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21 tháng 12 2015

a)(x-1)^3=125=5^3

=>x-1=5=>x=6

b)2^x+2-2^x=96

=>2^x.2^2-2^x=96

=>2^x.(2^2-1)=96

=>2^x.3=96

=>2^x=96/3=32=2^5

=>x=5

c)(2^x +1)^3=343=7^3

=>2^x +1=7

=>2^x=6

sai đề ko?

 

a: =>12x-64=32

=>12x=96

=>x=8

b: =>x-1=5

=>x=6

c: =>2^x*3=96

=>2^x=32

=>x=5

a: =>3^x=3^4*3=3^5

=>x=5

b: =>\(2^{x+1}=2^5\)

=>x+1=5

=>x=4

c: \(\Leftrightarrow3^{x+2-3}=3\)

=>x-1=1

=>x=2

d: \(\Leftrightarrow x^2=\dfrac{32}{2}=16\)

=>x=4 hoặc x=-4

e: (2x-1)^4=81

=>2x-1=3 hoặc 2x-1=-3

=>2x=4 hoặc 2x=-2

=>x=-1 hoặc x=2

f: (2x-6)^4=0

=>2x-6=0

=>x-3=0

=>x=3

18 tháng 8 2023

a) \(3^x=81\cdot3\)

\(\Rightarrow3^x=3^4\cdot3\)

\(\Rightarrow3^x=3^5\)

\(\Rightarrow x=5\)

b) \(2^{x+1}=32\)

\(\Rightarrow2^{x+1}=2^5\)

\(\Rightarrow x+1=5\)

\(\Rightarrow x=4\)

c) \(3^{x+2}:27=3\)

\(\Rightarrow3^{x+2}:3^3=3\)

\(\Rightarrow3^{x+2-3}=3\)

\(\Rightarrow3^{x-1}=3\)

\(\Rightarrow x-1=1\)

\(\Rightarrow x=2\)

d) \(2x^2=32\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x^2=4^2\)

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

e) \(\left(2x-1\right)^4=81\)

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

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

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

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

f)  \(\left(2x-6\right)^4=0\)

\(\Rightarrow2x-6=0\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=6:2\)

\(\Rightarrow x=3\)

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

12 tháng 7 2021

ai giúp mik vs

5 tháng 5 2015

a/2x+2-2x=96

2x(22-1)=96

2x3=96

2x=96:3

2x=32

2x=25

x=5

b/(x-1)3=125

(x-1)3=53

x-1=5

x=6

c/(2x+1)3=343

(2x+1)3=73

2x+1=7

2x=6

x=3

 

 

2 tháng 10 2016

1/2.2x+4.2x=9.25

a: -2x(x+3)+x(2x-1)=10

=>-2x^2-6x+2x^2-x=10

=>-7x=10

=>x=-10/7

b: Sửa đề: 2/3x(9/2x+1/4)-(3x^2+2)=3

=>3x^2+1/6x-3x^2-2=3

=>1/6x-2=3

=>x=30

28 tháng 8 2023

sao sửa, đề nó vậy á

`#3107.101107`

`1.`

`a,`

`(2x - 3)^2 = |3 - 2x|`

`=> (2x - 3)^2 = |2x - 3|`

`=>`\(\left[{}\begin{matrix}2x-3=\left(2x-3\right)^2\\2x-3=-\left(2x-3\right)^2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x-3-\left(2x-3\right)^2=0\\2x-3+\left(2x-3\right)^2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}\left(2x-3\right)\left(1-2x+3\right)=0\\\left(2x-3\right)\left(1+2x-3\right)=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x-3=0\\4-2x=0\\2x-2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)

Vậy, `x \in {3/2; 2; 1}`

`b,`

`(x - 1)^2 + (2x - 1)^2 = 0`

`=>`\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(2x-1\right)^2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy, `x \in {1; 1/2}`

`c,`

`5 - x^2 = 1`

`=> x^2 = 4`

`=> x^2 = (+-2)^2`

`=> x = +-2`

Vậy, `x \in {-2; 2}`

`d,`

`x - 2\sqrt{x} = 0`

`=> x^2 - (2\sqrt{x})^2 = 0`

`=> x^2 - 4x = 0`

`=> x(x - 4) = 0`

`=>`\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy, `x \in {0; 4}`

`g,`

`(x - 1) + 1/7 = 0`

`=> x - 1 + 1/7 = 0`

`=> x - 6/7 = 0`

`=> x = 6/7`

Vậy, `x = 6/7.`

14 tháng 9 2023

\(a,\dfrac{3}{7}-x=\dfrac{1}{2}x-3\)

\(\Rightarrow-x-\dfrac{1}{2}x=-3-\dfrac{3}{7}\)

\(\Rightarrow-\dfrac{3}{2}x=-\dfrac{24}{7}\)

\(\Rightarrow x=-\dfrac{24}{7}:\left(-\dfrac{3}{2}\right)\)

\(\Rightarrow x=\dfrac{16}{7}\)

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

\(\Rightarrow5x+2x=\dfrac{5}{3}+\dfrac{2}{3}\)

\(\Rightarrow7x=\dfrac{7}{3}\)

\(\Rightarrow x=\dfrac{7}{3}:7\)

\(\Rightarrow x=\dfrac{1}{3}\)

#Toru

a: 3/7-x=1/2x-3

=>-3/2x=-3+3/7

=>-1/2x=-1+1/7=-6/7

=>1/2x=6/7

=>x=6/7*2=12/7

b: =>5x+2x=5/3+2/3

=>7x=7/3

=>x=1/3

a: \(\left|3x-2\right|=4\)

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

b: Ta có: \(\left|5x-3\right|=\left|x-7\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-3=x-7\\5x-3=7-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-4\\6x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)

8 tháng 8 2018

a) \(2^x\cdot2^2-2^x=96\)

\(2^x\cdot\left(4-1\right)=96\)

\(2^x=32=2^5\)

=> x = 5

b) \(\left(2x+1\right)^3=7^3\)

=> 2x +1 = 7

=> 2x = 6

=> x = 3

Vậy,..........

8 tháng 8 2018

a. 2x+2 - 2x = 96

2x . 22 - 2x  = 96

2x. (22 - 1)  = 96

2x                = 96 : 3

2x                = 32

2x                = 25

x                  = 5

b. (2x + 1)3 = 343

(2x + 1)3     = 73

2x + 1         = 7

2x               = 7 - 1

x                 = 6 : 2

x                 = 3

23 tháng 9 2023

a)19 - (x + 23)=24- 6

   19 - (x + 23) = 16 - 6 

    19 - (x + 23) = 10

     (x + 23) = 19 - 10

      x + 23= 9

      x + 2= 33

      x + 2 = 3

      x= 3-2

       x= 1

23 tháng 9 2023

x=1

x=-1