2(x-5)+3=-7

=>2x-10=-7+3=-4

=>2x=-4+10=6

=>x=6:2=3

4x+(27+2^3)=1

=>4x+(27+8)=1

=>4x+35=1

=>4x=1-35=-34

=>x=-34:4=8.5

lx-3l+5=-27

=>lx-3l=-27-5=-32

mà lx-3l>0 với mọi x

=>k co gt x t/m

-3lx-1l=-27

=>lx-1l=-27:(-3)=9

=>x-1=+9

+)x-1=9=>x=10

+)x-1=-9=>x=-8

a/\(x:27=3,6\)
\(\Rightarrow x=97,2\)
b/\(\dfrac{2x+1}{-27}=\dfrac{-3}{2x+1}\)
\(\Rightarrow\left(2x+1\right)^2=81\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{4;-5\right\}\)

a) x=3,6\(\times\) 27=97,2

a,x:27=3,6

x=3,6:27

x\(\approx\)0,13

\(\Rightarrow\)x=0,1(3)

x^4-1=27

=>x^4-1=3³

^=>x³=3³(vì 4-1=3)

=> x=3

3^x-1=27

=>3^x-1=3³

=>x-1=3

=>x=2

a) Ta có: x^4-1 = 27

Mà 3³ = 27

  x³ = 3³

Vậy x = 3

b) Ta có: 3^x-1 = 27

Mà 3³ = 27

=> 3^x-1 = 3³

=> x - 1 = 3

=>    x    = 3+1=4

Vậy x = 4

x^{4 - 1} = 27

x³ = 27

x³ = 3³

=> x = 3

3^{x - 1} = 27

3^{x - 1} = 3³

x - 1 = 3

     x = 3 + 1

      x = 4

Ta có: \(4\left(x+3\right)\left(3x-2\right)-3\left(x-1\right)\left(4x-1\right)=-27\)

\(\Leftrightarrow4\left(3x^2+7x-6\right)-3\left(4x^2-5x+1\right)=-27\)
\(\Leftrightarrow12x^2+28x-24-12x^2+15x-3=-27\)

\(\Leftrightarrow43x=-27+24+3=0\)

hay x=0

a)\(4\left(x+3\right)\left(3x-2\right)-3\left(x-1\right)\left(4x-1\right)=-27\)

\(\Leftrightarrow12x^2+28x-24-12x^2+15x-3=-27\)

\(\Leftrightarrow43x=0\\ \Leftrightarrow x=0\)

đề sai thì phải

\(a,3^{x+1}=27\)

\(3^{x+1}=3^3\)

\(x+1=3\)

\(x=3-1=2\)

\(b,6^{x-1}=36\)

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

\(x-1=2\)

\(x=2+1=3\)

\(c,3^{2x+1}=243\)

\(3^{2x+1}=3^5\)

\(2x+1=5\)

\(2x=5+1=6\)

\(x=6:2=3\)

Mình hong hiểu câu d :<

\(x^{50}=x\\ \Rightarrow x^{50}-x=0\\ \Rightarrow x\left(x^{49}-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^{49}=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(a,\\ \left(x+\dfrac{1}{2}\right)^3=\dfrac{8}{125}=\dfrac{2^3}{5^3}\\ \left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow\left(x+\dfrac{1}{2}\right)=\dfrac{2}{5}\\ x=\dfrac{2}{5}-\dfrac{1}{2}\\ x=-\dfrac{1}{10}\)

\(b,3\left|x\right|-27=\dfrac{1}{5}\\ 3\left|x\right|=\dfrac{1}{5}+27\\ 3\left|x\right|=\dfrac{136}{5}\\ \left|x\right|=\dfrac{136}{5}:3\\ \left|x\right|=\dfrac{136}{15}\\ Vậy:x=\dfrac{136}{15}.or.x=-\dfrac{136}{15}\)

\(c,\\ \dfrac{1}{2}.\left(x+\dfrac{1}{3}\right)=\dfrac{1}{16}\\ \left(x+\dfrac{1}{3}\right)=\dfrac{1}{16}:\dfrac{1}{2}\\ \left(x+\dfrac{1}{3}\right)=\dfrac{1}{8}\\ x=\dfrac{1}{8}-\dfrac{1}{3}\\ x=-\dfrac{5}{24}\)

a) \(4^n=4096\Rightarrow4^n=4^6\Rightarrow n=6\)

b) \(5^n=15625\Rightarrow5^n=5^6\Rightarrow n=6\)

c) \(6^{n+3}=216\Rightarrow6^{n+3}=6^3\Rightarrow n+3=3\Rightarrow n=0\)

d) \(x^2=x^3\Rightarrow x^3-x^2=0\Rightarrow x^2\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

e) \(3^{x-1}=27\Rightarrow3^{x-1}=3^3\Rightarrow x-1=3\Rightarrow x=4\)

f) \(3^{x+1}=9\Rightarrow3^{x+1}=3^2\Rightarrow x+1=2\Rightarrow x=1\)

g) \(6^{x+1}=36\Rightarrow6^{x+1}=6^2\Rightarrow x+1=2\Rightarrow x=1\)

h) \(3^{2x+1}=27\Rightarrow3^{2x+1}=3^3\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\)

i) \(x^{50}=x\Rightarrow x^{50}-x=0\Rightarrow x\left(x^{49}-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}=1=1^{49}\end{matrix}\right.\)  \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

4ⁿ  =  4096 

4ⁿ = 2¹²

n = 12

5ⁿ = 15625 

5ⁿ = 5⁶

n   = 6

6ⁿ⁺³ = 216

6ⁿ⁺³ = 2³.3³

6ⁿ⁺³ = 6³

n + 3 = 3

4ⁿ = 4096

4ⁿ = 4⁶

n = 6 (nhận)

Vậy n = 6

--------------------

5ⁿ = 15625

5ⁿ = 5⁶

n = 6 (nhận)

Vậy n = 6

--------------------

4ⁿ⁻¹ = 1024

4ⁿ⁻¹ = 4⁵

n - 1 = 5

n = 6 (nhận)

Vậy n = 6

-------------------

6ⁿ⁺³ = 216

6ⁿ⁺³ = 6³

n + 3 = 3

n = 0  (nhận)

Vậy n = 0

--------------------

x² = x³

x³ - x² = 0

x(x² - 1) = 0

x = 0 (nhận) hoặc x² - 1 = 0

*) x² - 1 = 0

x² = 1

x = 1 (nhận) hoặc x = -1 (loại)

Vậy x = 0; x = 1 

--------------------

3ˣ⁻¹ = 27

3ˣ⁻¹ = 3³

x - 1 = 3

x = 3 + 1

x = 4 (nhận)

Vậy x = 4

---------------------

3ˣ⁺¹ = 9

3ˣ⁺¹ = 3²

x + 1 = 2

x = 2 - 1

x = 1 (nhận)

Vậy x = 1

--------------------

6ˣ⁺¹ = 36

6ˣ⁺¹ = 6²

x + 1 = 2

x = 2 - 1

x = 1 (nhận)

Vậy x = 1

--------------------

3²ˣ⁺¹ = 27

3²ˣ⁺¹ = 3³

2x + 1 = 3

2x = 3 - 1

2x = 2

x = 1 (nhận)

Vậy x = 1

--------------------

x⁵⁰ = x

x⁵⁰ - x = 0

x(x⁴⁹ - 1) = 0

x = 0 (nhận) hoặc x⁴⁹ - 1 = 0

*) x⁴⁹ - 1 = 0

x⁴⁹ = 1

x = 1 (nhận)

Vậy x = 0; x = 1

ai k mình k lại [ chỉ 3 người đầu tiên mà trên 10 điểm hỏi đáp ]

3(1-4x)(x-1) + 4(3x-2)(x+3) = -27

<=>  (3-12x)(x-1) + (12x-8)(x+3) = -27

<=>  3x-3-12x²+12x+12x²+36x-8x-24+27 = 0

<=>  (-12x²+12x²)+(3x+12x+36x)+(27-3+36) = 0

<=>  51x+60 =0 

<=>  51x = -60

<=>  x= -60 : 51

<=>  x=  -20/17