căn -x^2+3x-2 + căn x+1 = căn 2 giúp e với ạ 27/08/2021 Bởi Eden căn -x^2+3x-2 + căn x+1 = căn 2 giúp e với ạ
Đáp án: \({\left[ {\begin{array}{*{20}{l}}{x = 1}\\{x = 1,906795303}\end{array}} \right.}\) Giải thích các bước giải: \(\begin{array}{*{20}{l}}{DK:\left\{ {\begin{array}{*{20}{l}}{ – {x^2} + 3x – 2 \ge 0}\\{x + 1 \ge 0}\end{array}} \right.}\\{ \to \left\{ {\begin{array}{*{20}{l}}{1 \le x \le 2}\\{x \ge {\rm{ \;}} – 1}\end{array}} \right.}\\{ \to 1 \le x \le 2}\\{\sqrt { – {x^2} + 3x – 2} {\rm{ \;}} + \sqrt {x + 1} {\rm{ \;}} = \sqrt 2 }\\{{\rm{\;}} \to \sqrt { – {x^2} + 3x – 2} {\rm{ \;}} = \sqrt 2 {\rm{ \;}} – \sqrt {x + 1} }\\{ \to {\rm{ \;}} – {x^2} + 3x – 2 = 2 – 2\sqrt {2x + 2} {\rm{ \;}} + x + 1}\\{ \to {\rm{ \;}} – {x^2} + 2x – 5 = {\rm{ \;}} – 2\sqrt {2x + 2} }\\{ \to {x^2} – 2x + 5 = 2\sqrt {2x + 2} }\\{ \to {x^4} + 4{x^2} + 25 – 4{x^3} + 10{x^2} – 20x = 4\left( {2x + 2} \right)}\\{ \to {x^4} – 4{x^3} + 14{x^2} – 28x + 17 = 0}\\{ \to {x^4} – {x^3} – 3{x^3} + 3{x^2} + 11{x^2} – 11x – 17x + 17 = 0}\\{ \to {x^3}\left( {x – 1} \right) – 3{x^2}\left( {x – 1} \right) + 11x\left( {x – 1} \right) – 17\left( {x – 1} \right) = 0}\\{ \to \left( {x – 1} \right)\left( {{x^3} – 3{x^2} + 11x – 17} \right) = 0}\\{ \to \left[ {\begin{array}{*{20}{l}}{x = 1\left( {TM} \right)}\\{{x^3} – 3{x^2} + 11x – 17 = 0}\end{array}} \right.}\\{ \to \left[ {\begin{array}{*{20}{l}}{x = 1}\\{x = 1,906795303\left( {TM} \right)}\end{array}} \right.}\\{KL:\left[ {\begin{array}{*{20}{l}}{x = 1}\\{x = 1,906795303}\end{array}} \right.}\end{array}\) Bình luận
Đáp án:
\({\left[ {\begin{array}{*{20}{l}}
{x = 1}\\
{x = 1,906795303}
\end{array}} \right.}\)
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{DK:\left\{ {\begin{array}{*{20}{l}}
{ – {x^2} + 3x – 2 \ge 0}\\
{x + 1 \ge 0}
\end{array}} \right.}\\
{ \to \left\{ {\begin{array}{*{20}{l}}
{1 \le x \le 2}\\
{x \ge {\rm{ \;}} – 1}
\end{array}} \right.}\\
{ \to 1 \le x \le 2}\\
{\sqrt { – {x^2} + 3x – 2} {\rm{ \;}} + \sqrt {x + 1} {\rm{ \;}} = \sqrt 2 }\\
{{\rm{\;}} \to \sqrt { – {x^2} + 3x – 2} {\rm{ \;}} = \sqrt 2 {\rm{ \;}} – \sqrt {x + 1} }\\
{ \to {\rm{ \;}} – {x^2} + 3x – 2 = 2 – 2\sqrt {2x + 2} {\rm{ \;}} + x + 1}\\
{ \to {\rm{ \;}} – {x^2} + 2x – 5 = {\rm{ \;}} – 2\sqrt {2x + 2} }\\
{ \to {x^2} – 2x + 5 = 2\sqrt {2x + 2} }\\
{ \to {x^4} + 4{x^2} + 25 – 4{x^3} + 10{x^2} – 20x = 4\left( {2x + 2} \right)}\\
{ \to {x^4} – 4{x^3} + 14{x^2} – 28x + 17 = 0}\\
{ \to {x^4} – {x^3} – 3{x^3} + 3{x^2} + 11{x^2} – 11x – 17x + 17 = 0}\\
{ \to {x^3}\left( {x – 1} \right) – 3{x^2}\left( {x – 1} \right) + 11x\left( {x – 1} \right) – 17\left( {x – 1} \right) = 0}\\
{ \to \left( {x – 1} \right)\left( {{x^3} – 3{x^2} + 11x – 17} \right) = 0}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x = 1\left( {TM} \right)}\\
{{x^3} – 3{x^2} + 11x – 17 = 0}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x = 1}\\
{x = 1,906795303\left( {TM} \right)}
\end{array}} \right.}\\
{KL:\left[ {\begin{array}{*{20}{l}}
{x = 1}\\
{x = 1,906795303}
\end{array}} \right.}
\end{array}\)