a) |x-3,5| = 7,5

\(\Rightarrow\hept{\begin{cases}x-3,5=7,5\\x-3,5=-7,5\end{cases}}\Rightarrow\hept{\begin{cases}x=11\\x=-4\end{cases}}\)

b) 9^x:3^x = 81

=> (3²)^x:3^x = 3⁴

3^2x : 3^x = 3⁴

<=> 3^2x-x = 3⁴

2x-x = 4

<=> x = 4

c) \(x:\left(\frac{-1}{3}\right)^3=\frac{-1}{3}\)

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

\(x=\left(\frac{-1}{3}\right)^4=\frac{\left(-1\right)^4}{3^4}=\frac{1}{81}\)

d) \(\frac{-2}{x}=\frac{-x}{\frac{8}{25}}\)

=> -2x = (-2).\(\frac{8}{25}\)

-2x = \(\frac{-16}{25}\)

=> 2x = \(-\left(\frac{-16}{25}\right)\)

\(x=\frac{16}{25}:2\)

x= \(\frac{8}{25}\)

\(a,12,3\cdot4,7+5,3\cdot12,3-1,4\cdot7,5+7,5\cdot5,4\)

\(=12,3\cdot\left(4,7+5,3\right)-7,5\cdot\left(1,4+5,4\right)\)

\(=12,3\cdot10-7,5\cdot6,8\)

\(=123-51\)

\(=72.\)

\(b,3\cdot x-1=\frac{3}{4}:\frac{15}{16}\)

\(\Leftrightarrow3\cdot x-1=\frac{3}{4}\cdot\frac{16}{15}\)

\(\Leftrightarrow3\cdot x-1=\frac{4}{5}\)

\(\Leftrightarrow3\cdot x=\frac{4}{5}+1\)

\(\Leftrightarrow3\cdot x=\frac{9}{5}\)

\(\Leftrightarrow x=\frac{9}{5}\div3\)

\(\Leftrightarrow x=\frac{9}{5}\cdot\frac{1}{3}\)

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

a) \(12,3\times4,7+5,3\times12,3-1,4\times7,5+7,5\times5,4\)

\(=12,3\times\left(4,7+5,3\right)+7,5\times\left(5,4-1,4\right)\)

\(=12,3\times10+7,5\times4\)

\(=123+30\)

\(=153\)

b) \(3.x-1=\frac{3}{4}:\frac{15}{16}\)

\(\Rightarrow3.x-1=\frac{4}{5}\)

\(\Rightarrow3.x=\frac{9}{5}\Rightarrow x=\frac{3}{5}\)

c,  Trong phép nhân 1 x 3 x 5 x...x 19 có chứa thừa số 5 nên tích là một số chia hết cho 5

=>chữ số tận cùng của tích là 0 hoặc 5

Vì các thừa số là số lẻ nên tích là số lẻ nên chữ số tận cùng của tích là 5

Trong phép nhân 1 x 2 x 3 x...x 8 x 9 có chứa tích 2 x 5 = 10 nên phép nhân 1 x 2 x 3 x...x 9 có chữ số tận cùng là 0

=> 1 x 3 x 5 x 7 x 9 x ....x 13 x 15 x 17 x 19 + 1 x 2 x 3 x ....x 8 x 9 = ...5 + ....0 = ....5

\(a,\frac{x+15}{x}=\frac{4}{3}\Rightarrow4x=3x+45\Leftrightarrow x=45\)

\(b,\frac{7,5-x}{3,5+x}=\frac{5}{6}\Rightarrow17,5+5x=45-6x\Leftrightarrow11x=27,5\Rightarrow x=2,5\)

\(c,\frac{x-20}{x-10}=\frac{x+40}{x+70}\Rightarrow\left(x-20\right)\left(x+70\right)=\left(x-10\right)\left(x+40\right)\)

\(\Leftrightarrow x^2+50x-1400=x^2+30x-400\)

\(\Leftrightarrow20x=1000\)

\(\Rightarrow x=50\)

a. \(\frac{\left(x+15\right)}{x}=\frac{4}{3}\Leftrightarrow4x=3\left(x+15\right)\Leftrightarrow4x=3x+45\Leftrightarrow x=45\)

Vậy x=45

b. \(\frac{7,5-x}{3,5+x}=\frac{5}{6}\Leftrightarrow5\left(3,5+x\right)=6\left(7,5-x\right)\Leftrightarrow17,5+5x=45-6x\Leftrightarrow11x=27,5\Leftrightarrow x=2,5\)

Vậy x=2,5

c. \(\frac{x+20}{x-10}=\frac{x+40}{x+70}\Leftrightarrow\left(x+40\right)\left(x-10\right)=\left(x+20\right)\left(x+70\right)\)

\(\Leftrightarrow x^2+30x-400=x^2+90x+1400\Leftrightarrow-60x=-30\Leftrightarrow x=-30\)

Vậy x=-30

 a) \(1\frac{2}{7} = 1 + \frac{2}{7} = \frac{9}{2}\)

\(\begin{array}{l}x:1\frac{2}{7} =  - 3,5\\x:\frac{9}{7} =  - \frac{7}{2}\\x =  - \frac{7}{2}.\frac{9}{7}\\x =  - \frac{9}{2}\end{array}\)

b) \(0,4.x - \frac{1}{5}.x = \frac{3}{4}\)

\(\begin{array}{l}\frac{2}{5}.x - \frac{1}{5}.x = \frac{3}{4}\\\left( {\frac{2}{5} - \frac{1}{5}} \right).x = \frac{3}{4}\\\frac{1}{5}.x = \frac{3}{4}\\x = \frac{3}{4}:\frac{1}{5}\\x = \frac{3}{4}.5\\x = \frac{{15}}{4}\end{array}\)

mình chưa tìm ra câu trả lời xin lỗi

Anh chỉ giải câu a thôi, câu b anh thấy nó bình thường mà.

Cộng vào mỗi phân số thêm 1 đơn vị được:

\(\frac{x+2013}{2009}+\frac{x+2013}{2010}=\frac{x+2013}{2011}+\frac{x+2013}{2012}\).

Tới đây tự làm tiếp nhá.

.a, \(\frac{x+1}{999}+\frac{x+2}{998}=\frac{x+3}{997}+\frac{x+4}{996}\)

.\(< =>\frac{x+1}{999}+1+\frac{x+2}{998}+1=\frac{x+3}{997}+1+\frac{x+4}{996}+1\)

.\(< =>\frac{x+1}{999}+\frac{999}{999}+\frac{x+2}{998}+\frac{998}{998}=\frac{x+3}{997}+\frac{997}{997}+\frac{x+4}{996}+\frac{996}{996}\)

.\(< =>\frac{x+1+999}{999}+\frac{x+2+998}{998}=\frac{x+3+997}{997}+\frac{x+4+996}{996}\)

.\(< =>\frac{x+1000}{999}+\frac{x+1000}{998}-\frac{x+1000}{997}-\frac{x+1000}{996}=0\)

.\(< =>\left(x+1000\right)\left(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\right)=0\)

.Do \(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\ne0\)

.Suy ra \(x+1000=0\Leftrightarrow x=-1000\)

.b, \(\frac{x+1}{1001}+\frac{x+2}{1002}=\frac{x+3}{1003}+\frac{x+4}{1004}\)

.\(< =>\frac{x+1}{1001}-1+\frac{x+2}{1002}-1=\frac{x+3}{1003}-1+\frac{x+4}{1004}-1\)

.\(< =>\frac{x+1}{1001}-\frac{1001}{1001}+\frac{x+2}{1002}-\frac{1002}{1002}=\frac{x+3}{1003}-\frac{1003}{1003}+\frac{x+4}{1004}-\frac{1004}{1004}\)

.\(< =>\frac{x+1-1001}{1001}+\frac{x+2-1002}{1002}=\frac{x+3-1003}{1003}+\frac{x+4-1004}{1004}\)

.\(< =>\frac{x-1000}{1001}+\frac{x+1000}{1002}-\frac{x+1000}{1003}-\frac{x+1000}{1004}=0\)

.\(< =>\left(x-1000\right)\left(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\right)=0\)

.Do \(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\ne0\)

.Suy ra \(x-1000=0\Leftrightarrow x=1000\)

cảm ơn

mình làm luộn 2 câu còn lại nhé ^^

.c,\(|x|-\frac{15}{2}=\frac{15}{4}\)

.\(< =>|x|=\frac{15}{4}+\frac{15}{2}=\frac{15}{4}+\frac{30}{4}=\frac{45}{4}\)

.\(< =>\orbr{\begin{cases}x=\frac{45}{4}\\x=-\frac{45}{4}\end{cases}}\)

.d,\(|\frac{3}{4}-x|+1=\frac{3}{2}\)

.\(< =>|\frac{3}{4}-x|=\frac{3}{2}-1=\frac{3}{2}-\frac{2}{2}=\frac{1}{2}\)

.\(< =>\orbr{\begin{cases}\frac{3}{4}-x=\frac{1}{2}\\\frac{3}{4}-x=-\frac{1}{2}\end{cases}< =>\orbr{\begin{cases}x=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\\x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\end{cases}}}\)

b. (x+1)(1/10+1/11+1/12-1/13-1/14)=0

x+1=0 (vì : 1/10+1/11+1/12-1/13-1/14>0)

x=-1