\(\frac{\left(0,8^5\right)}{\left(0,4^6\right)}=80\)

nha bạn 

\(\frac{\left(0,8^5\right)}{\left(0,4^6\right)}=80\)nha  Linh Nguyễn                               ^_^

\(\frac{0,8^5}{0,4^6}=\frac{\frac{1024}{3125}}{\frac{64}{16525}}=80\)

\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)

\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)

-_-

A=(3− 

4

1

​

 + 

2

3

​

 )−(5+ 

3

1

​

 − 

6

5

​

 )−(6− 

4

7

​

 + 

3

2

​

 )

⇒A=3− 

4

1

​

 + 

2

3

​

 −5− 

3

1

​

 + 

6

5

​

 −6+ 

4

7

​

 − 

3

2

​

⇒A=(3−5−6)−( 

4

1

​

 + 

4

7

​

 )+( 

2

3

​

 + 

6

5

​

 − 

3

2

​

 )

⇒A=−8− 

2

3

​

 + 

3

5

​

=− 

6

47

​

 .

B=0,5+ 

3

1

​

 +0,4+ 

7

5

​

 + 

6

1

​

 − 

35

4

​

 + 

41

1

​

\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.⇒B=(0,5+0,4)+( 

3

1

​

 + 

6

1

​

 )+( 

7

5

​

 − 

35

4

​

 )+ 

41

1

​

⇒B= 

10

9

​

 + 

2

1

​

 + 

5

3

​

 + 

41

1

​

⇒B=2+ 

41

1

​

⇒B= 

41

83

​

 .

\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)

\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)

\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)

\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)

\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)

\(=7-\dfrac{26}{5}\)

\(=\dfrac{9}{5}\)

\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)

\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)

\(=\dfrac{2}{3}+\dfrac{21}{8}\)

\(=\dfrac{79}{24}\)

\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)

\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)

\(=\dfrac{31}{4}:\dfrac{49}{8}\)

\(=\dfrac{62}{49}\)

\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)

\(a,\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(3^2\right)^{10}.5^{10}.5^{20}}{\left(5^2\right)^{15}.3^{15}}=\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)

\(b,\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}=\dfrac{\left(0,4.2\right)^5}{\left(0.4\right)^6}=\dfrac{\left(0.4\right)^5.2^5}{\left(0,4\right)^6}=\dfrac{2^5}{0,4}=\dfrac{32}{0,4}=80\)

a,\(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{3^{20}.5^{30}}{3^{15}.3^{30}}=3^5=243\)

b,\(\dfrac{\left(0,8\right)^5}{0,4^6}=\dfrac{0,4^5.2^5}{0,4^6}=\dfrac{2^5}{0,4}=80\)

a/ \(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(5.3^2\right)^{10}.5^{20}}{\left(5^2.3\right)^{15}}=\dfrac{5^{10}.3^{20}.5^{20}}{5^{30}.3^{15}}=\dfrac{5^{30}.3^{20}}{5^{30}.3^{15}}=3^5=243\)

b/ \(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}=\dfrac{\left(0,4.2\right)^5}{\left(0,4\right)^6}=\dfrac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}=\dfrac{2^5}{0,4}=80\)

Bài 1: 

a) Ta có: \(A=-1.7\cdot2.3+1.7\cdot\left(-3.7\right)-1.7\cdot3-0.17:0.1\)

\(=1.7\cdot\left(-2.3\right)+1.7\cdot\left(-3.7\right)+1.7\cdot\left(-3\right)+1.7\cdot\left(-1\right)\)

\(=1.7\cdot\left(-2.3-3.7-3-1\right)\)

\(=-10\cdot1.7=-17\)

b) Ta có: \(B=2\dfrac{3}{4}\cdot\left(-0.4\right)-1\dfrac{2}{3}\cdot2.75+\left(-1.2\right):\dfrac{4}{11}\)

\(=\dfrac{11}{4}\cdot\left(-0.4\right)-\dfrac{5}{3}\cdot\dfrac{11}{4}+\left(-1.2\right)\cdot\dfrac{11}{4}\)

\(=\dfrac{11}{4}\left(-0.4-\dfrac{5}{3}-1.2\right)\)

\(=-\dfrac{539}{60}\)

c) Ta có: \(C=\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)

\(=\dfrac{2^3\cdot5^3\cdot7^4}{2^2\cdot5^2\cdot7^4}\)

\(=10\)

Ta có: \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}\)

\(=\frac{2^5\cdot\left(0,4\right)^5}{\left(0,4\right)\cdot\left(0,4\right)^5}=\frac{2^5}{0,4}=\frac{32}{0,4}=80\)

=\(\left[\dfrac{\left(0,4.2\right)^5}{\left(0,4\right)^6}+\dfrac{2^9.2^6.3^8}{\left(3.2\right)^6.2^9}\right]=\left[\dfrac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}+\dfrac{2^6.3^8}{3^6.2^6}\right]\)

=\(\left[\dfrac{2^5}{0,4}+3^2\right]\)

=\(\left[80+9\right]=89\)

\(\left[\dfrac{\left(2.0,4\right)^5}{0,4,0,4^5}+\dfrac{2^{15}.3^8}{3^6.2^6.2^9}\right]\div\dfrac{3^{20}.5^{30}}{3^{15}.5^{30}}\)

\(=\left[\dfrac{2^5.0.4^5}{0,4.0,4^5}+\dfrac{2^{15}.3^8}{3^6.2^{15}}\right]\div3^5\)

\(=\left[\dfrac{2^5}{0,4}+3^2\right]\div243\)

\(=80+\left(3^5\div3^2\right)\)

\(=80+3^3\)

\(=80+27\)

\(=107\)