\(\dfrac{1}{20}\) x 22 x 33 :125

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27 tháng 8 2023

\(5^5.\dfrac{1}{20}.2^2.3^3:125\)

\(=5^5.\dfrac{1}{2^2.5}.2^2.3^3:5^3\)

\(=\dfrac{2^2.3^3.5^5}{2^2.5^4}=3^3.5=27.5=135\)

27 tháng 8 2023

= 3125 x 1/20 x 4 x 27 : 125

= 625/4 x 4 x 27 : 125

= 625 x 27 : 125

= 16875 : 125

= 135

3 tháng 7 2018

1)
\(=\dfrac{\left(2.3\right)^{20}.\left(5^2\right)^{19}}{\left(2^3\right)^7.\left(3^2\right)^{10}.\left(5^3\right)^{13}}\)
\(=\dfrac{2^{20}.3^{20}.5^{38}}{2^{21}.3^{20}.5^{39}}\)
\(=\dfrac{1}{2.5}\)
\(=\dfrac{1}{10}\)

a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)

\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)

=>x=10

b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)

\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)

hay \(x\in\left\{0;1;2\right\}\)

c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)

\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)

\(\Leftrightarrow6-x=0\)

hay x=6

Bài 1: 

a: =>13x+8=9x+20

=>4x=12

hay x=3

b: \(\Leftrightarrow5x-7=-8-11-3x\)

=>5x-7=-3x-19

=>8x=-12

hay x=-3/2

c: \(\Leftrightarrow\left[{}\begin{matrix}12x-7=5\\12x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)

e: =>3x+1=-5

=>3x=-6

hay x=-2

10 tháng 7 2017

a)

\(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=255\\ x+x-1+x-2+...+x-50=255\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+50\right)\\ 51x-1275=255\\ 51x=1530\\ x=30\)

e)

\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\\ x+x+1+x+2+...+x+30=1240\\ \left(x+x+x+...+x\right)+\left(1+2+3+...+30\right)=1240\\ 31x+465=1240\\ 31x=775\\ x=25\)

f)

\(\left(x-1\right)+\left(x-2\right)+...+\left(x-19\right)+\left(x-20\right)=-610\\ x-1+x-2+...+x-19+x-20=-610\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+20\right)=-610\\ 20x-210=-610\\ 20x=-400\\ x=-20\)

10 tháng 7 2017

b) Ta có : \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)

\(\Rightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)

\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\\\left(x-\frac{1}{3}\right)^2=\left(-\frac{1}{2}\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=-\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=-\frac{1}{6}\end{cases}}\)

10 tháng 7 2017

b) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{4}\\x-\frac{1}{3}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{1}{12}\end{cases}}\)

d) \(\frac{x+5}{2}=\frac{8}{x+5}\)

\(\Rightarrow\left(x+5\right)^2=16\)

\(\Rightarrow\orbr{\begin{cases}x+5=16\\x+5=-16\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=-21\end{cases}}}\)

17 tháng 8 2017

\(\dfrac{x-2}{5}=\dfrac{x}{3}\)

\(\Leftrightarrow\left(x-2\right)3=5x\)

\(\Leftrightarrow3x-6=5x\)

\(\Leftrightarrow5x-3x=-6\)

\(\Leftrightarrow2x=-6\)

\(\Leftrightarrow x=-3\)

Vậy .....

b, \(B=1+2+2^2+..........+2^{2017}\)

\(\Leftrightarrow2B=2+2^2+.......+2^{2018}\)

\(\Leftrightarrow2B-B=\left(2+2^2+......+2^{2018}\right)-\left(1+2+......+2^{2017}\right)\)

\(\Leftrightarrow B=2^{2018}-1\)

c, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)

\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)

\(\Leftrightarrow4x+92=3x+120\)

\(\Leftrightarrow4x-3x=120-92\)

\(\Leftrightarrow x=28\)

17 tháng 8 2017

a, \(\dfrac{x-2}{5}=\dfrac{x}{3}\)

\(\Leftrightarrow3\left(x-2\right)=5x\)

\(\Leftrightarrow3x-6=5x\)

\(\Leftrightarrow5x-3x=6\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

b, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)

\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)

\(\Leftrightarrow4x+92=2x+80\)

\(\Leftrightarrow4x-2x=80-92\)

\(\Leftrightarrow2x=-12\)

\(\Leftrightarrow x=-6\)

c, \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...........+\dfrac{1}{2^{2017}}\)

\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...........+\dfrac{1}{2^{2016}}\)

\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2016}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2017}}\right)\)

\(\Leftrightarrow A=1-\dfrac{1}{2^{2017}}\)

d, \(B=1+2+2^2+........+2^{2017}\)

\(\Leftrightarrow2B=2+2^2+2^3+......+2^{2018}\)

\(\Leftrightarrow2B-B=\left(2+2^2+.....+2^{2018}\right)-\left(1+2+....+2^{2017}\right)\)

\(\Leftrightarrow B=2^{2018}-1\)

17 tháng 8 2017

\(\dfrac{x-2}{5}=\dfrac{x}{3}=>3\left(x-2\right)=5x\)

\(< =>3x-6=5x=>x=-3\)

\(\dfrac{x+23}{x+40}=\dfrac{3}{4}=>4\left(x+23\right)=3\left(x+40\right)\)

\(4x+92=3x+120=>x=28\)

a: =>5x=3x-6

=>2x=-6

hay x=-3

b: \(\Leftrightarrow\left(x-3\right)^2=4\cdot5^2=100\)

=>x-3=10 hoặc x-3=-10

=>x=13 hoặc x=-7

c: \(\left|x^3+1\right|+2\ge2\forall x\)

Dấu '=' xảy ra khi x=-1

4 tháng 8 2017

Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)

\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)

\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)

\(\Rightarrow24A=5^{51}-5\)

\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)

Vậy ............................................................

4 tháng 8 2017

1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)

\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)

Vậy \(x=3\)

b) \(\left(4x-1\right)^3=-27.125\)

\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)

\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)

Vậy \(x=-3,5\)

c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)

\(\Rightarrow4x+4=4x+12\)

\(\Rightarrow4x=4x+8\)

\(\Rightarrow x\in\varnothing\)

d) \(\left(x-5\right)^7=\left(x-5\right)^9\)

\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)

\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)

14 tháng 10 2017

\(\left(\dfrac{3}{5}\right)^5.x=\left(\dfrac{9}{25}\right)^3.\left(\dfrac{3}{5}\right)^2\)

\(\Rightarrow\left(\dfrac{3}{5}\right)^3.x=\left(\dfrac{9}{25}\right)^3\)

\(\Rightarrow x=\left(\dfrac{3}{5}\right)^3\)

\(\Rightarrow x=\dfrac{27}{125}\)

14 tháng 10 2017

Bạn sai rồi nhé như thế này mới đúng :

Lời giải :

\(\left(\dfrac{3}{5}\right)^5.x=\left(\dfrac{9}{25}\right)^3.\left(\dfrac{3}{5}\right)^2\)

\(\left(\dfrac{3}{5}\right)^2.\left(\dfrac{3}{5}\right)^3.x=\left(\dfrac{9}{25}\right)^3.\left(\dfrac{3}{5}\right)^2\)

Vì : \(\left(\dfrac{3}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\)

Nên : \(\left(\dfrac{3}{5}\right)^3.x=\left(\dfrac{9}{25}\right)^3\)

\(\Rightarrow x=\left(\dfrac{9}{25}\right)^3:\left(\dfrac{3}{5}\right)^3\)

\(\Rightarrow x=\left(\dfrac{9}{25}:\dfrac{3}{5}\right)^3\)

\(\Rightarrow x=\left(\dfrac{3}{5}\right)^3\)

\(\Rightarrow x=\dfrac{3.3.3}{5.5.5}\)

\(\Rightarrow x=\dfrac{3^3}{5^3}\)

\(\Rightarrow x=\left(3:5\right)^3\)

\(\Rightarrow x=0,6^3\)

\(\Rightarrow x=0,216\)

Vậy \(x=0,216\)