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18 tháng 12 2016

Tìm x:

a)-23+0,5x=1,5

-8+0,5x = 1,5

0,5x = 1,5-(-8) = 9,5

x = 9,5:0,5 = 19

b)(3)x81=27

(-3)x:81 =-27

(-3)x = -27.81 = -2187

(-3)x = (-3)7

=> x=7

 

c)112.x4=0,5

1,5.x = 0,5+4 = 4,5

x = 4,5:1,5 = 3

 

d)123:x4=6:0,3

\(\frac{5}{3}\):\(\frac{x}{4}\) = 20

\(\frac{x}{4}\) = \(\frac{5}{3}\):20 = \(\frac{1}{12}\)

=> x:4 = \(\frac{1}{12}\)

x = \(\frac{1}{12}\).4 = \(\frac{1}{3}\)

 

 

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

25 tháng 8 2017

a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)

\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)

\(x=\left(\frac{3}{7}\right)^2\)

\(x=\frac{9}{49}\)

Vậy...

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

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

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

\(x=-\frac{1}{3}\)

Vậy...

c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

=>\(x-\frac{1}{2}=\frac{1}{3}\)

\(x=\frac{1}{3}+\frac{1}{2}\)

\(x=\frac{5}{6}\)

Vậy...

d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)

=>\(x+\frac{1}{4}=\frac{2}{3}\)

\(x=\frac{2}{3}-\frac{1}{4}\)

\(x=\frac{5}{12}\)

Vậy...

Phù, mãi mới xong, tk cho mk nha bn

24 tháng 6 2020

bạn tự làm đi tính toán thôi mà

20 tháng 5 2017

a: ~12.58

b: 10

c:-1

d:55/4

20 tháng 3 2018

dkr68

29 tháng 3 2020

Câu 6 :

a, Ta có : \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)

=> \(\frac{15x}{15}+\frac{5\left(2x+\frac{x-1}{5}\right)}{15}=\frac{15}{15}-\frac{3\left(3x-\frac{1-2x}{3}\right)}{15}\)

=> \(15x+5\left(2x+\frac{x-1}{5}\right)=15-3\left(3x-\frac{1-2x}{3}\right)\)

=> \(15x+10x+\frac{5\left(x-1\right)}{5}=15-9x+\frac{3\left(1-2x\right)}{3}\)

=> \(15x+10x+x-1=15-9x+1-2x\)

=> \(15x+10x+x-1-15+9x-1+2x=0\)

=> \(37x-17=0\)

=> \(x=\frac{17}{37}\)

Vậy phương trình trên có nghiệm là \(S=\left\{\frac{17}{37}\right\}\)

Bài 7 :

a, Ta có : \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)

=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)

=> \(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)

=> \(x-23=0\)

=> \(x=23\)

Vậy phương trình trên có nghiệm là \(S=\left\{23\right\}\)

c, Ta có : \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)

=> \(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)

=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)

=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)

=> \(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)

=> \(x+2005=0\)

=> \(x=-2005\)

Vậy phương trình trên có nghiệm là \(S=\left\{-2005\right\}\)

e, Ta có : \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)

=> \(\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)

=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)

=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)

=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)

=> \(x-100=0\)

Vậy phương trình trên có nghiệm là \(S=\left\{100\right\}\)

23 tháng 7 2016

\(\frac{1}{3}x:\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)

\(\Rightarrow\frac{1}{3}x:\frac{2}{3}=\frac{35}{8}\)

\(\Rightarrow\frac{1}{3}x=\frac{35}{12}\)

\(\Rightarrow x=\frac{35}{4}\)

a) \(\left(\frac{1}{3}\cdot x\right):\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)

\(=\left(\frac{1}{3}\cdot x\right):\frac{2}{3}=\frac{35}{8}\)

\(\Rightarrow\frac{1}{3}\cdot x=\frac{35}{8}\cdot\frac{2}{3}\)

\(\Rightarrow x=\frac{35}{12}:\frac{1}{3}=\frac{35}{4}\)

b) \(4,5:0,3=2,5:\left(0,1\cdot x\right)\)

\(=15=2,5:\left(0,1\cdot x\right)\)

\(\Rightarrow0,1\cdot x=2,25:15\)

\(\Rightarrow x=\frac{3}{20}:0,1=\frac{3}{2}=1,5\)

c) \(8:\left(\frac{1}{4}\cdot x\right)=2:0,02\)

\(8:\left(\frac{1}{4}\cdot x\right)=100\)

\(\Rightarrow\frac{1}{4}\cdot x=8:100\)

\(\Rightarrow x=\frac{2}{25}:\frac{1}{4}=\frac{8}{25}=0,32\)

d) \(3:2\frac{1}{4}=\frac{3}{4}:\left(6\cdot x\right)\)

\(=\frac{4}{3}=\frac{3}{4}:\left(6:x\right)\)

\(\Rightarrow6\cdot x=\frac{3}{4}:\frac{4}{3}\)

\(\Rightarrow x=\frac{9}{16}:6=\frac{3}{32}=0,09375\)

29 tháng 11 2016

a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)

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

\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)

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

\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)

\(\Rightarrow x+3=-3\)

\(\Rightarrow x=-6\)

b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)

\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)

\(\Rightarrow2x+2=-2\)

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\)

c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)

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

\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)

\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)

\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)

\(\Rightarrow x+5=-4\)

\(\Rightarrow x=-9\)

29 tháng 11 2016

d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)

\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)

\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)

\(\Rightarrow2x=18\)

\(\Rightarrow x=9\)

e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)

\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)

\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)

\(\Rightarrow2x=18\)

\(\Rightarrow x=9\)