tìm x
2.(x+1)^2 – 1 =49 3^2.3^x =3^5
32<2^x<128 (x-1 )^3 =125
2^x+2-2^x = 96
tìm x
2.(x+1)^2 – 1 =49 3^2.3^x =3^5
32<2^x<128 (x-1 )^3 =125
2^x+2-2^x = 96
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$
$⇒ 2^x = 2^5$
$⇒ x = 5$
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:
$$\eqalign{
& a)\,\,2{\left( {x + 1} \right)^2} – 1 = 49 \cr
& \,\,\,\,\,\,2{\left( {x + 1} \right)^2}\,\,\,\,\,\,\,\, = 49 + 1 \cr
& \,\,\,\,\,\,2{\left( {x + 1} \right)^2}\,\,\,\,\,\,\,\, = 50 \cr
& \,\,\,\,\,\,\,\,\,\,\,{\left( {x + 1} \right)^2}\,\,\,\,\,\, = 50:2 \cr
& \,\,\,\,\,\,\,\,\,\,\,{\left( {x + 1} \right)^2}\,\,\,\,\,\, = 25 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,x + 1\,\,\,\,\,\,\,\,\,\,\, = 5 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5 – 1 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4 \cr
& b)\,\,{3^2}{.3^x} = {3^5} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,{3^x} = {3^5}:{3^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,{3^x} = {3^{5 – 2}} \cr
& \,\,\,\,\,\,\,\,\,\,\,{3^x} = {3^3} \cr
& \,\,\,\,\,\,\,\,\,\,\,x = 3 \cr
& c)\,\,32 < {2^x} < 128 \cr & \,\,\,\,\,\,\,{2^5} < {2^x} < {2^7} \cr & \,\,\,\,\,\,\,\,5 < x < 7 \cr & \,\,\,\,\,\,\,\,x = 6 \cr & d)\,\,{\left( {x - 1} \right)^3} = 125 \cr & \,\,\,\,\,\,\,\,{\left( {x - 1} \right)^3} = {5^3} \cr & \,\,\,\,\,\,\,\,\,\,x - 1 = 5 \cr & \,\,\,\,\,\,\,\,\,\,x = \,\,5 + 1 \cr & \,\,\,\,\,\,\,\,\,\,x = 6 \cr & e)\,\,{2^{x + 2}} - {2^x} = 96 \cr & \,\,\,\,\,{2^x}{.2^2} - {2^x} = 96 \cr & \,\,\,\,\,{2^x}\left( {4 - 1} \right) = 96 \cr & \,\,\,\,\,{2^x}.3 = 96 \cr & \,\,\,\,\,{2^x}\,\,\,\,\, = 96:3 \cr & \,\,\,\,\,{2^x}\,\,\,\,\, = 32 \cr & \,\,\,\,\,{2^x}\,\,\,\,\, = {2^5} \cr & \,\,\,\,\,x\,\,\,\,\,\,\, = 5 \cr} $$