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tìm x biết
a) |2x−1|=x+4|2x−1|=x+4
* \(2x-1=x+4\)
\(<=> 2x-x=4+1\)
\(<=> x=5\)
* \(-2x-1=x+4\)
\(<=> -2x-x=4+1\)
\(<=> -3x=5\)
\(<=> x=\dfrac{-3}{5}\) (loại)
Vậy \(x=5\)
b) (3x−1)4=81
\(<=> (3x-1)^4=3^4\)
\(<=> 3x-1=4\)
\(<=> 3x=5\)
\(<=> x=\dfrac{5}{3}\)
Vậy \(x=\dfrac{5}{3}\)
(3x−1)4=8c) c,(x−2)3=−64(x−2)3=−64
\(<=> (x-2)^3=(-4)^3\)
\(<=> x-2=-4\)
\(<=> x=-2\)
Vậy \( x=-2\)
h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)
\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)
\(x=\left(\frac{3}{7}\right)^2\)
\(x=\frac{9}{49}\)
Vậy...
b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)
\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)
\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)
\(x=-\frac{1}{3}\)
Vậy...
c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
=>\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
Vậy...
d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)
=>\(x+\frac{1}{4}=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{1}{4}\)
\(x=\frac{5}{12}\)
Vậy...
Phù, mãi mới xong, tk cho mk nha bn
a, \(\left|2x-1\right|=x+4\)
\(\orbr{\begin{cases}2x-1=x+4\\-2x+1=x+4\end{cases}\Rightarrow\orbr{\begin{cases}x-5=0\\-3x-3=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)
b, \(\left(3x-1\right)^4=81\)
\(\left(3x-1\right)^4=3^4\Leftrightarrow\orbr{\begin{cases}3x-1=3\\3x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}3x-4=0\\3x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{2}{3}\end{cases}}}\)
c, \(\left(x-2\right)^3=-64\)
\(\left(x-2\right)^3=\left(-4\right)^3\Leftrightarrow x-2=-4\Leftrightarrow x=-2\)
d, chia 2 TH làm như phần a đó, chắc vậy :v
Bài làm
a) \(\left|2x-1\right|=x+4\)
\(\Rightarrow\orbr{\begin{cases}2x-1=x+4\\2x-1=-x-4\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)
Vậy x = { 5; -1 }
b) \(\left(3x-1\right)^4=81\)
\(\Rightarrow\left(3x-1\right)^4=\left(\pm3\right)^8\)
\(\Rightarrow\orbr{\begin{cases}3x-1=3\\3x-1=-3\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\3x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{2}{3}\end{cases}}}\)
Vậy x = { -4/2; -2/3 }
c) \(\left(x-2\right)^3=-64\)
\(\Rightarrow\left(x-2\right)^3=-4^3\)
\(\Rightarrow x-2=-4\)
\(\Rightarrow x=-2\)
d) \(\left|x-3\right|-\left|2x-1\right|=0\)
\(\Rightarrow\left|x-3\right|=\left|2x-1\right|\)
\(\Rightarrow\orbr{\begin{cases}x-3=2x-1\\x-3=-2x+1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-x=2\\3x=4\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{4}{3}\end{cases}}}\)
Vậy x = { -2; 4/3 }
a) ( x + 5 )3 = -64
x + 5 = - 4
x = - 4 - 5
x = -9
b) (2x - 3)2=9
2x - 3 = 3
2x = 3+3
2x = 6
x = 6 : 2
x = 3
e) \(\dfrac{8}{2x}=4\)
=> 4 . 2x = 8
8x =8
x = 8 : 8
x = 1
g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)
\(\left(\dfrac{1}{2}\right)^{2x}:\left(\dfrac{1}{2}\right)^1=\dfrac{1}{8}\)
\(\left(\dfrac{1}{2}\right)^{2x}:\dfrac{1}{2}=\dfrac{1}{8}\)
\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{8}.\dfrac{1}{2}\)
\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{16}\)
\(\left(\dfrac{1}{2}\right)^{2x}=\left(\dfrac{1}{2}\right)^{2.2}\)
=> x = 2
h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)
\(\dfrac{1}{4}.x=\dfrac{1}{32}\)
x = \(\dfrac{1}{32}:\dfrac{1}{4}\)
x = \(\dfrac{1}{8}\)
i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)
\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)
\(x=\dfrac{-1}{27}\)
a) (x + 5)3 = -64
=> (x + 5)3 = (-4)3
x + 5 = -4
x = -4 - 5
x = -9
b) (2x - 3)2 = 9
=> (2x - 3)2 = (\(\pm\)3)2
=> 2x - 3 = 3 hoặc 2x - 3 = -3
*2x - 3 = 3
2x = 3 + 3
2x = 9
x = \(\dfrac{9}{2}\)
*2x - 3 = -3
2x = -3 + 3
2x = 0
x = 0 : 2
x = 0
Vậy x \(\in\left\{\dfrac{9}{2};0\right\}\)
c) \(\dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}\)
=> \(x.\dfrac{x}{2}=4.\dfrac{4}{2}\)
\(\dfrac{x}{2}=8\)
x = 8 : 2
x = 4
d) \(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
=> (-2)n . (-2)2= (-2)5
(-2)n = (-2)5 : (-2)2
(-2)n = (-2)3
Vậy n = 3
e) \(\dfrac{8}{2x}=4\)
=> 2x . 4 = 8
2x = 8 : 4
2x = 2
x = 1
g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^3\)
2x - 1 = 3
2x = 3 + 1
2x = 4
x = 4 : 2
x = 2
h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)
\(x=\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{2}\right)^2\)
\(x=\left(\dfrac{1}{2}\right)^3\)
\(x=\dfrac{1}{8}\)
i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)
\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)
\(x=\left(\dfrac{-1}{3}\right)^4:\left(\dfrac{-1}{3}\right)\)
\(x=\left(\dfrac{-1}{3}\right)^3\)
\(x=\dfrac{-1}{27}\).
\(\left(1-2x\right)^4=81\Leftrightarrow\left[{}\begin{matrix}1-2x=\sqrt[4]{81}\\1-2x=-\sqrt[4]{81}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1-3\\2x=1+3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=-2\\2x=4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\) vậy \(x=-1;x=2\)