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( 1/3 ) x + ( 1 / 3 ) x - 2 = 10 / 243
( 1 / 3 ) x ( 1 + ( 1/3 ) -2 = 10 / 243
( 1 / 3 ) x . 10 = 10 / 243
( 1 / 3 ) x = 1/243
( 1 / 3 ) x = ( 1/3 ) 5
=> x = 5
\(\left(x-1,2\right)^2=4\)
⇔\(x^2-2.x.1,2+1,2^2=4\)
⇔\(x^2-2,4x+1,44=4\)
⇔\(x^2-2,4x=4-1,44\)
⇔\(x\left(x-2,4\right)=2,56\)
⇔\(x=2,56\) hoặc \(x-2,4=2,56\)
⇔\(x=2,56\) hoặc \(x=4,96\)
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) \(\left(x+1,5\right)^8+\left(2,7-y\right)^{10}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\2,7-y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,7\end{matrix}\right.\)
\(\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^{x+2}=\dfrac{10}{243}\)
\(\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^x.\left(\dfrac{1}{3}\right)^2=\dfrac{10}{243}\)
\(\left(\dfrac{1}{3}\right)^x.\left[1+\left(\dfrac{1}{3}\right)^2\right]=\dfrac{10}{243}\)
\(\left(\dfrac{1}{3}\right)^x.\dfrac{10}{9}=\dfrac{10}{243}\)
\(\left(\dfrac{1}{3}\right)^x=\dfrac{10}{243}:\dfrac{10}{9}\)
\(\left(\dfrac{1}{3}\right)^x=\dfrac{1}{27}\)
\(\left(\dfrac{1}{3}\right)^x=\left(\dfrac{1}{3}\right)^3\)
\(\Rightarrow x=3\)
\(8;a,3^2.\frac{1}{243}.81^2.\frac{1}{3^3}\)
\(=\frac{3^2.\left(3^4\right)^2}{243.3^3}\)
\(=\frac{3^2.3^8}{3^5.3^3}\)
\(=\frac{3^{10}}{3^8}=3^2=9\)
\(b,\frac{4.2^5}{2^3.\frac{1}{16}}\)
\(=\frac{2^2.2^5}{2^3.\frac{1}{2^4}}\)
\(=\frac{2^7}{\frac{1}{2}}=2^7.2=2^8\)
a, \(3^2.\frac{1}{243}.81^2.\frac{1}{3}^3\)
\(=3^2.\frac{1}{243}.\left(3^4\right)^2.\frac{1}{27}\)
\(=3^2.\frac{1}{243}.3^8.\frac{1}{27}\)
\(=\frac{3^2.3^8}{243.27}\)
\(=\frac{3^2.3^8}{3^5.3^3}\)
\(=\frac{3^{10}}{3^8}=3^2=9\)
b, \(\left(4.2^5\right):\left(2^3.\frac{1}{16}\right)\)
\(=\left(2^2.2^5\right):\left(8.\frac{1}{16}\right)\)
\(=2^7:\frac{1}{2}\)
\(=2^8\)
phần b nhé
(X+0,7)3=-27
(X+0,7)3=(-3)3
X+0,7=-3
X=3-0,7=2,3
Vậy...
(3^x)^2=1/243.3^3
(3^x)^2=1/9
3x=1/3 hoặc 3x=-1/3
Suy ra x=1/9 hoặc x=-1/9
3^ x -1 = 1/243
3^x =1/243 +1
3^x = 244 / 243
Ta thấy đây ko phải lũy thừa của 3 => Ko có x thỏa mãn
81^-2x . 27^x =9^5
81^-2 . 81^x . 27^x =9^5
1/9^4 . (81.27)^x =9 ^5
3^6x = 9^5 : 1/9^4
3^6x = 9^9
3^6x = 3^18
=> 6x =18
x=3
2^x +2^x +3 =144
2.(2^x) =141
2^x+1 = 141
Ta thấy 141 ko phải lũy thừa của 2 => ko có x thỏa mãn
1) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2=1^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
2) \(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-2+1\)
\(\Leftrightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
3) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow x+2=x+4\)
\(\Leftrightarrow2=4\)
\(\Rightarrow x\in\varnothing\)
4) \(\left(2x-3\right)^5=-243\)
\(\Leftrightarrow\left(2x-3\right)^5=\left(-3\right)^5\)
\(\Leftrightarrow2x-3=-3\)
\(\Leftrightarrow2x=-3+3\)
\(\Leftrightarrow2x=0\)
\(\Rightarrow x=0\)
\(\left(x-2\right)^2=1\)
\(=>\left(x-2\right).\left(x-2\right)=1\)
\(+TH1\)\(\left(x-2\right)=1\)
\(=>1.1=1\left(TM\right)\)
\(+TH2\)\(\left(x-2\right)=-1\)
\(=>\left(-1\right).\left(-1\right)=1\left(TM\right)\)
\(=>x=1;-1\)
\(\left(x-2\right)^2=1\)
=> \(\left(x-2\right)^2=1^2\)
=> \(\left(x-2\right)^2=\left(-1\right)^2\)
=> \(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=1+2\\x=\left(-1\right)+2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{3;1\right\}.\)
\(\left(2x-1\right)^3=-8\)
=> \(\left(2x-1\right)^3=\left(-2\right)^3\)
=> \(2x-1=-2\)
=> \(2x=\left(-2\right)+1\)
=> \(2x=-1\)
=> \(x=\left(-1\right):2\)
=> \(x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}.\)
\(\left(2x-3\right)^5=-243\)
=> \(\left(2x-3\right)^5=\left(-3\right)^5\)
=> \(2x-3=-3\)
=> \(2x=\left(-3\right)+3\)
=> \(2x=0\)
=> \(x=0:2\)
=> \(x=0\)
Vậy \(x=0.\)
Chúc bạn học tốt!
a) (x - 2)2 = 1
=> x - 2 = 1 hoặc x - 2 = -1
x = 3 ; x = 1
Vậy x = 3; x = 1
b) (2x - 1)3 = -8
=> 2x - 1 = -2
2x = -1
x = \(\frac{-1}{2}\)
Vậy x = \(\frac{-1}{2}\)
c) (2x - 3)5 = -243
=> (2x - 3)5 = (-3)5
=> 2x - 3 = -3
2x = 0
x = 0
Vậy x = 0