a)(x+15)^200+|y-3|=0 b)(x-1)^x+2=(x-1)^-x+6 c)|5-3x|+2/5=1/6 d)1/5-|1/5-x|1 e)(x+1/5)^4=81 f)(x+2)^2=25 g)3^x+5=81 h)17-|2x-3|=3 Các bạn nhanh giúp mì

0 bình luận về “a)(x+15)^200+|y-3|=0 b)(x-1)^x+2=(x-1)^-x+6 c)|5-3x|+2/5=1/6 d)1/5-|1/5-x|1 e)(x+1/5)^4=81 f)(x+2)^2=25 g)3^x+5=81 h)17-|2x-3|=3 Các bạn nhanh giúp mì”

1. Đáp án:

   \(\eqalign{
   & a)\,\,\left( { – 15;3} \right) \cr
   & b)\,\,x = 2 \cr
   & c)\,\,vo\,\,nghiem \cr
   & d)\,\,vo\,\,nghiem \cr
   & e)\,\,\left[ \matrix{
   x = {{14} \over 5} \hfill \cr
   x = – {{16} \over 5} \hfill \cr} \right. \cr
   & f)\,\,\left[ \matrix{
   x = 3 \hfill \cr
   x = – 7 \hfill \cr} \right. \cr
   & g)\,\,x = – 1 \cr
   & h)\,\,\left[ \matrix{
   x = {{17} \over 2} \hfill \cr
   x = – {{11} \over 2} \hfill \cr} \right. \cr} \)

   Giải thích các bước giải:

   \(\eqalign{
   & a)\,\,{\left( {x + 15} \right)^{200}} + \left| {y – 3} \right| = 0 \cr
   & Ta\,\,co:\,\,\left\{ \matrix{
   {\left( {x + 15} \right)^{200}} \ge 0 \hfill \cr
   \left| {y – 3} \right| \ge 0 \hfill \cr} \right. \Rightarrow {\left( {x + 15} \right)^{200}} + \left| {y – 3} \right| \ge 0 \cr
   & Dau\,\,” = ”\,\,xay\,\,ra \Leftrightarrow \left\{ \matrix{
   {\left( {x + 15} \right)^{200}} = 0 \hfill \cr
   \left| {y – 3} \right| = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
   x + 15 = 0 \hfill \cr
   y – 3 = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
   x = – 15 \hfill \cr
   y = 3 \hfill \cr} \right. \cr
   & \Rightarrow Nghiem:\,\,\left( {x;y} \right) = \left( { – 15;3} \right) \cr
   & b)\,\,{\left( {x – 1} \right)^{x + 2}} = {\left( {x – 1} \right)^{ – x + 6}} \cr
   & \Leftrightarrow {{{{\left( {x – 1} \right)}^{x + 2}}} \over {{{\left( {x – 1} \right)}^{ – x + 6}}}} = 1 \cr
   & \Leftrightarrow {\left( {x – 1} \right)^{x + 2 + x – 6}} = 1 \cr
   & \Leftrightarrow {\left( {x – 1} \right)^{2x – 4}} = 1 \cr
   & \Leftrightarrow \left[ \matrix{
   x – 1 = 1 \hfill \cr
   2x – 4 = 0 \hfill \cr} \right. \Leftrightarrow x = 2 \cr
   & c)\,\,\left| {5 – 3x} \right| + {2 \over 5} = {1 \over 6} \cr
   & \Leftrightarrow \left| {5 – 3x} \right| = {1 \over 6} – {2 \over 5} = – {7 \over {30}} < 0\,\,\left( {vo\,\,nghiem} \right) \cr & d)\,\,{1 \over 5} - \left| {{1 \over 5} - x} \right| = 1 \cr & \Leftrightarrow \left| {{1 \over 5} - x} \right| = {1 \over 5} - 1 = - {4 \over 5} < 0\,\,\left( {Vo\,\,nghiem} \right) \cr & e)\,\,{\left( {x + {1 \over 5}} \right)^4} = 81 \cr & \Leftrightarrow {\left( {x + {1 \over 5}} \right)^4} = {3^4} \cr & \Leftrightarrow \left[ \matrix{ x + {1 \over 5} = 3 \hfill \cr x + {1 \over 5} = - 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = 3 - {1 \over 5} = {{14} \over 5} \hfill \cr x = - 3 - {1 \over 5} = - {{16} \over 5} \hfill \cr} \right. \cr & f)\,\,{\left( {x + 2} \right)^2} = 25 \cr & \Leftrightarrow {\left( {x + 2} \right)^2} = {5^2} \cr & \Leftrightarrow \left[ \matrix{ x + 2 = 5 \hfill \cr x + 2 = - 5 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = 3 \hfill \cr x = - 7 \hfill \cr} \right. \cr & g)\,\,{3^{x + 5}} = 81 \cr & \Leftrightarrow {3^{x + 5}} = {3^4} \cr & \Leftrightarrow x + 5 = 4 \cr & \Leftrightarrow x = - 1 \cr & h)\,\,17 - \left| {2x - 3} \right| = 3 \cr & \Leftrightarrow \left| {2x - 3} \right| = 17 - 3 = 14 \cr & \Leftrightarrow \left[ \matrix{ 2x - 3 = 14 \hfill \cr 2x - 3 = - 14 \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ 2x = 17 \hfill \cr 2x = - 11 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = {{17} \over 2} \hfill \cr x = - {{11} \over 2} \hfill \cr} \right. \cr} \)

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