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 =

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

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  1. 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:

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

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