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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 \(x = \frac{{ - 9}}{{10}}\).
a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)
=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
=> x + 1 = 0
=> x = -1
b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)
=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)
=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)
=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
=> x - 2021 = 0
=> x = 2021
c) \(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)
=> \(-\frac{1}{12}x+6=7\)
=> \(-\frac{1}{12}x=1\)
=> x = -12
7/4.x+3/2=-4/5
7/4.x=-4/5-3/2
7/4.x=-23/10
x=-23/10:7/4
x=-46/35
vậy x=-46/35
1/4+3/4.x=3/4
1.x=3/4
x=3/4:1
x=3/4
vậy x=3/4
x.(1/4+1/5)-(1/7+1/8)=0
x.9/20-15/56=0
x.51/280=0
x=0:51/280
x=0
vậy x=0
3/35-(3/5+x)=2/7
(3/5+x)=3/35-2/7
(3/35+x)=-1/5
x=-1/5-3/5
x=-4/5
vậy x=-4/5
\(a,1\frac{3}{4}.x+1\frac{1}{2}=\frac{4}{5}\)
\(\frac{7}{4}.x=\frac{4}{5}-\frac{3}{2}\)
\(\frac{7}{4}.x=\frac{-7}{10}\)
\(x=\frac{-7}{10}:\frac{7}{4}\)
\(x=\frac{-2}{5}\)
\(b,\frac{1}{4}+\frac{3}{4}.x=\frac{3}{4}\)
\(\frac{3}{4}.x=\frac{3}{4}-\frac{1}{4}\)
\(\frac{3}{4}.x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{3}{4}\)
\(x=\frac{2}{3}\)
\(c,x.\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)
\(x.\frac{9}{20}-\frac{15}{56}=0\)
\(x.\frac{9}{20}=\frac{15}{56}\)
\(x=\frac{15}{56}:\frac{9}{20}\)
\(x=\frac{25}{42}\)
\(d,\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)
\(\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)
\(\frac{3}{5}+x=\frac{-1}{5}\)
\(x=\frac{-1}{5}-\frac{3}{5}\)
\(x=\frac{-4}{5}\)
Học tốt
Câu a) số lớn lắm
b) \(3^{-3}\cdot3^5\cdot3^x=3^8\)
=> \(\frac{1}{27}\cdot3^5\cdot3^x=3^8\)
=> \(\frac{1}{27}\cdot3^x=3^3\)
=> \(3^x=3^3:\frac{1}{27}=3^3:\left(\frac{1}{3}\right)^3=3^3:\frac{1^3}{3^3}=3^3\cdot3^3=3^6\)
=> x = 6
b) \(\left(7x+2\right)^{-1}=3^{-2}\)
=> \(\frac{1}{7x+2}=\frac{1}{9}\)
=> 7x + 2 = 9
=> 7x = 7
=> x = 1
Bài 2:
a) \(3^4\cdot\frac{1}{729}\cdot81^3\cdot\frac{1}{9^2}\)
\(=3^4\cdot\left(\frac{1}{3}\right)^6\cdot\left(3^4\right)^3\cdot\left(\frac{1}{3}\right)^4\)
\(=3^4\cdot\left(\frac{1}{3}\right)^6\cdot3^{12}\cdot\left(\frac{1}{3}\right)^4=3^{16}\cdot\left(\frac{1}{3}\right)^{10}=\frac{3^{16}}{3^{10}}=3^6\)
b) \(\left(8\cdot2^5\right):\left(2^4\cdot\frac{1}{32}\right)=\left(2^3\cdot2^5\right):\left(2^4\cdot\left(\frac{1}{2}\right)^5\right)\)
\(=2^8:\left(2^4\cdot\frac{1^5}{2^5}\right)=2^8:\left(\frac{2^4}{2^5}\right)=2^8:2^{-1}=512\)
c) \(12^8\cdot9^{12}=\left(2^2\cdot3\right)^8\cdot\left(3^2\right)^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)
d) Tương tự
Câu b thôi các bạn nhé, câu a mình ko cần nx với cả mình ghi sai dữ liệu câu a r
a, \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot...\cdot\frac{30}{62}\cdot\frac{31}{64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{2\cdot2\cdot3\cdot2\cdot4\cdot2\cdot5\cdot2\cdot....\cdot31\cdot2\cdot32\cdot2}=2x\)
\(\Leftrightarrow\frac{1}{2\cdot2\cdot2\cdot2\cdot....\cdot2\cdot2\cdot32}=2x\)
Có : (31 - 1) : 1 + 1 = 31 (thừa số 2)
\(\Rightarrow\frac{1}{2^{31}.32}=2x\)
\(\Rightarrow x=\frac{1}{2^{31}.32}\div2\)
b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow x+1=x+4\)
\(\Leftrightarrow0=3\text{ (vô lý) }\)
a)
\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)
Vậy \(x = \frac{{ - 1}}{{15}}\).
b)
\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)
Vậy \(x = 3\).
c)
\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)
Vậy \(x = \frac{8}{5}\)
d)
\(\begin{array}{l}3,2:x = - \frac{6}{{11}}\\\frac{{16}}{5}:x = - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)
Vậy \(x = \frac{{ - 88}}{{15}}\).
a)\(\left(\frac{-3}{4}\right)^{10}\cdot x=\left(-\frac{3}{4}\right)^{12}\)
\(x=\left(\frac{-3}{4}\right)^{12}:\left(-\frac{3}{4}\right)^{10}\)
\(x=\left(-\frac{3}{4}\right)^{12-10}\)
\(x=\left(-\frac{3}{4}\right)^2=\frac{9}{16}\)
b)\(x:\left(\frac{2}{3}\right)^8=\left(\frac{9}{4}\right)^4\)
\(x=\left(\frac{9}{4}\right)^4\cdot\left(\frac{2}{3}\right)^8\)
\(x=\left(\frac{9}{4}\right)^4\cdot\left[\left(\frac{2}{3}\right)^2\right]^4\)
\(x=\left(\frac{9}{4}\right)^4\cdot\left(\frac{4}{9}\right)^4\)
\(x=\left(\frac{9}{4}\cdot\frac{4}{9}\right)^4\)
\(x=1^4=1\)
c)\(\left(x-1\right)^3=-64\)
\(\Rightarrow\left(x-1\right)^3=\left(-4\right)^3\)
\(\Rightarrow x-1=-4\)
\(\Rightarrow x=-3\)
d)\(\left(x+1\right)^4=81\)
\(\Rightarrow\left(x+1\right)^4=\left(\pm3\right)^4\)
\(\Rightarrow\orbr{\begin{cases}x+1=3\\x+1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)
a)\(\left(\frac{-3}{4}\right)^{10}\cdot x=\left(-\frac{3}{4}\right)^{12}\)
\(\Leftrightarrow x=\left(\frac{-3}{4}\right)^{12}:\left(-\frac{3}{4}\right)^{10}\)
\(\Leftrightarrow x=\left(-\frac{3}{4}\right)^{12-10}\)
\(\Leftrightarrow x=\left(-\frac{3}{4}\right)^2=\frac{9}{16}\)
b)\(x:\left(\frac{2}{3}\right)^8=\left(\frac{9}{4}\right)^4\)
\(\Leftrightarrow x=\left(\frac{9}{4}\right)^4\cdot\left(\frac{2}{3}\right)^8\)
\(\Leftrightarrow x=\left(\frac{9}{4}\right)^4\cdot\left[\left(\frac{2}{3}\right)^2\right]^4\)
\(\Leftrightarrow x=\left(\frac{9}{4}\right)^4\cdot\left(\frac{4}{9}\right)^4\)
\(\Leftrightarrow x=1^4\Leftrightarrow x=1\)
c) \(\left(x-1\right)^3=-64\)
\(\Leftrightarrow\left(x-1\right)^3=\left(-4\right)^3\)
\(\Leftrightarrow x-1=-4\Leftrightarrow x=-3\)
d)\(\left(x+1\right)^4=81\)
\(\Leftrightarrow\left(x+1\right)^4=\left(3\right)^4\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=3\\x+1=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)