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Giải:
a) \(5< 5^x < 625\)
\(\Leftrightarrow5< 5^x< 5^4\)
Vì \(5=5=5\)
Nên \(1< x< 4\)
\(\Leftrightarrow x\in\left\{2;3\right\}\)
Vậy ...
b) \(2^{x-1}=16\)
\(\Leftrightarrow2^{x-1}=2^4\)
Vì \(2=2\)
Nên \(x-1=4\)
\(\Leftrightarrow x=4+1=5\)
Vậy ...
c) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)^{x+2}}{\left(x-1\right)^{x+2}}=\dfrac{\left(x-1\right)^{x+6}}{\left(x-1\right)^{x+2}}\)
\(\Leftrightarrow1=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=1+1=2\)
Vậy ...
Bài 1:
\((1-2x)^2=9=3^2=(-3)^2\)
\(\Rightarrow \left[\begin{matrix} 1-2x=3\\ 1-2x=-3\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-1\\ x=2\end{matrix}\right.\)
Bài 2:
\((x+5)^3=-64=(-4)^3\)
\(\Rightarrow x+5=-4\Rightarrow x=-9\)
Bài 3:
\((3x-5)^2=16=4^2=(-4)^2\)
\(\Rightarrow \left[\begin{matrix} 3x-5=4\\ 3x-5=-4\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=\frac{1}{3}\end{matrix}\right.\)
Bài 4:
\((x-1)^3=27=3^3\)
\(\Rightarrow x-1=3\Rightarrow x=4\)
Bài 5:
\(x^2+x=0\Leftrightarrow x(x+1)=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ x+1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=0\\ x=-1\end{matrix}\right.\)
Bài 6:
\(5^{x+2}=625=5^4\)
\(\Rightarrow x+2=4\Rightarrow x=2\)
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
1, x2 = 0
=> x=0
2,x2=1
=> x= 1 hoặc x=-1
3,x2=3
=>\(x=\sqrt{3}\)
4,x2=6
=>\(x=\sqrt{6}\)
5,x2=7
=>\(x=\sqrt{7}\)
a)
Ta thấy \(\left\{\begin{matrix} |x+\frac{19}{5}|\geq 0\\ |y+\frac{1890}{1975}|\geq 0\\ |z-2005|\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{Z}\)
\(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|\geq 0\)
Do đó, để \(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|=0\) thì :
\(\left\{\begin{matrix} |x+\frac{19}{5}|= 0\\ |y+\frac{1890}{1975}|= 0\\ |z-2005|=0\end{matrix}\right.\Rightarrow x=\frac{-19}{5}; y=\frac{-1890}{1975}; z=2005\)
b) Giống phần a, vì trị tuyệt đối của một số luôn không âm nên để tổng các trị tuyệt đối bằng $0$ thì:
\(\left\{\begin{matrix} |x+\frac{3}{4}|=0\\ |y-\frac{1}{5}|=0\\ |x+y+z|=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=-\frac{3}{4}\\ y=\frac{1}{5}\\ z=-(x+y)=\frac{11}{20}\end{matrix}\right.\)
c) \(\frac{16}{2^x}=1\Rightarrow 16=2^x\)
\(\Leftrightarrow 2^4=2^x\Rightarrow x=4\)
d) \((2x-1)^3=-27=(-3)^3\)
\(\Rightarrow 2x-1=-3\)
\(\Rightarrow 2x=-2\Rightarrow x=-1\)
e) \((x-2)^2=1=1^2=(-1)^2\)
\(\Rightarrow \left[\begin{matrix} x-2=1\\ x-2=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=1\end{matrix}\right.\)
f) \((x+\frac{1}{2})^2=\frac{4}{25}=(\frac{2}{5})^2=(\frac{-2}{5})^2\)
\(\Rightarrow \left[\begin{matrix} x+\frac{1}{2}=\frac{2}{5}\\ x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-1}{10}\\ x=\frac{-9}{10}\end{matrix}\right.\)
g) \((x-1)^2=(x-1)^6\)
\(\Leftrightarrow (x-1)^6-(x-1)^2=0\)
\(\Leftrightarrow (x-1)^2[(x-1)^4-1]=0\)
\(\Rightarrow \left[\begin{matrix} (x-1)^2=0\\ (x-1)^4=1=(-1)^4=1^4\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x-1=-1\\ x-1=1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x=0\\ x=2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=\left\{0;1;2\right\}\)
1)
\(2^{x-1}=16\\ 2^{x-1}=2^4\\ \Rightarrow x-1=4\\ x=4+1\\ x=5\)
5)
\(\left(x-1\right)^2=25\Rightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
6)
\(\left|2x-1\right|=5\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
5) (x-1)2 = 25
(x-1)2 = 52
x-1 = 5
x = 5+1
x = 6
6) \(\left|2x-1\right|=5 \)
\(TH1:\) \(2x-1=5\)
\(\Leftrightarrow2x=5+1\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=6:2\)
\(\Leftrightarrow x=3\)
\(TH2:2x-1=-5\)
\(\Leftrightarrow2x=-5+1\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-4:2\)
\(\Leftrightarrow x=-2\)
Vậy x = 3 hoặc x = -2.
Tick nha!