Câu 5:

\(D\left(2\right)=21a+9b-6a-4b\)

\(D\left(2\right)=\left(21a-6a\right)+\left(9b-4b\right)\)

\(D\left(2\right)=15a+5b\)

Mà: \(3a+b=18\Rightarrow b=18-3b\)

\(\Rightarrow D\left(2\right)=15a+5\left(18-3b\right)\)

\(D\left(2\right)=15a+90-15a\)

\(D\left(2\right)=90\)

Vậy: ...

Câu 4:

\(D\left(1\right)=4a+10b-b+2a\)

\(D\left(1\right)=\left(4a+2a\right)+\left(10b-b\right)\)

\(D\left(1\right)=6a+9b\)

Mà: \(2a+3b=12\Rightarrow a=\dfrac{12-3b}{2}\)

\(\Rightarrow D\left(1\right)=6\left(\dfrac{12-3b}{2}\right)+9b\)

\(D\left(1\right)=\dfrac{6\left(12-3b\right)}{2}+9b\)

\(D\left(1\right)=3\left(12-3b\right)+9b\)

\(D\left(1\right)=36-9b+9b\)

\(D\left(1\right)=36\)

Vậy: ...

Câu 3:

Sửa đề: \(C=5a-4b+7a-8b\)

\(C=\left(5a+7a\right)-\left(4b+8b\right)\)

\(C=12a-12b\)

\(C=12\left(a-b\right)\)

\(C=12\cdot8\)

\(C=96\)

Vậy: ...

4:

D=6a+9b=3(2a+3b)=36

5: 

D=15a+5b=5(3a+b)=90

nhớ cho cả hướng dẫn giải

a: \(A=\dfrac{16^5\cdot15^5}{2^{10}\cdot3^5\cdot5^4}=\dfrac{2^{20}\cdot3^5\cdot5^5}{2^{10}\cdot3^5\cdot5^4}=2^{10}\cdot5=5120\)

b: \(B=\dfrac{2^{15}\cdot3+2^{19}\cdot10}{2^{12}\cdot26}=\dfrac{2^{15}\left(3+2^4\cdot10\right)}{2^{13}\cdot13}=2^2\cdot\dfrac{163}{13}=\dfrac{652}{13}\)

Tính bình thường dc ko bn

Bài 1: 

\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)

\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)

\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)

Bài 2: 

a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)

\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)

\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)

b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)

Bài 3:

\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)

\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)

Bài 4:

 \(\dfrac{3}{4}-x=1\)

\(\Rightarrow-x=1-\dfrac{3}{4}\)

\(\Rightarrow x=-\dfrac{1}{4}\)

Vậy: \(x=-\dfrac{1}{4}\)

\(x+4=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{5}-4\)

\(\Rightarrow x=-\dfrac{19}{5}\)

Vậy: \(x=-\dfrac{19}{5}\)

\(x-\dfrac{1}{5}=2\)

\(\Rightarrow x=2+\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{11}{5}\)

Vậy: \(x=\dfrac{11}{5}\)

\(x+\dfrac{5}{3}=\dfrac{1}{81}\)

\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)

\(\Rightarrow x=-\dfrac{134}{81}\)

Vậy: \(x=-\dfrac{134}{81}\)

Theo đầu bài ta có:
\(\frac{a-1}{2}=\frac{b+3}{4}=\frac{c-5}{6}\)
\(\Rightarrow\frac{5c-25}{30}=\frac{3a-3}{6}=\frac{4b+12}{16}\)
\(=\frac{\left(5c-25\right)-\left(3a-3\right)-\left(4b+12\right)}{30-6-16}\)
\(=\frac{\left(5c-3a-4b\right)-\left(25-3+12\right)}{8}\)
\(=\frac{50-34}{8}=\frac{16}{8}=2\)
\(\Rightarrow\hept{\begin{cases}a=2\cdot2+1=5\\b=2\cdot4-3=5\\c=2\cdot6+5=17\end{cases}}\)
 

làm sao ra đc 50--34 thế bài mình với