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c=39/64

d=913/105

3) C thiếu đề

4) \(D=\frac{1}{9}-\left|\frac{-5}{23}\right|-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{4}-\frac{7}{30}\)

\(D=\frac{1}{9}-\frac{5}{23}+\frac{5}{23}-\frac{1}{9}-\frac{25}{7}+\frac{50}{4}-\frac{7}{30}\)

\(D=\frac{1}{9}-\frac{1}{9}-\frac{5}{23}+\frac{5}{23}+\frac{-25}{7}+\frac{50}{4}-\frac{7}{30}\)

\(D=0+0+\frac{125}{14}-\frac{7}{30}\)

\(D=\frac{913}{105}\)

Trả lời:

\(B=\frac{0,75-30\%+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+1\frac{4}{7}+3\frac{2}{3}}\)

\(B=\frac{\frac{3}{4}-\frac{3}{10}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{3}}\)

\(B=\frac{2019}{1820}\div\frac{2431}{420}\)

\(B=\frac{2019}{1820}\times\frac{420}{2431}\)

\(B=\frac{6057}{31603}\)

a) 25¹⁵ và 8¹⁰. 3³⁰

25¹⁵ = (5² ) ¹⁵ = 5³⁰

8¹⁰. 3³⁰ = (2³ )¹⁰. 3³⁰ = 2³⁰. 3³⁰ = 6³⁰

Vì 5³⁰< 6³⁰

nên 25¹⁵< 8¹⁰. 3³⁰

b) \(\frac{4^{15}}{7^{30}}\)và \(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)

\(\frac{4^{15}}{7^{30}}=\frac{\left(2^2\right)^{15}}{7^{30}}=\frac{2^{30}}{7^{30}}\)

\(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}=\frac{\left(2^3\right)^{10}.3^{30}}{7^{30}.\left(2^2\right)^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{3^{30}}{7^{30}}\)

Vì \(\frac{2^{30}}{7^{30}}< \frac{3^{30}}{7^{30}}\)

nên \(\frac{4^{15}}{7^{30}}< \frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)

a)\(25^{15}=5^{2^{15}}=5^{30}\)

\(8^{10}.3^{30}=2^{3^{10}}.3^{30}=\left(2.3\right)^{30}=6^{30}\)

\(5^{30}< 6^{30}=>25^{15}< 8^{10}.3^{30}\)

b)\(\frac{4^{15}}{7^{30}}=\frac{2^{2^{15}}}{7^{30}}=\frac{2^{30}}{7^{30}}=\left(\frac{2}{7}\right)^{30}\)

\(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{6^{30}}{14^{30}}=\left(\frac{6}{14}\right)^{30}=\left(\frac{3}{7}\right)^{30}\)

Vì hai số có mũ bằng 30 nên ta so sánh :\(\frac{2}{7}< \frac{3}{7}\)

=>\(\frac{4^{15}}{7^{30}}< \frac{8^{10}.3^{30}}{7^{30}.4^{15}}\).

trời ơi không ai giúp mình hu hu

c, \(\frac{-32}{-2^n}=4\)

\(\Rightarrow-2^n=-32:4\)

\(\Rightarrow-2^n=-8\)

\(\Rightarrow-2^n=-2^3\Rightarrow n=3\)

d, \(\frac{8}{2^n}=2\)

\(\Rightarrow2^n=8:2\)

\(\Rightarrow2^n=4\)

\(\Rightarrow2^n=2^2\Rightarrow n=2\)

e, \(\frac{25^3}{5^n}=25\)

\(\Rightarrow5^n=25^3:25\)

\(\Rightarrow5^n=25^2\)

\(\Rightarrow5^n=5^4\Rightarrow n=4\)

i , \(8^{10}:2^n=4^5\)

\(\Rightarrow2^n=8^{10}:4^5\)

\(\Rightarrow2^n=\left(2^3\right)^{10}:\left(2^2\right)^5\)

\(\Rightarrow2^n=2^{30}:2^{10}\)

\(\Rightarrow2^n=2^{20}\Rightarrow n=20\)

k, \(2^n.81^4=27^{10}\)

\(\Rightarrow2^n=27^{10}:81^4\)

\(\Rightarrow2^n=\left(3^3\right)^{10}:\left(3^4\right)^4\)

\(\Rightarrow2^n=3^{30}:3^{16}\)

\(\Rightarrow2^n=3^{14}\)

\(\Rightarrow2^n=4782969\)Không chia hết cho 2 nên ko có Gt n thỏa mãn 

bài 1:

\(a,-\frac{3}{25}+\frac{8}{35}=-\frac{21}{175}+\frac{40}{175}=\frac{19}{175}\)

\(b,\frac{4}{25}+\frac{-2}{15}=\frac{12}{75}-\frac{10}{75}=\frac{2}{75}\)

\(c,-\frac{1}{21}+\frac{-1}{14}=-\frac{2}{42}-\frac{3}{42}=-\frac{5}{42}\)

\(d,-\frac{4}{20}+0,6=-\frac{4}{20}+\frac{12}{20}=\frac{8}{20}=\frac25\)

bài 2:

\(a,-\frac35-\left(-\frac{8}{24}\right)=-\frac35+\frac{8}{24}=-\frac{72}{120}+\frac{40}{120}=-\frac{32}{120}=-\frac{4}{15}\)

\(b,-\frac45-\frac29=-\frac{36}{45}-\frac{10}{45}=-\frac{46}{45}\)

\(c,-\frac19-\frac{5}{12}=-\frac{4}{36}-\frac{15}{36}=-\frac{19}{36}\)

\(d,4,5-\left(-\frac25\right)=\frac92+\frac25=\frac{45}{10}+\frac{4}{10}=\frac{49}{10}\)

f, \(\dfrac{2^9.4^{10}}{8^8}=\dfrac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\dfrac{2^9.2^{20}}{2^{24}}=\dfrac{2^{29}}{2^{24}}=2^5=32\)

a: \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{14}{25}+\dfrac{11}{25}+\dfrac{2}{7}=\dfrac{2}{7}\)

b: \(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{3}{2}-\dfrac{2}{5}=1-1-1=-1\)

c: \(=\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{5}{2}-2=\dfrac{4}{25}+\dfrac{7}{2}-2=\dfrac{83}{50}\)

1.a) Sửa lại đề: \(\frac{11}{17}\)ở mẫu chuyển thành \(\frac{11}{7}\)

\(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)\(=\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}=\frac{3}{11}\)

( vì \(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\ne0\))

2.a) \(\frac{3}{5}+\frac{3}{2}.x=\frac{-5}{7}\)\(\Leftrightarrow\frac{3}{2}.x=\frac{-5}{7}-\frac{3}{5}\)

\(\Leftrightarrow\frac{3}{2}.x=\frac{-46}{35}\)\(\Leftrightarrow x=\frac{-46}{35}:\frac{3}{2}\)\(\Leftrightarrow x=\frac{-92}{105}\)

Vậy \(x=\frac{-92}{105}\)

b) \(\left(4x-\frac{1}{3}\right).\left(\frac{3}{2}x+\frac{5}{6}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{3}=0\\\frac{3}{2}x+\frac{5}{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=\frac{1}{3}\\\frac{3}{2}x=\frac{-5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-5}{9}\end{cases}}\)

Vậy \(x=\frac{-5}{9}\)hoặc \(x=\frac{1}{12}\)