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26 tháng 11 2018
a) x = 8
Vì khi cơ số là 0 thì có mũ mấy lên bao nhiêu cũng = 0
=>( 2.8-16)^8-(2.8-16)^3=(16-16)^8-(16-16)^3=0^8-0^3=0-0=0
b) x = 2
Vì khi cơ số =1 thì mũ lên bao nhiêu cũng =1
Mỏi tay quá , chắc đến đây đã hiểu rồi tự làm nha ! Nhớ ks nhé !
30 tháng 4 2019
c) \(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{2};3\right\}\)
e) \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(\Leftrightarrow2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}:2=\frac{7}{4}.\frac{1}{2}=\frac{7}{8}\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\left(-\frac{7}{8}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
30 tháng 4 2019
Mấy bài này ko quá khó, tải MathPhoto trong đt về nó tự lm
DH
8 tháng 9 2017
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!!!
MV
8 tháng 8 2017
a,
\(\left(x-\dfrac{1}{2}\right)^2=0\\ \Rightarrow x-\dfrac{1}{2}=0\\ \Rightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
b,
\(\left(x-2\right)^2=1\\ \Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy \(x=3\text{ hoặc }x=1\)
c,
\(\left(2x-1\right)^3=-8\\ \Rightarrow2x-1=-2\\ \Rightarrow2x=-1\\ \Rightarrow x=\dfrac{-1}{2}\)
Vậy \(x=\dfrac{-1}{2}\)
d,
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{-1}{4}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
Vậy \(x=\dfrac{-1}{4}\text{ hoặc }x=\dfrac{-3}{4}\)
LG
8 tháng 8 2017
a) \(\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow\left(x-\dfrac{1}{2}\right)^2=0^2\)
\(\Rightarrow x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
b) \(\left(x-2\right)^2=1\Rightarrow\left(x-2\right)^2=1^2\)
\(\Rightarrow\left[{}\begin{matrix}x-2=-1\\x-2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1+2\\x=1+2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
c) \(\left(2x-1\right)^3=-8\Rightarrow\left(2x-1\right)^3=-2^3\)
\(\Rightarrow2x-1=-3\Rightarrow2x=-3+1\)
\(\Rightarrow2x=2\Rightarrow x=1\)
Vậy \(x=1\)
d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=-\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}-\dfrac{1}{2}\\x=\dfrac{1}{4}-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
29 tháng 6 2019
\(a,(2x-1)(y-2)=13\)
\(\Leftrightarrow(2x-1)(y-2)\inƯ(13)\)
\(\Leftrightarrow(2x-1)(y-2)\in\left\{\pm1;\pm13\right\}\)
Lập bảng :
| 2x - 1 | 1 | -1 | 13 | -13 |
| y - 2 | -13 | 13 | -1 | 1 |
| x | 1 | 0 | 7 | -6 |
| y | -11 | 15 | 1 | 3 |
22 tháng 7 2016
\(f\)) \(32^{-x}.16^x=1024\)
\(\left(2\right)^{-5x}.2^{4x}=2^{10}\)
\(\Leftrightarrow2^{4x-5x}=2^{10}\)
\(\Leftrightarrow2^{-x}=2^{10}\)
\(\Leftrightarrow-x=10\)
\(\Leftrightarrow x=-10\)
\(g\)) \(3^{x-1}.5+3^{x-1}=162\)
\(3^{x-1}.\left(5+1\right)=162\)
\(3^{x-1}.6=162\)
\(3^{x-1}=162:6\)
\(3^{x-1}=27\)
\(\Leftrightarrow3^{x-1}=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
\(h\)) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^6.\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(2x-1\right)^6.\left[1-\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^2=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=1\\\left(2x-1\right)^2=\left(1,-1\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x-1=-1\\2x-1=1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x=0\\2x=2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=0\\x=1\end{cases}}\)
\(i\)) \(5^x+5^{x+2}=650\)
\(5^x.\left(1+5^2\right)=650\)
\(5^x.26=650\)
\(5^x=650:26\)
\(5^x=25\)
\(\Leftrightarrow5^x=5^2\)
\(\Leftrightarrow x=2\)