Đề bài
Tìm số tự nhiên x, biết :
\(\eqalign{ & a)\;\left( {x + 83} \right) - 37 = 56 \cr & b)\;128 - 3\left( {x - 5} \right) = 17 \cr & c)\;20 + 3x = {5^6}:{5^3} \cr & d)\;\left[ {\left( {2x + 1} \right).3 + 55} \right]:4 = 25 \cr & e)\;\left( {4x - {4^2}} \right){.7^3} = {4.7^4} \cr & g)\;{6^2}{.2^2}.5:\left[ {3.12 - \left( {2x - 6} \right)} \right] = {2^3}.5 \cr & h)\;8x - 3x = {6^{27}}:{6^{25}} + 44:11 \cr & i)\;3 \;\vdots \left( {x + 1} \right). \cr} \)
Lời giải chi tiết
\(\eqalign{ & a)\left( {x + 83} \right) - 37 = 56 \cr & x + 83 = 56 + 37 \cr & x + 83 = 93 \cr & x = 93 - 83 \cr & x = 10 \cr & b)128 - 3\left( {x - 5} \right) = 17 \cr & 3(x - 5) = 128 - 17 \cr & 3(x - 5) = 111 \cr & x - 5 = 111:3 \cr & x - 5 = 37 \cr & x = 37 + 5 \cr & x = 42 \cr & c)20 + 3x = {5^6}:{5^3} \cr & 20 + 3x = {5^{6 - 3}} \cr & 20 + 3x = {5^3} \cr & 20 + 3x = 125 \cr & 3x = 125 - 20 \cr & 3x = 105 \cr & x = 105:3 \cr & x = 35 \cr & \cr} \)
\(\eqalign{ & d)\left[ {\left( {2x + 1} \right).3 + 55} \right]:4 = 25 \cr & (2x + 1).3 + 55 = 25.4 \cr & (2x + 1).3 + 55 = 100 \cr & (2x + 1).3 = 100 - 55 \cr & (2x + 1).3 = 45 \cr & 2x + 1 = 45:3 \cr & 2x + 1 = 15 \cr & 2x = 15 - 1 \cr & 2x = 14 \cr & x = 14:2 \cr & x = 7 \cr & e)\left( {4x - {4^2}} \right){.7^3} = {4.7^4} \cr & 4x - 16 = ({4.7^4}):{7^3} \cr & 4x - 16 = 4.7 \cr & 4x - 16 = 28 \cr & 4x = 28 + 16 \cr & 4x = 44 \cr & x = 44:4 \cr & x = 11 \cr} \)
\(\eqalign{ & \cr & g){6^2}{.2^2}.5:\left[ {3.12 - \left( {2x - 6} \right)} \right] = {2^3}.5 \cr & 3.12 - (2x - 6) = ({6^2}{.2^2}.5):({2^3}.5) \cr & 36 - (2x - 6) = {6^2}:2 \cr & 36 - (2x - 6) = 36:2 \cr & 36 - (2x - 6) = 18 \cr & 2x - 6 = 36 - 18 \cr & 2x - 6 = 18 \cr & 2x = 18 + 6 \cr & 2x = 24 \cr & x = 24:2 \Leftrightarrow x = 12 \cr & h)8x - 3x = {6^{27}}:{6^{25}} + 44:11 \cr & (8 - 3)x = {6^{27 - 25}} + 4 \cr & 5x = {6^2} + 4 \cr & 5x = 36 + 4 \cr & 5x = 40 \cr & x = 40:5 \Leftrightarrow x = 8 \cr & i)\;3 \;\vdots \left( {x + 1} \right) \cr & \Rightarrow (x + 1) \in U(3) = {\rm{\{ }}1;3\} \cr & \Rightarrow x \in {\rm{\{ 0;2\} }} \cr} \)
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