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Ta có: 1/3 + −2/5+ 1/6 + −1/5 ≤ x < −3/4+2/7+-1/4+3/5+5/7
⇒10-12+5-6/30≤ x< -105+40-35+84+100/140
⇒-3/30≤ x <84/140
⇒-0,1≤ x < 0,6
⇒x=0
a, 22x+1 = 83.165
=> 22x+1 = (23)3 . (24)5
=> 22x+1 = 29 . 220
=> 22x+1 = 229
=> 2x+1 = 29 ( vì cơ số 2 > 0 )
=> 2x = 29 - 1
=> 2x = 28
=> x = 28 : 2
=> x = 14
Vậy x = 14
Ta có : \(6.2^{12}+2^{13}=3.2.2^{12}+2^{13}=3.2^{13}+2^{13}=2^{13}\left(3+1\right)=2^{13}.4=2^{13}.2^2=2^{15}=\left(2^3\right)^5=8^5\)\(3^{10}=\left(3^2\right)^5=9^5\)
Vì 8 < 9
=> 85 < 95 ( vì 5 > 0 )
=> 6.212 + 213 < 310
Vậy 6.212 + 213 < 310
1: \(=-\dfrac{7}{80}\cdot\dfrac{4}{7}-\dfrac{2}{9}:\dfrac{16}{3}+\dfrac{5}{24}\cdot\left(\dfrac{-50+38}{15}\right)^2\)
\(=\dfrac{-1}{20}-\dfrac{2}{9}\cdot\dfrac{3}{16}+\dfrac{5}{24}\cdot\dfrac{16}{25}\)
\(=\dfrac{-1}{20}-\dfrac{1}{24}+\dfrac{2}{15}\)
\(=\dfrac{-6-5+16}{120}=\dfrac{5}{120}=\dfrac{1}{24}\)
2: \(=1500-\left\{10^3-11\cdot\left[49-5\cdot8\right]\right\}\)
\(=1500-\left\{1000-11\cdot9\right\}\)
\(=1500-1000+99=599\)
9) \(\dfrac{x}{4}=\dfrac{9}{x}\)
Theo định nghĩa về hai phân số bằng nhau, ta có:
\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
8)
\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)
7)
\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)
6)
\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)
5)
\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)
4)
\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
3)
\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)
2)
\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)
1)
\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\Leftrightarrow\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ....
Bài 1:
a, \(\left(x-2\right)^2=9\)
\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)
b, \(\left(3x-1\right)^3=-8\)
\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)
\(\Rightarrow x=-\dfrac{1}{3}\)
c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)
\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)
d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)
Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)
e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)
Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)
f, \(\left(\dfrac{1}{2}\right)^{2x-1}=8\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^{-3}\) Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(2x-1=-3\) \(\Rightarrow2x=-2\Rightarrow x=-1\) Chúc bạn học tốt!!!
Câu 1:
a) \(-\dfrac{2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Rightarrow-\dfrac{2}{3x}+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}x+\dfrac{2}{3}x=\dfrac{1}{6}+\dfrac{1}{3}\)
\(\Rightarrow x.\left(\dfrac{2}{3}+\dfrac{2}{3}\right)=\dfrac{1}{2}\)
\(\Rightarrow x.\dfrac{4}{3}=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}:\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{3}{8}\)
a) \(\dfrac{-3}{8}-\dfrac{1}{8}:x=-5\dfrac{1}{4}-2\dfrac{1}{4}\)
\(\dfrac{-3}{8}-\dfrac{1}{8}:x=\dfrac{-21}{4}-\dfrac{9}{4}\)
\(\dfrac{-3}{8}-\dfrac{1}{8}:x=\dfrac{-15}{2}\)
\(\dfrac{1}{8}:x=\dfrac{-3}{8}-\dfrac{-15}{2}\)
\(\dfrac{1}{8}:x=\dfrac{57}{8}\)
\(x=\dfrac{1}{8}:\dfrac{57}{8}\)
\(x=\dfrac{1}{57}\)
b) \(\left(1\dfrac{1}{3}-25\%-\dfrac{5}{12}\right)-2x=1,6:\dfrac{3}{5}\)
\(\left(1\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{5}{12}\right)-2x=\dfrac{8}{5}:\dfrac{3}{5}\)
\(\left(\dfrac{4}{3}-\dfrac{1}{4}-\dfrac{5}{12}\right)-2x=\dfrac{8}{3}\)
\(\left(\dfrac{13}{12}-\dfrac{5}{12}\right)-2x=\dfrac{8}{3}\)
\(\dfrac{2}{3}-2x=\dfrac{8}{3}\)
\(2x=\dfrac{2}{3}-\dfrac{8}{3}\)
\(2x=-2\)
\(x=\left(-2\right):2\)
\(x=-1\)
\(\dfrac{2^{2x-3}}{4^{10}}=8^3.16^5\)
\(\Rightarrow2^{2x-3}=8^3.16^5.4^{10}\)
\(\Rightarrow2^{2x-3}=2^9.2^{20}.2^{20}\)
\(\Rightarrow2^{2x-3}=2^{49}\)
Vì \(2\ne\pm1;2\ne0\) nên \(2x-3=49\)
\(\Rightarrow2x=52\Rightarrow x=26\)
Vậy..........
Chúc bạn học tốt!!!