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Tìm x biết
\(\frac{0,1\left(6\right)+0,\left(3\right)}{0,\left(3\right)+1,1\left(6\right)}\)- x=0,(2)
\(\frac{0,1\left(6\right)+0,\left(3\right)}{0,\left(3\right)+1,1\left(6\right)}-x=0,\left(2\right)\)
\(\Rightarrow\frac{\frac{1}{6}+\frac{1}{3}}{\frac{1}{3}+\frac{7}{6}}-x=\frac{2}{9}\)
\(\Rightarrow\frac{\frac{1}{2}}{\frac{3}{2}}-x=\frac{2}{9}\)
\(\Rightarrow\frac{1}{3}-x=\frac{2}{9}\)
\(\Rightarrow x=\frac{1}{3}-\frac{2}{9}=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}\)
= \(\frac{1}{11}\cdot x=0.\left(2\right)\)
\(\Rightarrow x=0,\left(2\right):\frac{1}{11}\)
x = 0
\(\frac{0,1\left(6\right)+0,\left(03\right)}{0,\left(3\right)+1,1\left(6\right)}\times x=0,2\)
\(=\frac{1}{11}\times x=0,\left(2\right)\)
\(\Rightarrow x=0,\left(2\right)\div\frac{1}{11}\)
B1 :
\(\frac{0,1\left(6\right)+0,\left(3\right)}{0,\left(3\right)+1,1\left(6\right)}\) . x = 0,(2)
=\(\frac{0,5}{1,5}\).x=0,(2)
x=0,(2):\(\frac{0,5}{1,5}\)
x=0,(6)=\(\frac{2}{3}\)
b2:
[12,(1) - 2,3(6)] : 4,(21)
=9,7(4):4,(21)
=\(\frac{9,7\left(4\right)}{4,\left(21\right)}\)
a)\(\left(\frac{1}{3}\right)^{-1}-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^4.2^3=3-1+\frac{1}{16}.8=3-1+\frac{1}{2}=\frac{5}{2}\\ \)
b)\(2^2.2^3.\left(\frac{2}{3}\right)^{-2}=2^5.\frac{9}{4}=72\)
c)\(\left(\frac{4}{3}\right)^{-2}.\left(\frac{3}{4}\right)^3:\left(\frac{-2}{3}\right)^{-3}=\left(\frac{3}{4}\right)^2.\left(\frac{3}{4}\right)^3:\left(\frac{-2}{3}\right)^{-3}=\left(\frac{3}{4}\right)^5:\left(\frac{3}{2}\right)^3=\frac{9}{128}\)
2)
\(3^{x+1}=9^x\Leftrightarrow3^x.3=9^x\Rightarrow3=9^x:3^x\Rightarrow3=3^x\Rightarrow x=1\)
\(\left(x-0,1\right)^2=6,25\Leftrightarrow\left(x-0,1\right)^2=2,5^2\Rightarrow\left(x-0,1\right)=2,5\Rightarrow x=2,5+0,1=2,6\)
\(3^{2x-1}=243\Leftrightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow2x=6\Rightarrow x=3\)
\(\left(4x-3\right)^4=\left(4x-3\right)^2\Rightarrow x=1\)
a, \(\left(x-3\right)\left(x+2\right)>0\)
th1 : \(\hept{\begin{cases}x-3>0\\x+2>0\end{cases}\Rightarrow\hept{\begin{cases}x>3\\x>-2\end{cases}\Rightarrow}x>3}\)
th2 : \(\hept{\begin{cases}x-3< 0\\x+2< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 3\\x< -3\end{cases}\Rightarrow}x< -3}\)
vậy x > 3 hoặc x < -3
b, \(\left(x+5\right)\left(x+1\right)< 0\)
th1 : \(\hept{\begin{cases}x+5>0\\x+1< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-5\\x< -1\end{cases}\Rightarrow x\in\left\{-4;-3;-2\right\}}}\)
th2 : \(\hept{\begin{cases}x+5< 0\\x+1>0\end{cases}\Rightarrow\hept{\begin{cases}x< -5\\x>-1\end{cases}\Rightarrow}x\in\varnothing}\)
vậy x = -4; -3; -2
c, \(\frac{x-4}{x+6}\le0\)
xét \(\frac{x-4}{x+6}=0\)
\(\Rightarrow x-4=0;x\ne-6\)
\(\Rightarrow x=4\ne-6\)
xét \(\frac{x-4}{x+5}< 0\)
th1 : \(\hept{\begin{cases}x-4< 0\\x+5>0\end{cases}\Rightarrow\hept{\begin{cases}x< 4\\x>-5\end{cases}\Rightarrow}x\in\left\{3;2;1;0;-1;-2;-3;-4\right\}}\)
th2 : \(\hept{\begin{cases}x-4>0\\x+5< 0\end{cases}\Rightarrow\hept{\begin{cases}x>4\\x< -5\end{cases}\Rightarrow x\in\varnothing}}\)
d tương tự c
\(\frac{\left(x-6\right)}{x-7}\ge0\)
Th1: x - 6 < 0
<=> x - 6 + 6 < 0 + 6
<=> x - 6 + 6 > 0 + 6
=> x < 6
Th2: x - 7
<=> x - 7 + 7 < 0 + 7
<=> x - 7 + 7 > 0 + 7
=> x > 7
=> x < 6 hoặc x > 7