1)tìm x
a)(1/3-5/6.x)^3+21/54=5/6
b)1/12-1/3(1/2.x-1)^4=1/16
c)(3/5)^x-1+5.(3/5)^x-1=54/25
d)(2/3)^x+(2/3)^x+2=101/243
2) so sánh
a)1/27^11 và 1/81^8
b)1/3^99 và 1/11^21
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2) a) \(\frac{1}{27^{11}}=\frac{1}{\left(3^3\right)^{11}}=\frac{1}{3^{33}}\)
\(\frac{21}{81^8}=\frac{21}{\left(3^4\right)^8}=\frac{21}{3^{32}}=\frac{21.3}{3^{33}}=\frac{63}{3^{33}}>\frac{1}{3^{33}}\)
=> \(\frac{21}{81^8}>\frac{1}{27^{11}}\)
b) Rõ ràng : 399 < 1121 => \(\frac{1}{399}>\frac{1}{11^{21}}\)
a) \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{5}{6}-\frac{21}{54}\)=> \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{24}{54}=\frac{4}{9}\)
=> \(\frac{1}{3}-\frac{5}{6}x=\sqrt[3]{\frac{4}{9}}\) => \(\frac{5}{6}x=1-\sqrt[3]{\frac{4}{9}}\)
=> x = \(\frac{6}{5}-\frac{6}{5}.\sqrt[3]{\frac{4}{9}}\)
b) => \(\frac{1}{13}\left(\frac{1}{2}x-1\right)^4=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}\)
=> \(\left(\frac{1}{2}x-1\right)^4=\frac{13}{48}\)
=> \(\frac{1}{2}x-1=\sqrt[4]{\frac{13}{48}}\) hoặc \(\frac{1}{2}x-1=-\sqrt[4]{\frac{13}{48}}\)
=> \(x=2+2\sqrt[4]{\frac{13}{48}}\) hoặc \(x=2-2\sqrt[4]{\frac{13}{48}}\)
\(\dfrac{11}{8}:x-\dfrac{2}{5}+-\dfrac{1}{6}=-\dfrac{1}{5}\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{1}{5}-\left(-\dfrac{1}{6}\right)\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{1}{5}+\dfrac{1}{6}\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{6}{30}+\dfrac{5}{30}\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{1}{30}\\ =>\dfrac{11}{8}:x=-\dfrac{1}{30}+\dfrac{2}{5}\\ =>\dfrac{11}{8}:x=-\dfrac{1}{30}+\dfrac{12}{30}\\ =>\dfrac{11}{8}:x=\dfrac{11}{30}\\ =>x=\dfrac{11}{8}:\dfrac{11}{30}\\ =>x=\dfrac{11}{8}.\dfrac{30}{11}\\ =>x=\dfrac{30}{8}\\ =>x=\dfrac{15}{4}\\ \dfrac{4}{7}x-\dfrac{1}{3}x+\left(-\dfrac{16}{21}\right)=-\dfrac{2}{3}\\ =>\left(\dfrac{4}{7}-\dfrac{1}{3}\right)x=-\dfrac{2}{3}-\left(-\dfrac{16}{21}\right)\\ =>\left(\dfrac{12}{21}-\dfrac{7}{21}\right)x=-\dfrac{2}{3}+\dfrac{16}{21}\\ =>\dfrac{5}{21}x=-\dfrac{14}{21}+\dfrac{16}{21}\\ =>\dfrac{5}{21}x=\dfrac{2}{21}\\ =>x=\dfrac{2}{21}:\dfrac{5}{21}\)
\(=>x=\dfrac{2}{21}.\dfrac{21}{5}\\ =>x=\dfrac{2}{5}\\ -\dfrac{11}{12}x+\dfrac{15}{2}\left(x+-\dfrac{1}{5}\right)=\dfrac{67}{8}\\ =>-\dfrac{11}{12}x+\dfrac{15}{2}.x-\dfrac{1}{5}=\dfrac{67}{8}\\ =>\left(-\dfrac{11}{12}+\dfrac{15}{2}\right)x=\dfrac{67}{8}+\dfrac{1}{5}\\ =>\left(-\dfrac{11}{12}+\dfrac{90}{12}\right)x=\dfrac{335}{40}+\dfrac{8}{40}\\ =>\dfrac{79}{12}x=\dfrac{343}{40}\\ =>x=\dfrac{343}{40}:\dfrac{79}{12}\\ =>x=\dfrac{343}{40}.\dfrac{12}{79}\\ =>x=\dfrac{343.12}{40.79}\\ =>x=\dfrac{343.3}{10.79}\\ =>x=\dfrac{1029}{790}\)
1:
a: Vì \(\dfrac{-4}{3}=\dfrac{-4\cdot3}{3\cdot3}=\dfrac{-12}{9}=\dfrac{12}{9}\\ \Rightarrow\dfrac{-4}{3}=\dfrac{12}{9}\)
b: Vì : \(-2\cdot3=-6\\ -6\cdot8=-48\)
nên 2 p/s ko bằng nhau
1.
a, => 21-x+3 < 0
=> 24-x < 0
=> x < 24
b, => 7+x > 0
=> x > -7
c, => x-1 < 0 ; x+2 > 0 ( vì x-1 < x+2 )
=> x < 1 ; x > -2
=> -2 < x < 1
Tk mk nha
a: \(x+6\dfrac{1}{8}=8\)
=>\(x+\dfrac{49}{8}=\dfrac{64}{8}\)
=>\(x=\dfrac{64}{8}-\dfrac{49}{8}=\dfrac{15}{8}\)
b: \(\dfrac{11}{2}\cdot x=\dfrac{1}{5}:\dfrac{1}{3}\)
=>\(x\cdot\dfrac{11}{2}=\dfrac{1}{5}\cdot3=\dfrac{3}{5}\)
=>\(x=\dfrac{3}{5}:\dfrac{11}{2}=\dfrac{3}{5}\cdot\dfrac{2}{11}=\dfrac{6}{55}\)
c: \(x\cdot\dfrac{3}{5}+\dfrac{2}{5}\cdot x=\dfrac{4}{9}+\dfrac{1}{3}\)
=>\(x\left(\dfrac{3}{5}+\dfrac{2}{5}\right)=\dfrac{4}{9}+\dfrac{3}{9}\)
=>\(x\cdot1=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}\)
CÁC BN BIẾT ANH Meowpeo và chị Simmy, Kamui và Naobi Chan , Sammy không
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Nếu có , cho mình và kết bạn với mình nhé!
TYM TYM
Các bn ơi giúp mình câu này đc ko mong các bạn giúp đỡ 4(x-1)=2x-6(x-2) các bạn giải hộ mình với
a.\(\dfrac{1}{3}\) + x = \(\dfrac{5}{6}\)
x = \(\dfrac{5}{6}\) - \(\dfrac{1}{3}\)
x = \(\dfrac{1}{2}\)
b. | x-1| - \(\dfrac{2}{5}\) = \(\dfrac{11}{10}\)
| x-1| = \(\dfrac{11}{10}\) + \(\dfrac{2}{5}\)
|x-1| = \(\dfrac{3}{2}\)
\(\left[{}\begin{matrix}x-1=\dfrac{3}{2}\\x-1=-\dfrac{3}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{2}+1\\x=-\dfrac{3}{2}+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
c, \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = 1
\(\dfrac{2}{3}\) (\(\dfrac{x}{2}\) + 3) = 1 - \(\dfrac{1}{3}\)
\(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = \(\dfrac{2}{3}\)
\(\dfrac{x}{2}\) + 3 = 1
\(\dfrac{x}{2}\) = 1 - 3
\(\dfrac{x}{2}\) = -2
\(x\) = -4
d, \(\dfrac{x+2}{3}\) = \(\dfrac{27}{x+2}\)
(x+2)2 = 27.3
(x+2) =92
\(\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)
a) => \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{5}{6}-\frac{21}{54}=\frac{24}{54}=\frac{4}{9}\)
=> \(\frac{1}{3}-\frac{5}{6}x=\sqrt[3]{\frac{4}{9}}\) => \(\frac{5}{6}x=\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\) => \(x=\frac{6}{5}.\left(\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\right)\)
b) \(\frac{1}{3}\left(\frac{1}{2}x-1\right)^4=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}\) => \(\left(\frac{1}{2}x-1\right)^4=\frac{3}{48}=\frac{1}{16}\)
=> \(\frac{1}{2}x-1=\frac{1}{2}\) hoặc \(\frac{1}{2}x-1=-\frac{1}{2}\)
=> \(\frac{1}{2}x=\frac{3}{2}\) hoặc \(\frac{1}{2}x=\frac{1}{2}\) => x = 3 hoặc x = 1
c) \(\left(1+5\right).\left(\frac{3}{5}\right)^{x-1}=\frac{54}{25}\) => \(\left(\frac{3}{5}\right)^{x-1}=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
=> x - 1= 2 => x = 3
d) \(\left(1+\left(\frac{2}{3}\right)^2\right).\left(\frac{2}{3}\right)^x=\frac{101}{243}\) => \(\frac{13}{9}.\left(\frac{2}{3}\right)^x=\frac{101}{243}\)
=> \(\left(\frac{2}{3}\right)^x=\frac{101}{243}:\frac{13}{9}=\frac{101}{351}\) (có lẽ đề sai)
2) \(\frac{1}{27^{11}}=\frac{1}{\left(3^3\right)^{11}}=\frac{1}{3^{33}}\); \(\frac{1}{81^8}=\frac{1}{\left(3^4\right)^8}=\frac{1}{3^{32}}\)
Vì 333 > 332 => \(\frac{1}{3^{33}}\) < \(\frac{1}{3^{32}}\) => \(\frac{1}{27^{11}}\) < \(\frac{1}{81^8}\)
b) \(\frac{1}{3^{99}}=\frac{1}{\left(3^3\right)^{33}}=\frac{1}{27^{33}}