K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

18 tháng 7 2020

a) 187 - {[497 - ( 8 x X + 11) : X] : 3 - 78} = 150

=> {[497 - ( 8 x X + 11) : X] : 3 - 78} = 187 - 150

=> {[497 - (8 x X + 11) : X] : 3 - 78} = 37

=> [497 - (8 x X +11): X ] : 3 - 78 = 37

=> [497 - (8 x X + 11) : X] : 3 = 115

=> 497 - ( 8 x X + 11) : X = 345

=> (8 x X  + 11) : X = 497 - 345 = 152

=> 8X + 11 = 152X

=> 152X - 8X = 11

=> 144X = 11

=> X = 11/144

b) 19,96 + 4,19 - 24,15 : \(\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)

=> 19,96 + 4,19 - 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)=23,15\)

=> 24,15 - 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)\)= 23,15

=> 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)\)= 1

=> \(x\cdot4-\frac{1}{4}=24,15\)

=> \(x\cdot4=24,15+\frac{1}{4}=24,4\)

=> x = 24,4 : 4 = 6,1

Còn câu c tương tự

AH
Akai Haruma
Giáo viên
28 tháng 9 2024

Lời giải:

$2\frac{2}{6}x+8\frac{2}{3}=3\frac{1}{3}$

$\frac{7}{3}x=3\frac{1}{3}-8\frac{2}{3}=\frac{-16}{3}$

$x=\frac{-16}{3}: \frac{7}{3}=\frac{-16}{7}$

----------------------

$3\frac{2}{7}x-\frac{1}{8}=2\frac{3}{4}$

$\frac{23}{7}x=2\frac{3}{4}+\frac{1}{8}$

$\frac{23}{7}x=\frac{23}{8}$

$x=\frac{23}{8}: \frac{23}{7}=\frac{7}{8}$

14 tháng 8 2016

\(19,96+4,19-24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)

\(24,15-24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)

\(24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=1\)

\(x:\frac{1}{4}-\frac{1}{4}=24,15\)

\(4x=24,4\)

\(x=6,1\)

14 tháng 8 2016

\(\text{Bài này cx đơn giản thôi!}\)

\(4.19-24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23.15-19.96\)

\(4.19-24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=3.19\)

\(24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=4.19-3.19\)

\(24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=1\)

\(x:\frac{1}{4}-\frac{1}{4}=24.15:1\)

\(x:\frac{1}{4}-\frac{1}{4}=24.15\)

\(x:\frac{1}{4}=24.15+\frac{1}{4}\)

\(x:\frac{1}{4}=24.4\)

\(x=24.4.\frac{1}{4}\)

\(x=6.1\)


 

9 tháng 8 2019

\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)

\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)

\(< =>\frac{-15+12-24}{63}\)

\(< =>\frac{-3}{7}\)

\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)

\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)

\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)

\(< =>\frac{7}{5}-\frac{4}{7}\)

\(< =>\frac{29}{35}\)

\(bai2:\)

\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)

\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)

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

\(< =>x=\frac{3}{5}:\frac{-3}{4}\)

\(< =>x=\frac{-4}{5}\)

\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)

\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)

\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)

\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)

9 tháng 8 2019

Bài 1:

a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\)                                 b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)

 \(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\)                                       \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\) 

  \(=\frac{3}{7}.\frac{-9}{9}\)                                                                  \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)

\(=\frac{-3}{7}\)                                                                           \(=\frac{7}{5}-\frac{4}{7}\)

                                                                                               \(=\frac{29}{35}\)

Bài 2:

a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\)                                               b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)

  \(\frac{-3}{4}x\)           \(=\frac{1}{5}+\frac{4}{10}\)                                     \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)

\(\frac{-3}{4}x\)             \(=\frac{3}{5}\)                                            \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)     

         \(x\)              \(=\frac{3}{5}:\frac{-3}{4}\)                                        \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)                                         

         \(x\)              \(=\frac{4}{-5}\)                                                   \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)

                                                                                                             \(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\) 

                                                                                                                          \(x.\frac{10}{3}=\frac{13}{12}\) 

                                                                                                                                    \(x=\frac{13}{12}:\frac{10}{3}\) 

                                                                                                                                     \(x=\frac{13}{40}\)                             

16 tháng 2 2015

c. khi và chỉ khi \(\frac{3}{5}\cdot\frac{1}{x}=\frac{1}{3}-\frac{4}{15}=\frac{1}{15}\)

                        \(\frac{1}{x}=\frac{1}{15}\cdot\frac{5}{3}=\frac{1}{9}\)

suy ra x=9

16 tháng 2 2015

a,7/6x=3/5-1/30

7/6x=17/30

x=17/30:7/6

x=17/35

b,(-9/2-2x).11/7=11/4

-9/2-2x=11/4:11/7

-9/2-2x=7/4

2x=-9/2-7/4

2x=-25/4

x=-25/4:2

x=-25/8

c,9

d,x=-2

e,2.8=x.x=x2

16=x2=>x=4 hoặc -4

5 tháng 7 2018

a) x=7 và 8

b) x+1 và 2

c) 1/4

Quên rùi!!!

19 tháng 4 2017

a) x = 1

19 tháng 4 2017

a.X=\(\frac{25}{7}\)

16 tháng 4 2019

a) \(\frac{2}{5}x-x=\frac{\left(-2018\right)^0}{5^2}\\ x\left(\frac{2}{5}-1\right)=\frac{1}{25}\\ x\left(\frac{2}{5}-\frac{5}{5}\right)=\frac{1}{25}\\ x\cdot\frac{-3}{5}=\frac{1}{25}\\ x=\frac{1}{25}:\frac{-3}{5}\\ x=\frac{1}{25}\cdot\frac{-5}{3}\\ x=\frac{-1}{15}\)Vậy \(x=\frac{-1}{15}\)

b) \(\left|-1\frac{1}{2}x+2x\right|-\frac{7}{4}=0,5\\ \left|x\left(-1\frac{1}{2}+2\right)\right|-\frac{7}{4}=\frac{1}{2}\\ \left|x\cdot\frac{1}{2}\right|=\frac{1}{2}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{2}{4}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x\cdot\frac{1}{2}=\frac{9}{4}\\x\cdot\frac{1}{2}=\frac{-9}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}:\frac{1}{2}\\x=\frac{-9}{4}:\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}\cdot2\\x=\frac{-9}{4}\cdot2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{2}\\x=\frac{-9}{2}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{9}{2};\frac{-9}{2}\right\}\)

c) \(x+\left(x+\frac{2}{7}\right)+\frac{-5}{11}=\frac{4}{11}\\ x+x+\frac{2}{7}=\frac{4}{11}-\frac{-5}{11}\\ 2x+\frac{2}{7}=\frac{4}{11}+\frac{5}{11}\\ 2x+\frac{2}{7}=\frac{9}{11}\\ 2x=\frac{9}{11}-\frac{2}{7}\\ 2x=\frac{63}{77}-\frac{22}{77}\\ 2x=\frac{41}{77}\\ x=\frac{41}{77}:2\\ x=\frac{41}{77\cdot2}\\ x=\frac{41}{154}\)Vậy \(x=\frac{41}{154}\)

d) \(\left|0,25x-20\%\right|+\frac{3}{8}=1\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\frac{3}{8}-\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\\ \Rightarrow\left[{}\begin{matrix}\frac{1}{4}x-\frac{1}{5}=1\\\frac{1}{4}x-\frac{1}{5}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=1+\frac{1}{5}\\\frac{1}{4}x=\left(-1\right)+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{5}{5}+\frac{1}{5}\\\frac{1}{4}x=\frac{-5}{5}+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{6}{5}\\\frac{1}{4}x=\frac{-4}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}:\frac{1}{4}\\x=\frac{-4}{5}:\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}\cdot4\\x=\frac{-4}{5}\cdot4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{24}{5}\\x=\frac{-16}{5}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{24}{5};\frac{-16}{5}\right\}\)