tìm x ∈ Q biết : \(\dfrac{3\left(x+1\right)}{2}\)=\(\dfrac{8}{27\left(x-1\right)}\)
Tìm x
1) \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\) 2) \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\) 3) \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
4) \(\left(\dfrac{4}{9}\right)^x=\left(\dfrac{8}{27}\right)^{10}\) 5) \(2^x=4^5.4^3\) help me !!!!
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
Tìm x biết:
a) \(\dfrac{-32}{\left(-2\right)^x}=4\) f) \(\left(3x-1\right)^3=\dfrac{-8}{27}\)
b) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\) g) \(\left(2x+3\right)^2=\dfrac{9}{121}\)
c) \(\dfrac{1}{9}.27^x=3^x\) h) \(5^x+5^{x+2}=650\)
d) \(9^x:3^3=\dfrac{1}{243}\) i) \(\left(x-7\right)^{x+1}-\left(x-7\right)=0\)
e) \(\dfrac{x7}{81}=27\) m) \(\left(\dfrac{-3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
\(\left(3-x\right)^3=-\dfrac{27}{64};\left(x-5\right)^3=\dfrac{1}{-27};\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8};\left(2x-1\right)^2=\dfrac{1}{4};\left(2-3x\right)^2=\dfrac{9}{4};\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)
\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) hoặc \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(=>2x-1=\dfrac{1}{2}\) \(2x-1=\dfrac{-1}{2}\)
\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\) \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)
\(2x=\dfrac{3}{2}\) \(2x=\dfrac{1}{2}\)
\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\) \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)
\(x=\dfrac{3}{4}\) \(x=\dfrac{1}{4}\)
____________
\(\left(2-3x\right)^2=\dfrac{9}{4}\)
\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)
\(=>2-3x=\dfrac{3}{2}\) \(2-3x=\dfrac{-3}{2}\)
\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\) \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)
\(3x=\dfrac{1}{2}\) \(3x=\dfrac{7}{2}\)
\(x=\dfrac{1}{2}.\dfrac{1}{3}\) \(x=\dfrac{7}{2}.\dfrac{1}{3}\)
\(x=\dfrac{1}{6}\) \(x=\dfrac{7}{6}\)
______________
\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này
(3-x)3=(-\(\dfrac{3}{4}\))3
3-x=-\(\dfrac{3}{4}\)
x=3-(-\(\dfrac{3}{4}\))
x=\(\dfrac{15}{4}\)
Tìm x:
\(a\)) \(\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)
\(b\)) \(\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{27}{8}\right)^3=\dfrac{81}{16}\)
\(c\)) \(\dfrac{1}{2}.2^x+4.2^x=9.2^5\)
\(d\)) \(\text{12 - (2x +1)}^2=-69\)
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)
\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)
\(x=\dfrac{1}{2}+\dfrac{1}{3}\)
\(x=\dfrac{1}{5}\)
Tìm x biết:
a) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
b) \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\left(x\ne0\right)\)
c) \(\left(2x+3\right)^2=\dfrac{9}{121}\)
d) \(\left(2x-1\right)^3=\dfrac{-8}{27}\)
c. \(^{ }\left(2x+3\right)^2=\dfrac{9}{121}\)
=> \(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
=> 2x +3 = \(\dfrac{3}{11}\) hoặc 2x+3 = \(\dfrac{-3}{11}\)
=> x= \(\dfrac{-15}{11}\) hoặc x = \(\dfrac{-18}{11}\)
d. \(\left(2x-1\right)^3=\dfrac{-8}{27}\)
=> \(\left(2x-1\right)^3=\left(\dfrac{-2}{3}\right)^3\)
=> 2x-1 = \(\dfrac{-2}{3}\)
=> x= \(\dfrac{1}{6}\)
a. \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
=> \(\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)
=> x + \(\dfrac{1}{2}\) = \(\dfrac{1}{4}\) hoặc \(x+\dfrac{1}{2}=\dfrac{-1}{4}\)
=> x = \(\dfrac{-1}{4}\) hoặc \(x=\dfrac{-3}{4}\)
1)Thực hiện phép tính
a) \(\dfrac{9^8\cdot5^6}{3^7\cdot27^3\cdot25^4}\) b) \(\dfrac{5^{20}\cdot6^{18}}{15^{20}\cdot4^{10}}\) c) \(\left(\dfrac{1}{3}\right)^{50}\cdot9^{25}-\left(\dfrac{2}{3}\right)^{100}:\left(\dfrac{8}{27}\right)^{33}\) d) \(\left(\dfrac{3}{8}\right)^{60}\cdot\left(\dfrac{2}{3}\right)^{60}:\left(\dfrac{1}{2}\right)^{119}\cdot\left(\dfrac{1}{2}\right)\)
2) Tìm x, biết:
a) \(\dfrac{x}{12}=\dfrac{5}{x}\) b)\(8^x=2^{2x+3}\) c)\(\dfrac{1}{2}\sqrt{\dfrac{1}{2}x-2}-\dfrac{2}{3}=\dfrac{1}{3}\)
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
Tìm x, biết:
\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=\dfrac{-1}{4}\)
\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-1}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-1}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}=\dfrac{-1}{4}\)
\(\Leftrightarrow\left(x-1\right)\left(-1\right)=4\)
\(\Leftrightarrow x-1=-4\)
\(\Leftrightarrow x=-3\left(tm\right)\)
Vậy ..............
Tìm x biết: a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\) b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\) d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}.\dfrac{10}{6}\)
a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
Th1 : \(x-\dfrac{1}{2}=0\)
\(x=0+\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\)
Th2 : \(-3-\dfrac{x}{2}=0\)
\(\dfrac{x}{2}=-3\)
\(x=\left(-3\right)\cdot2\)
\(x=-6\)
Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)
b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}+\dfrac{1}{8}\)
\(x=\dfrac{3}{4}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)
\(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)
\(\dfrac{3}{2}+x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}-\dfrac{3}{2}\)
\(x=0\)
d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)
\(x+\dfrac{1}{3}=-4\)
\(x=-4-\dfrac{1}{3}\)
\(x=-\dfrac{13}{3}\)
1)Tìm x, biết :
\(4.\left[3x-1\right]+\left[x\right]-2.\left[x-5\right]+7.\left[x-3\right]=12\)
\(\left[2\dfrac{1}{5}-x\right]+\left[x-\dfrac{1}{5}\right]+8\dfrac{1}{5}=1,2\)
\(3.\left[x+4\right]-\left[2.x+1\right]-5.\left[x-3\right]+\left[x-9\right]=5\)
\(2\left[x+3\dfrac{1}{2}\right]+\left[x\right]-3\dfrac{1}{2}=\left[2\dfrac{1}{5}-x\right]\)