1) \(\left(x+0,7\right)^3=-27\Leftrightarrow\left(x+0,7\right)^3=\left(-3\right)^3\Leftrightarrow x+0,7=-3\Leftrightarrow x=-3,7\)2) \(\left(2x-1\right)^{10}=49^5\Leftrightarrow\left(2x-1\right)^{10}=\left(7^2\right)^5\)

\(\Leftrightarrow\left(2x-1\right)^{10}=7^{10}\)

\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)

mấy câu khác tương tự

1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)

\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)

\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))

\(\Leftrightarrow x=-36\).

Vậy nghiệm của pt là x = -36.

2) x(x+1)(x+2)(x+3)= 24

⇔ x.(x+3)  .   (x+2).(x+1)  = 24

⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24

Đặt \(x^2\)+ 3x = b

⇒ b . (b+2)= 24

Hay: \(b^2\) +2b = 24

⇔\(b^2\) + 2b + 1 = 25

⇔\(\left(b+1\right)^2\)= 25

+ Xét b+1 = 5 ⇒ b=4 ⇒  \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0

⇒(x-1)(x+4)=0⇒x=1 và x=-4

+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0

⇒\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))² = - \(\dfrac{15}{4}\)  Hay ( \(x^2\) +\(\dfrac{3}{2}\) )²= -\(\dfrac{15}{4}\) (vô lí)

⇒x= 1 và x= 4

a: =>4^x=640

=>\(x\in\varnothing\)

b: =>\(3^{-2x}\cdot3^{3x}=243\)

=>3^x=243

=>x=5

a, (3x+1)³=-27                    b, (1/2.2^x+4.2^x ) = 9.25

=> ( 3x+1)³=(-3)³                    => [ 2^x.(1/2+4 ) ]=9.25                                c, ( 3x-2)⁵= - 243

=> 3x+1=-3                       => 2^x.9/2 = 225                                           => ( 3x-2)⁵=(-3)⁵

=>3x=-4                            => 2^x        = 225:9/2                                     => 3x-2=-3

=>  x = -4/3                       => 2^x       = 50 (ko có trường hợp nào )         => 3x= -1

                                                                                                            => x=-1/3

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

⇔\(x^2-2.x.1,2+1,2^2=4\)

⇔\(x^2-2,4x+1,44=4\)

⇔\(x^2-2,4x=4-1,44\)

⇔\(x\left(x-2,4\right)=2,56\)

⇔\(x=2,56\) hoặc \(x-2,4=2,56\)

⇔\(x=2,56\) hoặc \(x=4,96\)

a) \(\left(x-1,2\right)^2=4=2^2\)

\(\Leftrightarrow x-1,2=4\)

\(\Leftrightarrow x=5,2\)

b) \(\left(x+1\right)^3=-125=\left(-5\right)^3\)

\(\Leftrightarrow x+1=-5\)

\(\Leftrightarrow x=-6\)

c) \(\left(x+1,5\right)^8+\left(2,7-y\right)^{10}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\2,7-y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,7\end{matrix}\right.\)

3^ x -1 = 1/243

3^x =1/243 +1

3^x = 244 / 243

Ta  thấy đây ko phải lũy thừa của 3 => Ko có x thỏa mãn

81^-2x . 27^x =9^5

81^-2 . 81^x . 27^x =9^5

1/9^4 . (81.27)^x =9 ^5

            3^6x = 9^5 : 1/9^4

             3^6x = 9^9

            3^6x = 3^18

    => 6x =18

  x=3

2^x +2^x +3 =144

2.(2^x) =141

2^x+1 = 141

Ta thấy 141 ko phải lũy thừa của 2 => ko có x thỏa mãn

1)

a) \(0,25^x\cdot12^x=243\)

\(\Leftrightarrow\left(0,25\cdot12\right)^x=3^5\)

\(\Leftrightarrow3^x=3^5\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

b) \(38^y:19^y=512\)

\(\Leftrightarrow2y\cdot y=512\)

\(\Leftrightarrow2y^2=512\)

\(\Leftrightarrow y^2=256\)

\(\Leftrightarrow\left[{}\begin{matrix}y=16\\y=-16\end{matrix}\right.\)

Vậy \(y_1=-16;y_2=16\)

2)

a) \(3^x+3^{x+2}=2430\)

\(\Leftrightarrow\left(1+3^2\right)\cdot3^x=2430\)

\(\Leftrightarrow\left(1+9\right)\cdot3^x=2430\)

\(\Leftrightarrow10\cdot3^x=2430\)

\(\Leftrightarrow3^x=243\)

\(\Leftrightarrow3^x=3^5\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

b) \(2^{x+3}-2^x=224\)

\(\Leftrightarrow\left(2^3-1\right)\cdot2^x=224\)

\(\Leftrightarrow\left(8-1\right)\cdot2^x=224\)

\(\Leftrightarrow7\cdot2^x=224\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

3)

a) \(\left(x-\dfrac{1}{4}\right)^2=\dfrac{4}{9}\)

\(\Leftrightarrow x-\dfrac{1}{4}=\pm\dfrac{2}{3}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=-\dfrac{5}{12}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{11}{12};x_2=-\dfrac{5}{12}\)

b) \(\left(x+0,7\right)^3=-27\)

\(\Leftrightarrow\left(x+\dfrac{3}{10}\right)^3=\left(-3\right)^3\)

\(\Leftrightarrow x+\dfrac{3}{10}=-3\)

\(\Leftrightarrow x=-3-\dfrac{3}{10}\)

\(\Leftrightarrow x=-\dfrac{37}{10}\)

Vậy \(x=-\dfrac{37}{10}\)

4)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{1}{15};x_2=\dfrac{1}{3}\)

b) \(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)

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

\(\Leftrightarrow2x-1=1\)

\(\Leftrightarrow2x=1+1\)

\(\Leftrightarrow2x=2\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

1. a) \(0,25^x.12^x=243\)

\(\Rightarrow\left(0,25.12\right)^x=243\)

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

\(\Rightarrow x=5\)

Vậy \(x=5.\)

b) \(38^y:19^y=512\)

\(\Rightarrow\left(38:19\right)^y=512\)

\(\Rightarrow2^y=2^9\)

\(\Rightarrow y=9\)

Vậy \(y=9.\)

2) a) \(3^x+3^{x+2}=2430\)

\(\Rightarrow3^x\left(1+9\right)=2430\)

\(\Rightarrow3^x=243=3^5\)

\(\Rightarrow x=5\)

Vậy x=5.

b) \(2^{x+3}-2^x=224\)

\(\Rightarrow2^x\left(8-1\right)=224\)

\(\Rightarrow2^x=32=2^5\)

\(\Rightarrow x=5\)

Vậy x=5.

Bài 3: dễ tự làm.