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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 ...........................................................
BN
2
10 tháng 7 2019
\(\left(x-1,2\right)^2=4\)
⇔\(x^2-2.x.1,2+1,2^2=4\)
⇔\(x^2-2,4x+1,44=4\)
⇔\(x^2-2,4x=4-1,44\)
⇔\(x\left(x-2,4\right)=2,56\)
⇔\(x=2,56\) hoặc \(x-2,4=2,56\)
⇔\(x=2,56\) hoặc \(x=4,96\)
11 tháng 7 2019
a) \(\left(x-1,2\right)^2=4=2^2\)
\(\Leftrightarrow x-1,2=4\)
\(\Leftrightarrow x=5,2\)
b) \(\left(x+1\right)^3=-125=\left(-5\right)^3\)
\(\Leftrightarrow x+1=-5\)
\(\Leftrightarrow x=-6\)
c) \(\left(x+1,5\right)^8+\left(2,7-y\right)^{10}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\2,7-y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,7\end{matrix}\right.\)
NV
5 tháng 8 2023
a) \(2^x=8\)
⇔ \(2^x=2^3\)
⇒ \(x=3\)
b) \(3^x=27\)
⇔ \(3^x=3^3\)
⇒ \(x=3\)
c) \(\left(-\dfrac{1}{2}\right)x=\left(-\dfrac{1}{2}\right)^4\)
⇔ \(x=\left(-\dfrac{1}{2}\right)^4\div\left(-\dfrac{1}{2}\right)\)
⇔ \(x=\left(-\dfrac{1}{2}\right)^3\)
d) \(x\div\left(-\dfrac{3}{4}\right)=\left(-\dfrac{3}{4}\right)^2\)
⇔ \(x=\left(-\dfrac{3}{4}\right)^2\cdot\left(-\dfrac{3}{4}\right)\)
⇔ \(x=\left(-\dfrac{3}{4}\right)^3=-\dfrac{27}{64}\)
d) \(\left(x+1\right)^3=-125\)
⇔ \(\left(x+1\right)^3=\left(-5\right)^3\)
⇔ \(x+1=-5\)
⇔ \(x=-5-1=-6\)
5 tháng 8 2023
2:
a: (x-1,2)^2=4
=>x-1,2=2 hoặc x-1,2=-2
=>x=3,2(loại) hoặc x=-0,8(loại)
b: (x-1,5)^2=9
=>x-1,5=3 hoặc x-1,5=-3
=>x=-1,5(loại) hoặc x=4,5(loại)
c: (x-2)^3=64
=>(x-2)^3=4^3
=>x-2=4
=>x=6(nhận)
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
26 tháng 10 2021
a) \(\left|x\right|=2,5\Rightarrow\)\(\left[{}\begin{matrix}x=2,5\\x=-2,5\end{matrix}\right.\)
b) \(\left|x\right|=-1,2\left(VLý\right)\Rightarrow S=\varnothing\)
c) \(\left|x\right|+0,573=2\Rightarrow\left|x\right|=1,427\)
\(\Rightarrow\left[{}\begin{matrix}x=1,427\\x=-1,427\end{matrix}\right.\)
d) \(\left|x+\dfrac{1}{3}\right|-4=-1\Rightarrow\left|x+\dfrac{1}{3}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{100}{3}\end{matrix}\right.\)
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
\(x\) = -\(\dfrac{49}{10}\)
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
\(\left(x-1,2\right)^2=4\)
\(\left(x-1,2\right)^2=2^2\)
\(\Rightarrow\orbr{\begin{cases}x-1,2=2\\x-1,2=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=3,2\\x=-0,8\end{cases}}}\)
\(\left(x+1\right)^3=-125\)
\(\left(x+1\right)^3=\left(-5\right)^3\)
\(x+1=-5\)
\(x=-6\)
\(a,\left(x-1,2\right)^2=4\)
\(x-1,2=\pm2\)
\(\orbr{\begin{cases}x-1,2=2\\x-1,2=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=3,2\\x=-\frac{4}{5}\end{cases}}}\)
\(b,\left(x+1\right)^3=-125\)
\(x+1=-5\)
\(x=-6\)