Bài 1:

a) \(\frac{2}{5}+\frac{1}{5}.\left(\frac{3}{4}\right)\)

= \(\frac{2}{5}+\frac{3}{20}\)

= \(\frac{11}{20}\)

b) \(\frac{5}{12}.\left(-\frac{3}{4}\right)\) + \(\frac{7}{12}.\left(-\frac{3}{4}\right)\)

= \(\left(\frac{5}{12}+\frac{7}{12}\right).\left(-\frac{3}{4}\right)\)

= 1.\(\left(-\frac{3}{4}\right)\)

= \(-\frac{3}{4}\)

Còn câu c) đang nghĩ.

Bài 2:

a) \(\frac{5}{7}+\frac{2}{7}\)x = 1

1.x = 1

x = 1 : 1

x = 1

Vậy x = 1.

b) 0,2 + | x - 1, 3 = 1, 5|

0,2 + x = 1, 5 + 1, 3

0,2 + x = 2, 8

x = 2, 8 - 0, 2

x = 2, 6

Vậy x = 2, 6.

c) 2^x + 5 = 37

2^x = 37 - 5

2^x = 32

2^x = 2⁵

=> x = 5

Vậy x = 5.

d) 2^x + 2^{x + 1} = 48

2^x . 1 + 2^x . 2¹ = 48

2^x . ( 1 + 2) = 48

2^x . 3 = 48

2^x = 48 : 3

2^x = 16

2^x = 2⁴

=> x = 4

Vậy x = 4.

Chúc bạn học tốt!

làm bước trung gian giùm mình luôn nhé

thanks trước những bạn làm giùm nhé

mình đang cần gấp lắm sáng mai là mình cần ai đang on làm giùm mình nhé

thanks

              Bài làm :

a)\(=-\frac{3}{5}+\frac{28}{5}\times\frac{9}{14}=-\frac{3}{5}+\frac{18}{5}=3\)

b)\(=\frac{55}{126}+\frac{5}{42}+\frac{4}{9}=1\)

c)\(=-\frac{51}{13}-\frac{27}{13}=-6\)

d)\(=\frac{7}{3}-11\frac{1}{4}\times\frac{2}{15}=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\)

e)\(=1\times\frac{8}{3}\times0,25=\frac{2}{3}\)

\(=\frac{11}{-5}\cdot\frac{-9}{11}\cdot\frac{15}{-14}\cdot\frac{2}{5}+-\frac{2}{77}\cdot\frac{5}{-3}\)
\(=\frac{9}{5}\cdot-\frac{15}{14}\cdot\frac{2}{5}+\frac{10}{231}\)
\(=-\frac{841}{1155}\)

a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)

\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)

b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)

c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)

\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)

\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)

e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)

f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)

Dài đấy :))

a) \(\left|x-1\right|-\left(-2\right)^3=9\cdot\left(-1\right)^{100}\)

\(\Leftrightarrow\left|x-1\right|-\left(-8\right)=9\cdot1\)

\(\Leftrightarrow\left|x-1\right|+8=9\)

\(\Leftrightarrow\left|x-1\right|=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

b) \(\frac{x-2}{-4}=\frac{-9}{x-2}\)( ĐKXĐ : \(x\ne2\))

\(\Leftrightarrow\left(x-2\right)\left(x-2\right)=-4\cdot\left(-9\right)\)

\(\Leftrightarrow\left(x-2\right)^2=36\)

\(\Leftrightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}\left(tmđk\right)\)

c) \(\frac{x-5}{3}=\frac{-12}{5-x}\)( ĐKXĐ : \(x\ne5\))

\(\Leftrightarrow\frac{x-5}{3}=\frac{-12}{-\left(x-5\right)}\)

\(\Leftrightarrow\frac{x-5}{3}=\frac{12}{x-5}\)

\(\Leftrightarrow\left(x-5\right)\left(x-5\right)=3\cdot12\)

\(\Leftrightarrow\left(x-5\right)^2=36\)

\(\Leftrightarrow\left(x-5\right)^2=\left(\pm6\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=6\\x-5=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-1\end{cases}}\left(tmđk\right)\)

d) \(8x-\left|4x+\frac{3}{4}\right|=x+2\)

\(\Leftrightarrow8x-x-2=\left|4x+\frac{3}{4}\right|\)

\(\Leftrightarrow7x-2=\left|4x+\frac{3}{4}\right|\)(*)

\(\left|4x+\frac{3}{4}\right|\ge0\Leftrightarrow4x+\frac{3}{4}\ge0\Leftrightarrow x\ge-\frac{3}{16}\)

Vậy ta xét hai trường hợp sau :

1. \(x\ge-\frac{3}{16}\)

(*) <=>\(7x-2=4x+\frac{3}{4}\)

\(\Leftrightarrow7x-4x=\frac{3}{4}+2\)

\(\Leftrightarrow3x=\frac{11}{4}\)

\(\Leftrightarrow x=\frac{11}{12}\)(tmđk)

2. \(x< -\frac{3}{16}\)

(*) <=> \(7x-2=-\left(4x+\frac{3}{4}\right)\)

\(\Leftrightarrow7x-2=-4x-\frac{3}{4}\)

\(\Leftrightarrow7x+4x=-\frac{3}{4}+2\)

\(\Leftrightarrow11x=\frac{5}{4}\)

\(\Leftrightarrow x=\frac{5}{44}\left(ktmđk\right)\)

Vậy x = 11/12

e) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)

\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2020}\)

\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)

\(\Leftrightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4040}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{4040}\)

\(\Leftrightarrow x+1=4040\)

\(\Leftrightarrow x=4039\)

ĐKXD là gì vậy