4}\right)^5\times\frac{1}{\frac{4}{10}}=2^5\times\frac{5}{2}=2^4\times5=16\times5=80\)

c.

\(\frac{2^{15}\times9^4}{6^6\times8^3}=\frac{2^{15}\times\left(3^2\right)^4}{\left(2\times3\right)^6\times\left(2^3\right)^3}=\frac{2^{15}\times3^8}{2^6\times3^6\times2^9}=3^2=9\)

Chúc bạn học tốt ^^

cảm ơn bạn

\(\dfrac{0,8^5}{0,4^6}=\dfrac{\left(2^3\cdot\dfrac{1}{10}\right)^5}{\left(2^2\cdot\dfrac{1}{10}\right)^6}=\dfrac{2^{15}\cdot\dfrac{1}{10^5}}{2^{12}\cdot\dfrac{1}{10^6}}=\dfrac{2^3}{\dfrac{1}{10}}=2^3\cdot10=80\)

a) A = \(\frac{45^{10}.5^{10}}{75^{10}}=\frac{\left(5.3^2\right)^{10}.5^{10}}{\left(5^2.3\right)^{10}}=\frac{5^{10}.3^{20}.5^{10}}{5^{20}.3^{10}}=\frac{5^{20}.3^{10}}{5^{20}}=3^{10}=59049\)

b) B = \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,2.2^2\right)^5}{\left(0,2.2\right)^6}=\frac{\left(0,2\right)^5.2^{10}}{\left(0,2\right)^6.2^6}=\frac{2^4}{0,2}=\frac{16}{0,2}=80\)

c) C = \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{2^{15}}=3^2=9\)

\(A=\frac{\left(9^{10}×5^{10}×5^{10}\right)}{\left(25^{10}×3^{10}\right)}\)

\(A=\frac{\left(3^{20}×5^{20}\right)}{\left(5^{20}×3^{10}\right)}\)

\(A=\frac{3^{20}}{3^{10}}\)

\(A=3^{10}\)

Bài 3: 

a) Ta có: \(A=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)

d) Ta có: \(D=x^2-2x+2\)

\(=x^2-2x+1+1\)

\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)

Bài 1: 

a) Ta có: \(A=x^2-2x+5\)

\(=x^2-2x+1+4\)

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

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

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)