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VH
1
23 tháng 8 2023
\(\dfrac{8^2.125.9^2-32.5^3.81}{20^3.3^4-6^8.5^4}\)
\(=\dfrac{2^6.5^3.3^4-2^5.5^3.3^4}{4^3.5^3.3^4-2^8.3^8.5^4}\)
\(=\dfrac{2^6.5^3.3^4-2^5.5^3.3^4}{2^6.5^3.3^4-2^8.3^8.5^4}\)
\(=\dfrac{2^5.5^3.3^4\left(2-1\right)}{2^6.5^3.3^4\left(1-2^2.3^4.5\right)}\)
\(=\dfrac{2^5.5^3.3^4.1}{2^6.5^3.3^4\left(1-810\right)}\)
\(=\dfrac{1}{2.\left(-809\right)}\)
\(=-\dfrac{1}{1618}\)
PP
0
17 tháng 12 2022
a: \(\Leftrightarrow4^{x-5}\cdot17=68\)
=>4^x-5=4
=>x-5=1
=>x=6
b: \(\Leftrightarrow\dfrac{1}{3}:\left|2x-1\right|=\dfrac{1}{3}+\dfrac{2}{3}=1\)
=>|2x-1|=1/3
=>2x-1=1/3 hoặc 2x-1=-1/3
=>x=2/3 hoặc x=1/3
c: =>|2x-2|=|3x+15|
=>3x+15=2x-2 hoặc 3x+15=-2x+2
=>x=-17 hoặc x=-13/5
H9
HT.Phong (9A5)
CTVHS
1 tháng 11 2023
a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
1 tháng 11 2023
a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
NT
1
DH
19 tháng 2 2020
\(a,\frac{2}{3}+\frac{7}{4}:x=\frac{5}{6}\)
\(\Leftrightarrow\frac{7}{4}:x=\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow\frac{7}{4}:x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{21}{2}\)
\(b,\left(x+\frac{5}{3}\right).\left(x-\frac{5}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{5}{3}=0\\x-\frac{5}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{5}{3}\\\frac{5}{4}\end{matrix}\right.\)
\(c,\left(x-1,2\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,2=2\\x-1,2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3,2\\x=-0,8\end{matrix}\right.\)
\(d,\left(x+1\right)^3=-125\)
\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x+1=-5\)
\(\Leftrightarrow x=-6\)
Vậy ...........................................................
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 7 2023
a, (-0,2)2 \(\times\) 5 - \(\dfrac{2^{13}\times27^3}{4^6\times9^5}\)
= 0,04 \(\times\) 5 - \(\dfrac{2^{13}\times3^9}{2^{12}\times3^{10}}\)
= 0,2 - \(\dfrac{2}{3}\)
= \(\dfrac{2}{10}\) - \(\dfrac{2}{3}\)
= - \(\dfrac{7}{15}\)
b, \(\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\)
= \(\dfrac{5^6+4.5^6+8.5^6}{26.5^6}\)
= \(\dfrac{5^6.\left(1+4+8\right)}{26.5^6}\)
= \(\dfrac{1}{2}\)
16 tháng 2 2021
b) Ta có: \(-5+\left|3x-1\right|+6=\left|-4\right|\)
\(\Leftrightarrow\left|3x+1\right|+1=4\)
\(\Leftrightarrow\left|3x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=3\\3x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{3};-\dfrac{4}{3}\right\}\)
c) Ta có: \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-1-1\right)\left(x-1+1\right)=0\)
\(\Leftrightarrow x\cdot\left(x-1\right)^2\cdot\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;1;2\right\}\)
d) Ta có: \(5^{-1}\cdot25^x=125\)
\(\Leftrightarrow5^{-1}\cdot5^{2x}=5^3\)
\(\Leftrightarrow5^{2x-1}=5^3\)
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=4\)
hay x=2
Vậy: x=2
13 tháng 10 2021
\(5^6.20^{-1}.2^2.\left(-3\right)^2:125=15625.\dfrac{1}{20}.4.9:125=225\)