\(^3\)=-64

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28 tháng 7 2023

\(\left(x+5\right)^3=-64=\left(-4\right)^3\\ Vậy:x+5=-4\\Vậy:x=-4-5=-9 \)

28 tháng 7 2023

\(\left(x+5\right)^3=-64\)

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

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

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

\(\Rightarrow x=-9\)

Vậy: \(x=-9\)

19 tháng 3 2020

a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)

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

\(\Leftrightarrow3x-1=3\)

\(\Leftrightarrow3x=4\)

\(\Leftrightarrow x=\frac{4}{3}\)

b) Đề sai ! Sửa :

\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)

\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)

\(\Leftrightarrow2x+5=4\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=-\frac{1}{2}\)

c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)

\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)

\(\Leftrightarrow x+3=\frac{4}{3}\)

\(\Leftrightarrow x=-\frac{5}{3}\)

d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)

\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)

\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)

\(\Leftrightarrow x=\frac{4}{15}\)

24 tháng 7 2017

a. VP: \(\left(x+y\right)^{1999}\cdot\left(x-y\right)^{1999}=\left[\left(x+y\right)\left(x-y\right)\right]^{1999}\)

\(=\left(x^2-xy+xy-y^2\right)^{1999}=\left(x^2-y^2\right)^{1999}=VT\)

--> đpcm

b. VT: \(\dfrac{\left(5^4-5^3\right)^3}{125^4}=\dfrac{500^3}{125^4}=\dfrac{125^3\cdot4^3}{125^4}=\dfrac{4^3}{125}=\dfrac{64}{125}=VP\)

--> đpcm

28 tháng 7 2017

a. \(\left(x+5\right)^3=-64\)

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

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

=> x = -9

b. \(|2x-5|=8\)

\(\left[{}\begin{matrix}2x-5=8\\2x-5=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=13\\2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

c. \(\left|\dfrac{3}{4}x-\dfrac{1}{5}\right|=2\)

\(\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{1}{5}=2\\\dfrac{3}{4}x-\dfrac{1}{5}=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{11}{5}\\\dfrac{3}{4}x=\dfrac{-9}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{44}{15}\\x=\dfrac{-12}{5}\end{matrix}\right.\)

d. \(\left|3x-6\right|=x+4\)

\(\left[{}\begin{matrix}3x-6=x+4\\3x-6=-x-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x=4+6\\3x+x=-4+6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\4x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{2}\end{matrix}\right.\)

e. \(\left|x-3\right|=2x+1\)

\(\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-2x=1+3\\x+2x=-1+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}-x=4\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)

29 tháng 7 2017

thank bn nhayeu

30 tháng 7 2017

\(\frac{21}{x}=\frac{7}{-4}\Leftrightarrow7x=21.\left(-4\right)\Leftrightarrow7x=-84\Leftrightarrow x=-84:7\Leftrightarrow x=-12\)

\(\frac{114}{2x}=-\frac{8}{12}\Leftrightarrow\frac{57}{x}=-\frac{2}{3}\Leftrightarrow-2x=57.3\Leftrightarrow2x=171\Leftrightarrow x=\frac{171}{2}\)

10 tháng 8 2016

a) (x+5)3 = -64

   => (x+5)3 = (-4)3

   =>  x+5 = -4

   => x      = -4-5

   => x      = -9

b) (2x-3)2 = 9

   => (2x-3)2 = 32 hay (-3)2

    => 2x-3 = 3 hay 2x-3 = -3

   => 2x = 3+3 hay 2x = -3+3

   => 2x = 6     hay 2x = 0

   => x   = 6:2  hay x = 0:2

   => x   = 3     hay x = 0

   => 

10 tháng 8 2016

a)\(\left(x+5\right)^3=-64\)

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

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

    \(x=-9\)

b) \(\left(2x-3\right)^2=9\)

\(\left(2x-3\right)^2=3^2=\left(-3\right)^2\)

TH1 \(\left(2x-3\right)^2=3^2\)

        2x-3=3

        2x=6

           x =3

       

TH2\(\left(2x-3\right)^2=\left(-3\right)^2\)

        2x-3=-3

       2x = 0

      x = 0

10 tháng 2 2018

a) \(-5\cdot\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\cdot\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}\cdot x-\dfrac{5}{6}\\ -5\cdot x+1-\dfrac{1}{2}\cdot x-\dfrac{1}{3}=\dfrac{3}{2}\cdot x-\dfrac{5}{6}\\ x\cdot\left(-5-\dfrac{1}{2}\right)+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{3}{2}\cdot x\\ x\cdot\dfrac{-11}{2}+\dfrac{7}{6}=\dfrac{3}{2}\cdot x\\ \dfrac{3}{2}\cdot x-\dfrac{-11}{2}\cdot x=\dfrac{7}{6}\\ x\cdot\left(\dfrac{3}{2}-\dfrac{-11}{2}\right)=\dfrac{7}{6}\\ x\cdot7=\dfrac{7}{6}\\ x=\dfrac{7}{6}:7\\ x=\dfrac{1}{6}\)

Vậy x = \(\dfrac{1}{6}\)

11 tháng 2 2018

b, \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2^x\\ \dfrac{1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot30\cdot31}{2^{30}\cdot\left(1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot30\cdot31\right)\cdot64}=2^x\\ \dfrac{1}{2^{30}\cdot2^6}=2^x\\ \dfrac{1}{2^{36}}=2^x\\ 2^{-36}=2^x\\ \Rightarrow x=-36\)

a: |x-1/2|=7/2

=>x-1/2=7/2 hoặc x-1/2=-7/2

=>x=4 hoặc x=-3

b: \(x:\dfrac{3}{8}+\dfrac{5}{8}=x\)

=>8/3x-x=-5/8

=>5/3x=-5/8

hay x=-5/8:5/3=-5/8x3/5=-15/40=-3/8

c: \(\dfrac{5}{6}-\left|x-\dfrac{1}{2}\right|=\dfrac{15}{18}=\dfrac{5}{6}\)

=>|x-1/2|=0

=>x-1/2=0

hay x=1/2

e: \(\left(5x-3\right)^2-\dfrac{1}{64}=0\)

=>(5x-3)2=1/64

=>5x-3=1/8 hoặc 5x-3=-1/8

=>5x=25/8 hoặc 5x=23/8

=>x=5/8 hoặc x=23/40

25 tháng 9 2018

a/ \(4\dfrac{1}{3}:\dfrac{x}{4}=6:0,3\)

\(\Leftrightarrow\dfrac{13}{3}:\dfrac{x}{4}=20\)

\(\Leftrightarrow\dfrac{52}{3x}=20\)

\(\Leftrightarrow x=\dfrac{13}{15}\)

Vậy..

b/ \(\left(x-1\right)^5=-32\)

\(\Leftrightarrow\left(x-1\right)^5=\left(-2\right)^5\)

\(\Leftrightarrow x-1=-2\)

\(\Leftrightarrow x=-1\)

Vậy..

c/ \(\left(2^3:4\right).2^{x+1}=64\)

\(\Leftrightarrow2.2^{x+1}=64\)

\(\Leftrightarrow2^{x+2}=2^6\)

\(\Leftrightarrow x+2=6\)

\(\Leftrightarrow x=4\)

Vậy..

d/ \(\left|3-2x\right|-3=-3\)

\(\Leftrightarrow\left|3-2x\right|=0\)

\(\Leftrightarrow3-2x=0\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy..

e/ \(\left|x+\dfrac{4}{5}\right|-\dfrac{1}{7}=0\)

\(\Leftrightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{7}\\x+\dfrac{4}{5}=-\dfrac{1}{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{35}\\x=-\dfrac{33}{35}\end{matrix}\right.\)

Vậy..