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a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)
d) \(9x^2=1\)
\(\Leftrightarrow x^2=1:9\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Ta có \(x=\dfrac{2016}{x\times\left(x+1\right)\times\left(x+2\right)\times........\times\left(x+2016\right)}\)
\(\dfrac{1}{2015!}=\dfrac{2016}{2016!}=\dfrac{2016}{1\times2\times...........\times2016}\)
Vì x > 0=> \(\left(x+1\right)\times\left(x+2\right)\times...\times\left(x+2016\right)>1\times2\times...\times2016\)
\(\Rightarrow\dfrac{1}{\left(x+1\right)\times\left(x+2\right)\times.......\times\left(x+2016\right)}< \dfrac{1}{1\times2\times..........\times2016}\)\(\Rightarrow\dfrac{2016}{\left(x+1\right)\times\left(x+2\right)\times.......\times\left(x+2016\right)}< \dfrac{2016}{1\times2\times......\times2016}\)
\(\Leftrightarrow x< \dfrac{1}{2015!}\)(đpcm)
Ta có \(x=\dfrac{2016}{\left(x+1\right)\times\left(x+2\right)\times....\times\left(x+2016\right)}\)
\(\dfrac{1}{2015!}=\dfrac{2016}{2016!}=\dfrac{2016}{1\times2\times.....\times2016}\)
Vì x>0=>(x+1)×(x+2)×.............×(x+2016) >\(1\times2\times.....\times2016\)
\(\Rightarrow\dfrac{1}{\left(x+1\right)\times\left(x+2\right)\times......\times\left(x+2016\right)}>\dfrac{1}{1\times2\times......\times2016}\)
\(\Rightarrow\dfrac{2016}{\left(x+1\right)\times\left(x+2\right)\times......\times\left(x+2016\right)}>\dfrac{2016}{1\times2\times......\times2016}\)
\(\Leftrightarrow x< \dfrac{1}{2015!}\)(đpcm)
biết làm bài 1 thôi
\(\left(\frac{1}{2}+1\right)\times\left(\frac{1}{3}+1\right)\times\cdot\cdot\cdot\times\left(\frac{1}{999}+1\right)\)
= \(\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdot\cdot\cdot\times\frac{1000}{999}\)
lượt bỏ đi còn :
\(\frac{1000}{2}=500\)
a) tính bình thường thôi
b)\(\left(3x-4\right)\times\left(x-1\right)^3=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-4=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=1\end{cases}}\)
c) \(2^{2x-1}:4=8^3\)
\(\Leftrightarrow2^{2x-1}=2048\Leftrightarrow2^{2x-1}=2^{11}\Leftrightarrow2x-1=11\Leftrightarrow x=6\)
d) \(x^{17}=x\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
e) \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\x=6\\x=4\end{cases}}\)
vậy........
a) (x : 23 + 45) . 37 - 22 = 24.105
=> (x : 23 + 45).37 - 22 = 1680
=> (x : 23 + 45).37 = 1702
=> x : 23 + 45 = 46
=> x : 23 = 1
=> x = 23
b) (3x - 4).(x - 1)3 = 0
=> \(\orbr{\begin{cases}3x-4=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=1\end{cases}}\)
Vậy \(x\in\left\{\frac{4}{3};1\right\}\)
c) 22x - 1 : 4 = 83
=> 22x - 1 : 22 = (23)3
=> 22x - 1 : 22 = 29
=> 22x - 1 = 211
=> 2x - 1 = 11
=> 2x = 12
=> x = 6
d) x17 = x
=> x17 - x = 0
=> x(x16 - 1) = 0
=> \(\orbr{\begin{cases}x=0\\x^{16}-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^{16}=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
Vậy \(x\in\left\{0;1;-1\right\}\)
e) (x - 5)4 = (x - 5)6
=> (x - 5)6 - (x - 5)4 = 0
=> (x - 5)4[(x - 5)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1^2\end{cases}}\Rightarrow\orbr{\begin{cases}x-5=0\\x-5=\pm1\end{cases}}\Rightarrow x-5\in\left\{0;1;-1\right\}\)
=> \(x\in\left\{5;6;4\right\}\)
\(x-40\%x=3,6\)
\(\Rightarrow100\%x-40\%x=3,6\)
\(\Rightarrow60\%x=3,6\)
\(\Rightarrow\frac{60}{100}x=3,6\)
\(\Rightarrow x=6\)
\(3\frac{2}{7}x-\frac{1}{3}=-2\frac{3}{4}\)
\(\Rightarrow\frac{23}{7}x-\frac{1}{3}=-\frac{11}{4}\)
\(\Rightarrow\frac{23}{7}x=-\frac{33}{12}+\frac{4}{12}\)
\(\Rightarrow\frac{23}{7}x=\frac{29}{12}\)
\(\Rightarrow x=\frac{29}{12}:\frac{23}{7}=\frac{203}{276}\)
\(4\left(1+3x\right)-2\left(2-x\right)=0\)
\(\Leftrightarrow4+12x-4+2x=0\)
\(\Leftrightarrow14x=0\)
\(\Leftrightarrow x=0\)
Vậy .......
4(1+3x)−2(2−x)=0
4+12x−4+2x=0
14x=0
x=0
Vậy x=0