NM
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
6 tháng 12 2016
\(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left(x-2\right)x=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x-1\right)^{x+2}=0\\x-2=0\\x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\\x=0\end{array}\right.\)
16 tháng 9 2016
bài 12 :
a,\(\left(x-\frac{1}{2}\right)^2=0\)
Mà: 02=0
=> \(\left(x-\frac{1}{2}\right)^2=0^2\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{2}\)
b, \(\left(x-2\right)^2=1\)
Mà : 1=12
\(\Rightarrow\left(x-2\right)^2=1^2\)
=> x - 2 = 1
=> x = 3
c, \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)=-2\)
Vì -8 =-23
nên ...
=> 2x =-1
=> x=0.5
d.\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
cái này cũng như mấy cái trên thôi
21 tháng 9 2016
Bài 12:
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(x-2=\pm1\)
- Nếu \(x-2=1\)
\(x=3\)
- Nếu \(x-2=-1\)
\(x=1\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow2x-1=-2\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(x+\frac{1}{12}=\pm\frac{1}{4}\)
- Nếu \(x+\frac{1}{12}=\frac{1}{4}\)
\(x=\frac{1}{6}\)
- Nếu \(x+\frac{1}{12}=-\frac{1}{4}\)
\(x=-\frac{1}{3}\)
Bài 13: có người làm rồi
Bài 14:
a) \(25^3\div5^2\)
\(=\left(5^2\right)^3\div5^2\)
\(=5^6\div5^2=5^4\)
b) \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c) \(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
\(=3-1+\frac{1}{4}:2\)
\(=2+\frac{1}{8}=2\frac{1}{8}\)
18 tháng 3 2020
\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0^2\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy x = 1/2
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2=1^2\)
\(\Leftrightarrow x-2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy x = 3 hoặc x = 1
\(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\)
<=> 2x = -1
<=> x = -0,5
Vậy x = -0,5
18 tháng 3 2020
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+2\\x=-1+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Vậy\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-1\times2\)
\(x=-2\)
\(x\left(\frac{1}{2}\right)^2=\frac{1}{16}\)
\(x\left(\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
7 tháng 9 2016
\(a,\frac{8^{12}.5^{21}}{2^{17}.10^{19}}=\frac{\left(2^3\right)^{12}.5^{21}}{2^{17}.2^{19}.5^{19}}=\frac{2^{36}.5^{21}}{2^{36}.5^{19}}=25\)
\(b,\left(x-5\right).\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow x-5=0\)hoặc \(x+\frac{1}{2}=0\)
\(x=5\)hoặc \(x=-\frac{1}{2}\)
\(c,\left|x-6\right|-\frac{1}{2}=\frac{3}{2}\)
\(\left|x-6\right|=2\)
\(\Rightarrow x-6=2\)hoặc \(x-6=-2\)
\(x=8\)hoặc \(x=4\)
30 tháng 6 2016
\(x^2-y^2=1\)
Ta có : \(\left(\frac{x}{5}\right)^2=\left(\frac{y}{4}\right)^2\)
\(=>\frac{x^2}{25}=\frac{y^2}{16}\)
A/d dãy ............
\(\frac{x^2-y^2}{25-16}=\frac{1}{9}=>\frac{x}{5}=\frac{y}{4}=\frac{1}{3}\)
\(=>\frac{x}{5}=\frac{1}{3}=>x=\frac{5}{3}\)
\(=>\frac{y}{4}=\frac{1}{3}=>x=\frac{4}{3}\)
30 tháng 6 2016
\(\frac{x}{5}=\frac{y}{4}\)nên \(\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{1}{9}\)=> \(\frac{x}{5}=\sqrt{\frac{1}{9}};-\sqrt{\frac{1}{9}}=\frac{1}{3};\frac{-1}{3}\)
=> x = \(\frac{1}{3}.5;\frac{-1}{3}.5=\frac{5}{3};\frac{-5}{3}\)
16 tháng 9 2016
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
*\(x-2=1\Rightarrow x=3\)
*\(x-2=-1\Rightarrow x=1\)
Vậy x = 3; x = 1
c) \(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=-1\)
\(\Rightarrow x=\frac{-1}{2}\)
Vậy x = \(\frac{-1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{1}{2}\)
\(\Rightarrow x=\frac{-1}{4}\)
Vậy x = \(\frac{-1}{4}\)
PA
16 tháng 9 2016
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=-2+1\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
\(\left(x-2\right)^2=1\)
\(\left(x-2\right)^2=\left(\pm1\right)^2\)
\(\begin{cases}x-2=1\\x-2=-1\end{cases}\)
\(\begin{cases}x=1+2\\x=-1+2\end{cases}\)
\(\begin{cases}x=3\\x=1\end{cases}\)
\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{1}{4}\right)^2\)
\(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\)
\(\begin{cases}x=\frac{1}{4}-\frac{1}{2}\\x=-\frac{1}{4}-\frac{1}{2}\end{cases}\)
\(\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)
\(\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1}{8}\right)^{x-2}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1}{2}\right)^{3x-6}\)
\(\Leftrightarrow x=3x-6\)
\(\Leftrightarrow3x-x=6\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy ........
\(\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1}{8}\right)^{x-2}\\ \Rightarrow\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1^3}{2^3}\right)^{x-2}\\ \Rightarrow\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1}{2}\right)^{3\left(x-2\right)}\\ \Leftrightarrow3\left(x-2\right)=x\\ \Rightarrow3x-6=x\\ \Rightarrow3x-x=6\\ \Rightarrow x\left(3-1\right)=6\\ \Rightarrow2x=6\\ \Rightarrow x=6:2=3\)