Đáp án:

\(\eqalign{ & a)\,\,x = 4 \cr & b)\,\,\,x = 3 \cr & c)\,\,\,x = 6 \cr & d)\,\,x = 6 \cr & e)\,\,x = 5 \cr} \)

Giải thích các bước giải:

$$\eqalign{ & a)\,\,2{\left( {x + 1} \right)^2} - 1 = 49 \cr & \,\,\,\,\,\,2{\left( {x + 1} \right)^2}\,\,\,\,\,\,\,\, = 49 + 1 \cr & \,\,\,\,\,\,2{\left( {x + 1} \right)^2}\,\,\,\,\,\,\,\, = 50 \cr & \,\,\,\,\,\,\,\,\,\,\,{\left( {x + 1} \right)^2}\,\,\,\,\,\, = 50:2 \cr & \,\,\,\,\,\,\,\,\,\,\,{\left( {x + 1} \right)^2}\,\,\,\,\,\, = 25 \cr & \,\,\,\,\,\,\,\,\,\,\,\,x + 1\,\,\,\,\,\,\,\,\,\,\, = 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5 - 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4 \cr & b)\,\,{3^2}{.3^x} = {3^5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,{3^x} = {3^5}:{3^2} \cr & \,\,\,\,\,\,\,\,\,\,\,{3^x} = {3^{5 - 2}} \cr & \,\,\,\,\,\,\,\,\,\,\,{3^x} = {3^3} \cr & \,\,\,\,\,\,\,\,\,\,\,x = 3 \cr & c)\,\,32 < {2^x} < 128 \cr & \,\,\,\,\,\,\,{2^5} < {2^x} < {2^7} \cr & \,\,\,\,\,\,\,\,5 < x < 7 \cr & \,\,\,\,\,\,\,\,x = 6 \cr & d)\,\,{\left( {x - 1} \right)^3} = 125 \cr & \,\,\,\,\,\,\,\,{\left( {x - 1} \right)^3} = {5^3} \cr & \,\,\,\,\,\,\,\,\,\,x - 1 = 5 \cr & \,\,\,\,\,\,\,\,\,\,x = \,\,5 + 1 \cr & \,\,\,\,\,\,\,\,\,\,x = 6 \cr & e)\,\,{2^{x + 2}} - {2^x} = 96 \cr & \,\,\,\,\,{2^x}{.2^2} - {2^x} = 96 \cr & \,\,\,\,\,{2^x}\left( {4 - 1} \right) = 96 \cr & \,\,\,\,\,{2^x}.3 = 96 \cr & \,\,\,\,\,{2^x}\,\,\,\,\, = 96:3 \cr & \,\,\,\,\,{2^x}\,\,\,\,\, = 32 \cr & \,\,\,\,\,{2^x}\,\,\,\,\, = {2^5} \cr & \,\,\,\,\,x\,\,\,\,\,\,\, = 5 \cr} $$

Bạn tham khảo :

$2(x+1)^2 - 1 = 49$

$⇒ 2(x+1)^2 = 50$ 

$⇒ (x+1)^2 = 25$

$⇒ (x+1)^2 = 5^2$

$⇒ x+1 = 5$

$⇒ x = 4$ 

$3^2 . 3^x = 3^5$

$3^{2+x} = 3^{5}$

$⇒ 2+x =5$

$⇒ x  = 3$

$32< 2^x < 128$

⇒ $ 2^5 < 2^x < 2^7$

⇒ $x = 6$ 

$(x-1)^3 = 125$

⇒ $(x - 1)=5^3$

⇒ $x  - 1 = 5$

⇒ $x = 6$

$2^x + 2 - 2^x = 96$

$(4 - 1) . 2^x = 96$

$3 . 2^x = 96$

$ ⇒ 2^x = 32$

$⇒ 2^x = 2^5$
$⇒ x = 5$