\(\left(X-\frac{1}{2}\right)^3=\frac{1}{27}\)

AJ GIẢI HỘ MIK NHÉ

 

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2 tháng 12 2015

\(\Rightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\Rightarrow x-\frac{1}{2}=\frac{1}{3}\Rightarrow x=\frac{5}{6}\)

2 tháng 12 2015

(x-1/2)^3=1/27

(X-1/2)^3=(1/3)^3

X-1/2=1/3

X       =1/3+1/2

X        =2/6+3/6

X         =5/6

31 tháng 7 2016

a) Lm r nkoa!

b) \(\left(0,25x\right):3=\frac{5}{6}:0,125\)

\(=\left(0,25x\right):3=\frac{20}{3}\)

\(\Rightarrow0,25x=\frac{20}{3}\cdot3=20\)

\(\Rightarrow x=20:0,25=80\)

\(\Rightarrow x=80\)

c) \(0,01:2,5=\left(0,75x\right):0,75\)

\(=\frac{1}{250}=\left(0,75x\right):0,75\)

\(\Rightarrow0,75x=\frac{1}{250}\cdot0,75=\frac{3}{1000}\)

\(\Rightarrow x=\frac{3}{1000}:0,75=\frac{1}{250}\)

\(\Rightarrow x=\frac{1}{250}\)

d) \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)

\(=\frac{5}{3}=\frac{2}{3}:\left(0,1x\right)\)

\(\Rightarrow0,1x=\frac{2}{3}:\frac{5}{3}=\frac{2}{5}\)

\(\Rightarrow x=\frac{2}{5}:0,1=4\)

\(\Rightarrow4\)

7 tháng 1 2017

a)  2(x-1)+3(x-3)=-2                                                b)  x-1/3=x-2/2

2x-2+3x-9=-2                                                              2 (x-1)=3(x-2)

(2x+3x)+(-2-9)=-2                                                        2x-2=3x-6

5x+(-11)=-2                                                                 2x-3x=-6+2

5x=-2+11                                                                   -1x=-4

5x=9                                                                          x=4

x=1,8

Nhớ nha!

9 tháng 9 2016

a)n=1

b)n=4

c)n=1

d)n=6

e)n=-1

29 tháng 11 2016

a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)

\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)

\(\Rightarrow x+3=-3\)

\(\Rightarrow x=-6\)

b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)

\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)

\(\Rightarrow2x+2=-2\)

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\)

c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)

\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)

\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)

\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)

\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)

\(\Rightarrow x+5=-4\)

\(\Rightarrow x=-9\)

29 tháng 11 2016

d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)

\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)

\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)

\(\Rightarrow2x=18\)

\(\Rightarrow x=9\)

e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)

\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)

\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)

\(\Rightarrow2x=18\)

\(\Rightarrow x=9\)

28 tháng 6 2016

a) \(\left(\frac{1}{3}-\frac{1}{5}\right)^2:\left(\frac{1}{5}\right)^2=\left[\left(\frac{1}{3}-\frac{1}{5}\right):\frac{1}{5}\right]^2=\left(\frac{2}{15}:\frac{1}{5}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)

 

28 tháng 6 2016

c)\(7\frac{1}{23}+\frac{10}{27}-5\frac{1}{23}+\frac{17}{27}+2^3=\left(7\frac{1}{23}-5\frac{1}{23}\right)+\left(\frac{10}{27}+\frac{17}{27}\right)+2^3=2+1+8=11\)

d)\(5.\left(-\frac{5}{2}\right)^2+\frac{1}{5}.\left(-3\right)^2=5.\frac{25}{4}+\frac{1}{5}.9=\frac{125}{4}+\frac{9}{5}=\frac{661}{20}\)

23 tháng 9 2019

\(a,5,5-\left|x-0,4\right|=-1\frac{1}{5}\)

\(\Rightarrow5,5-\left|x-0,4\right|=-\frac{6}{5}\)

\(\Rightarrow-\left|x-0,4\right|=-\frac{6}{5}-5,5=-6,7\)

\(\Rightarrow\left|x-0,4\right|=6,7\)

\(\Rightarrow x-0,4=\pm6,7\)

\(\Rightarrow\orbr{\begin{cases}x-0,4=6,7\\x-0,4=-6,7\end{cases}\Rightarrow\orbr{\begin{cases}x=7,1\\x=-6,3\end{cases}}}\)

23 tháng 9 2019

\(a,5,5-\left|x-0,4\right|=-1\frac{1}{5}\)

=> \(\left|x-0,4\right|=5,5-\left[-\frac{6}{5}\right]=5,5+1,2=6,7\)

=> \(\left|x-0,4\right|=\pm6,7\)

Xét hai trường hợp :

TH1 : x - 0,4 = 6,7

=> x  = 6,7 + 0,4 = 7,1

TH2 : x - 0,4 = -6,7

=> x = -6,7 + 0,4 =-6,3

\(b,\left[1-\frac{3}{4}\left|x\right|\right]^2=\frac{16}{25}\)

=> \(\left[1-\frac{3}{4}\left|x\right|\right]=\pm\sqrt{\frac{16}{25}}\)

=> \(\left[1-\frac{3}{4}\left|x\right|\right]=\pm\frac{4}{5}\)

=> \(\orbr{\begin{cases}1-\frac{3}{4}\left|x\right|=\frac{4}{5}\\1-\frac{3}{4}\left|x\right|=-\frac{4}{5}\end{cases}}\)=> \(\orbr{\begin{cases}x=\pm\frac{4}{15}\\x=\pm\frac{12}{5}\end{cases}}\)

\(c,\left[0,1\left|x\right|-\frac{1}{2}\right]\left[0,5-\left|x\right|\right]=0\)

=> \(\orbr{\begin{cases}0,1\left|x\right|-\frac{1}{2}=0\\0,5-\left|x\right|=0\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{1}{10}\left|x\right|=\frac{1}{2}\\\left|x\right|=0,5\end{cases}}\)

=> \(\orbr{\begin{cases}\left|x\right|=5\\\left|x\right|=0,5\end{cases}}\)=> \(\orbr{\begin{cases}x\in\left\{5;-5\right\}\\x\in\left\{0,5;-0,5\right\}\end{cases}}\)

d, Xét hai trường hợp rồi ra kết quả thôi

31 tháng 8 2020

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

\(\left(\frac{4}{5}\right)^5\cdot x=\left(\frac{4}{5}\right)^7\)

=> \(x=\frac{\left(\frac{4}{5}\right)^7}{\left(\frac{4}{5}\right)^5}=\left(\frac{4}{5}\right)^2=\frac{16}{25}\)

\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}=\left(\pm\frac{1}{4}\right)^2\)

=> \(\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}\)

(3x + 1)3 = -27 => (3x + 1)3 = (-3)3 => 3x + 1 = -3 => 3x = -4 => x = -4/3

a)\(x:\left(\frac{-1}{3}\right)^3=\frac{-1}{3}\)

\(=>x:\frac{-1}{27}=\frac{-1}{3}\)

\(=>x=\frac{-1}{3}.\frac{-1}{27}=>x=\frac{1}{81}\)

b) \(\left(\frac{4}{5}\right)^5.x=\left(\frac{4}{5}\right)^7\)

\(=>x=\left(\frac{4}{5}\right)^7:\left(\frac{4}{5}\right)^5=>x=\left(\frac{4}{5}\right)^2=\frac{16}{25}\)

c)\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(=>\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\\\left(x+\frac{1}{2}\right)^2=\left(\frac{-1}{4}\right)^2\end{cases}}\)

\(=>\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=\frac{-1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=-1\end{cases}}}\)

d|) \(\left(3x+1\right)^3=-27\)

\(=>\left(3x+1\right)^3=\left(-3\right)^3\)

\(=>3x+1=-3\)

\(=>3x=-4=>x=\frac{-4}{3}\)

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