Câu hỏi của Mymom12345

Question

Tìm x:

a, 3^{2x + 1} + 3^2x = 36

b, (x + 1)⁵= ( x+ 1)³

Nhanh lên đi bạn, mai mình nộp rồi

Minh Thư · Accepted Answer

a) $3^{2x+1}+3^{2x}=36$

$\Leftrightarrow3^{2x}\left(3+1\right)=36$

$\Leftrightarrow3^{2x}.4=36$

$\Leftrightarrow3^{2x}=9$

$\Leftrightarrow3^{2x}=3^2$

$\Leftrightarrow2x=2\Leftrightarrow x=1$

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

$\Leftrightarrow\left(x+1\right)^5-\left(x+1\right)^3=0$

$\Leftrightarrow\left(x+1\right)^3\left[\left(x+1\right)^2-1\right]=0$

$\Leftrightarrow\orbr{\begin{cases}x+1=0\\left(x+1\right)^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\x\in\left\{-2;0\right\}\end{cases}}$

Câu trả lời:

a) $3^{2x+1}+3^{2x}=36$

$\Leftrightarrow3^{2x}\left(3+1\right)=36$

$\Leftrightarrow3^{2x}.4=36$

$\Leftrightarrow3^{2x}=9$

$\Leftrightarrow3^{2x}=3^2$

$\Leftrightarrow2x=2\Leftrightarrow x=1$

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

$\Leftrightarrow\left(x+1\right)^5-\left(x+1\right)^3=0$

$\Leftrightarrow\left(x+1\right)^3\left[\left(x+1\right)^2-1\right]=0$

$\Leftrightarrow\orbr{\begin{cases}x+1=0\\left(x+1\right)^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\x\in\left\{-2;0\right\}\end{cases}}$

Edogawa Conan · Answer

a)3^{2x + 1} + 3^2x = 36

=> 3^2x. 3 + 3^2x = 36

=> 3^2x . 4 = 36

=> 3^2x = 36 : 4

=> 3^2x = 9

=> 3^2x = 3²

=> 2x = 2

=> x = 1

b) (x + 1)⁵ = (x + 1)³

=> (x + 1)⁵ - (x + 1)³ = 0

=> (x + 1)³.[(x + 1)² - 1] = 0

=>(x + 1)³(x + 1 - 1)(x + 1 + 1) = 0

=> (x + 1)³.x.(x + 2) = 0

=> x = -1 hoặc x = 0 hoặc x = -2