cả 3 bài đều giống nhau nên mình làm 1 bài thôi nhé

\(\dfrac{2x}{32}=1\Rightarrow2x=32\)

x = 32 : 2

x = 16

\(\dfrac{2x}{32}=1\)

\(\Rightarrow2x.1=32.1\)

\(\Rightarrow2x=32\)

\(\Rightarrow x=16\)

\(\dfrac{2x}{32}=1\)

\(\Rightarrow2x=32\)

\(x=16\)

Chúc bn học tốt

\(2^{x+1}-2^x=32\)

=>2^x(2-1)=32

=>2^x=32

=>x=5

\(\left(2x+1\right)y-2x-1=-32\)

\(\Leftrightarrow2xy-2x+y-1=-32\)

\(\Leftrightarrow2x\left(y-1\right)+\left(y-1\right)=-32\)

\(\left(2x+1\right)\left(y-1\right)=-32\)

Đề bài nên cho là x,y thuộc Z; chia trường hợp rồi làm :vv

(2x + 1) y − 2x − 1 = −32

⇔2xy − 2x + y − 1 = −32

⇔2x (y − 1) + (y − 1) = −32

(2x + 1) (y − 1) = −32

=> 2x+1 và y-1 thuộc Ư(-32)

vậy.............................................................

                                                  Bài giải

\(\left(2x+1\right)y-2x-1=-32\)

\(\left(2x+1\right)y-\left(2x+1\right)=-32\)

\(\left(2x+1\right)\left(y-1\right)=-32\)

\(\Rightarrow\text{ }2x+1\text{ , }y-1\inƯ\left(-31\right)=\left\{\pm1\text{ ; }\pm31\right\}\)

Ta có bảng :

\(\Rightarrow\text{ }\left(x\text{ ; }y\right)=\left(-1\text{ ; }-30\right)\text{ ; }\left(-16\text{ ; }0\right)\text{ ; }\left(0\text{ ; }32\right)\text{ ; }\left(15\text{ ; }2\right)\)

a)19 - (x + 2³)=2⁴- 6

   19 - (x + 2³) = 16 - 6 

    19 - (x + 2³) = 10

     (x + 2³) = 19 - 10

      x + 2³= 9

      x + 2³ = 3³

      ^{x + 2 = 3}

      ^{x= 3-2}

       ^{x= 1}

sửa lại :

a)19 - (x + 2³)=2⁴- 6

   19 - (x + 2³) = 16 - 6 

    19 - (x + 2³) = 10

     (x + 2³) = 19 - 10

      x + 2³= 9

    =>  x + 8= 9             x= 1

      ^{=> x + 8 =-9                     x= -17}

sửa lại b) 4³ + 3² : (x + 1) = 65

Bài 2: 

a: Ta có: \(2^{x+1}\cdot3^y=12^x\)

\(\Leftrightarrow2^{x+1}\cdot3^y=2^{2x}\cdot3^x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

\(\dfrac{2x}{15}+\dfrac{2x}{35}+\dfrac{2x}{63}+...+\dfrac{2x}{195}=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{195}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{13\cdot15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\dfrac{4}{15}=\dfrac{4}{5}\\ x=\dfrac{4}{5}:\dfrac{4}{15}\\ x=3\)

Gọi \(D=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)

\(2D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\\ 2D+D=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\\ 3D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\\ 3D=1-\dfrac{1}{64}< 1\\ \Rightarrow D=\dfrac{1-\dfrac{1}{64}}{3}< \dfrac{1}{3}\)

Vậy \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)

\(\left(2x+1\right)^3=125\\ \Rightarrow\left(2x+1\right)^3=5^3\\ \Rightarrow2x+1=5\\ \Rightarrow2x=4\\ \Rightarrow x=2.\\ b,\left(2x-1\right)^4=16\\ \Rightarrow\left(2x-1\right)^4=2^4\\ \Rightarrow2x-1=2\\ \Rightarrow2x=3\\ \Rightarrow x=\dfrac{3}{2}.\\ c,6.3^x-2.3^x=36\\ \Rightarrow3^x.\left(6-2\right)=36\\ \Rightarrow3^x.4=36\\ \Rightarrow3^x=9\\ \Rightarrow3^x=3^2\\ \Rightarrow x=2.\\ d,2^{x+1}-2^x=32\\ \Rightarrow2^x.\left(2-1\right)=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5.\)

a. (2x-1)⁴=81

=>(2x-1)⁴=3⁴

=>2x-1=3

=>2x=3+1

=>2x=4

=>x=4:2

=>x=2

b.(x-1)⁵=-32

=>(x-1)⁵=(-2)⁵

=>x-1=-2

=>x=-2+1

=>x=-1

c.(2x-1)⁶=(2x-1)⁸

mà chỉ có: (-1)⁶=(-1)⁸; 0⁶=0⁸; 1⁶=1⁸

=> để (2x-1) \(\in\){-1;0;1} thì x \(\in\){0; 1/2; 1}

(2x-1)^4=3^4

=>2x-1=3

=>2x=4

=>x=2

ai h minh minh h lai cho

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35