\(a,\)\(-\frac{3}{5}\cdot x=\frac{1}{4}+0,75\)

\(-\frac{3}{5}\cdot x=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)

\(x=1\div\left(-\frac{3}{5}\right)\)

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

\(b,\)\(\left(\frac{1}{7}-\frac{1}{3}\right)\cdot x=\frac{28}{5}\times\left(\frac{1}{4}-\frac{1}{7}\right)\)

\(\left(\frac{3}{21}-\frac{7}{21}\right)\cdot x=\frac{28}{5}\cdot\left(\frac{7}{28}-\frac{4}{28}\right)\)

\(-\frac{4}{21}\cdot x=\frac{28}{5}\cdot\frac{3}{28}\)

\(-\frac{4}{21}\cdot x=\frac{3}{5}\)

\(x=\frac{3}{5}\div\left(-\frac{4}{21}\right)\)

\(x=-\frac{63}{20}\)

\(c,\)\(\frac{5}{7}\cdot x=\frac{9}{8}-0,125\)

\(\frac{5}{7}\cdot x=\frac{9}{8}-\frac{1}{8}\)

\(\frac{5}{7}\cdot x=1\)

\(x=1\div\frac{5}{7}\)

\(x=\frac{7}{5}\)

\(d,\)\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot36\)

\(\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot36\)

\(\frac{17}{33}\cdot x=\frac{1}{56}\cdot36\)

\(\frac{17}{33}\cdot x=\frac{9}{14}\)

\(x=\frac{9}{14}\div\frac{17}{33}\)

\(x=\frac{9}{14}\cdot\frac{33}{17}=\frac{297}{238}\)

a)\(\frac{3}{5}\times x=1\)                                        

\(x=1\div\frac{3}{5}\)

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

Mình chỉ làm duoc câu a thôi 

\(\frac{7}{5}+\left(\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right)\cdot x=\frac{16}{5}\)

\(\frac{2}{5}x=\frac{16}{5}-\frac{7}{5}\)

\(\frac{2}{5}x=\frac{9}{5}\)

 x      = \(\frac{9}{5}:\frac{2}{5}\)

x        = 9/2 

\(1\frac{2}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right).x=\frac{16}{5}\)

\(\frac{7}{5}+\left[\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right].x=\frac{16}{5}\)

\(\frac{7}{5}+\frac{2}{5}.x=\frac{16}{5}\)

\(\frac{2}{5}.x=\frac{16}{5}-\frac{7}{5}\)

\(\frac{2}{5}.x=\frac{9}{5}\)

\(x=\frac{9}{5}:\frac{2}{5}\)

\(x=\frac{9}{5}.\frac{5}{2}\)

\(x=\frac{9}{2}\)

\(\frac{7}{5}+\left(\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right).x=\frac{16}{5}\)

                                                \(\frac{2}{5}x=\frac{16}{5}-\frac{7}{5}\)

                                               \(\frac{2}{5}x=\frac{9}{5}\)

                                                    \(x=\frac{9}{5}\div\frac{2}{5}\)

                                                   \(x=\frac{9}{2}\)

\(x=\frac{903}{391}\)

Bài này sử dụng MTCT đó bạn!

903/391 mình nghĩ vậy

x = 903/391

a,(x-7)^x+1-(x-7)^x+11=0

=>(x-7)^x+1.[1-(x-7)¹⁰]=0

=>(x-7)^x+1=0

=>x-7=0

=>x=7

hoặc 1-(x-7)¹⁰=1

=>(x-7)¹⁰=1

=>x-7=-1;1

=>x=8;6

vậy x=6;7;8

b,(x-1)²=36/49

=>x-1=6/7;-6/7

=>x=13/7;1/7

vậy x=1/7;13/7

a. \(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\left|-\frac{16}{5}+\frac{2}{5}\right|-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\left|-\frac{14}{5}\right|-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\frac{14}{5}-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=2\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{5}{3}\end{cases}.}\)

Vậy \(x\in\left\{-\frac{5}{3};\frac{7}{3}\right\}.\)

b. \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)\(\Leftrightarrow\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}\times\left(x-7\right)^{10}=0\)\(\Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\Leftrightarrow\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}.}\)Xét 2 trường hợp:

Vậy \(x\in\left\{6;7;8\right\}.\)

ban nguyen nhat minh giang lai cho mk dong 2 cau b cai

mk cam on

Dòng 2 câu b là đặt nhân tử chung \(\left(x-7\right)^{x+1}\)ra ngoài

bằng 2

đáp án đúng = 2

ms làm hôm nay xong