Tìm x thỏa mãn |2x+3|+|2x-1|=8/[3(x+1)^2+2] 05/09/2021 Bởi Reagan Tìm x thỏa mãn |2x+3|+|2x-1|=8/[3(x+1)^2+2]
Đáp án: \(\left[ \begin{array}{l}x = \dfrac{{ – 3 + \sqrt 6 }}{3}\\x = – 1,579665221\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}\left| {2x + 3} \right| + \left| {2x – 1} \right| = \dfrac{8}{{3{{\left( {x + 1} \right)}^2} + 2}}\\ \to \left| {2x + 3} \right| + \left| {2x – 1} \right| = \dfrac{8}{{3{x^2} + 6x + 3 + 2}}\\ \to \left| {2x + 3} \right| + \left| {2x – 1} \right| = \dfrac{8}{{3{x^2} + 6x + 5}}\\TH1:x \ge \dfrac{1}{2}\\Pt \to 2x + 3 + 2x – 1 = \dfrac{8}{{3{x^2} + 6x + 5}}\\ \to 4x + 2 = \dfrac{8}{{3{x^2} + 6x + 5}}\\ \to 2x + 1 = \dfrac{4}{{3{x^2} + 6x + 5}}\\ \to 6{x^3} + 12{x^2} + 10x + 3{x^2} + 6x + 5 = 4\\ \to 6{x^3} + 15{x^2} + 16x + 1 = 0\\ \to x = – 0,06654041966\left( l \right)\\TH2:\dfrac{1}{2} > x \ge – \dfrac{3}{2}\\Pt \to 2x + 3 – 2x + 1 = \dfrac{8}{{3{x^2} + 6x + 5}}\\ \to 4 = \dfrac{8}{{3{x^2} + 6x + 5}}\\ \to 1 = \dfrac{4}{{3{x^2} + 6x + 5}}\\ \to 3{x^2} + 6x + 5 = 4\\ \to 3{x^2} + 6x + 1 = 0\\ \to \left[ \begin{array}{l}x = \dfrac{{ – 3 + \sqrt 6 }}{3}\left( {TM} \right)\\x = \dfrac{{ – 3 – \sqrt 6 }}{3}\left( l \right)\end{array} \right.\\TH3: – \dfrac{3}{2} > x\\Pt \to – 2x – 3 – 2x + 1 = \dfrac{8}{{3{x^2} + 6x + 5}}\\ \to – 2 – 4x = \dfrac{8}{{3{x^2} + 6x + 5}}\\ \to – 1 – 2x = \dfrac{4}{{3{x^2} + 6x + 5}}\\ \to 2x + 1 = \dfrac{{ – 4}}{{3{x^2} + 6x + 5}}\\ \to 6{x^3} + 12{x^2} + 10x + 3{x^2} + 6x + 5 = – 4\\ \to 6{x^3} + 16{x^2} + 16x + 9 = 0\\ \to x = – 1,579665221\left( {TM} \right)\end{array}\) \(KL:\left[ \begin{array}{l}x = \dfrac{{ – 3 + \sqrt 6 }}{3}\\x = – 1,579665221\end{array} \right.\) Bình luận
|2x+3|>0 |2x-1|>0 =>|2x+3|+|2x-1|>0 =>8/[3(x+1)^2+2]>0 =>3(x+1)^2+2 <8 =>3(x+1)^2<6 =>(x+1)^2<2 =>(x+1)<căn bậc 2 của 2 >x<2.41421356… =>x<2 Bình luận
Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{{ – 3 + \sqrt 6 }}{3}\\
x = – 1,579665221
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left| {2x + 3} \right| + \left| {2x – 1} \right| = \dfrac{8}{{3{{\left( {x + 1} \right)}^2} + 2}}\\
\to \left| {2x + 3} \right| + \left| {2x – 1} \right| = \dfrac{8}{{3{x^2} + 6x + 3 + 2}}\\
\to \left| {2x + 3} \right| + \left| {2x – 1} \right| = \dfrac{8}{{3{x^2} + 6x + 5}}\\
TH1:x \ge \dfrac{1}{2}\\
Pt \to 2x + 3 + 2x – 1 = \dfrac{8}{{3{x^2} + 6x + 5}}\\
\to 4x + 2 = \dfrac{8}{{3{x^2} + 6x + 5}}\\
\to 2x + 1 = \dfrac{4}{{3{x^2} + 6x + 5}}\\
\to 6{x^3} + 12{x^2} + 10x + 3{x^2} + 6x + 5 = 4\\
\to 6{x^3} + 15{x^2} + 16x + 1 = 0\\
\to x = – 0,06654041966\left( l \right)\\
TH2:\dfrac{1}{2} > x \ge – \dfrac{3}{2}\\
Pt \to 2x + 3 – 2x + 1 = \dfrac{8}{{3{x^2} + 6x + 5}}\\
\to 4 = \dfrac{8}{{3{x^2} + 6x + 5}}\\
\to 1 = \dfrac{4}{{3{x^2} + 6x + 5}}\\
\to 3{x^2} + 6x + 5 = 4\\
\to 3{x^2} + 6x + 1 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{ – 3 + \sqrt 6 }}{3}\left( {TM} \right)\\
x = \dfrac{{ – 3 – \sqrt 6 }}{3}\left( l \right)
\end{array} \right.\\
TH3: – \dfrac{3}{2} > x\\
Pt \to – 2x – 3 – 2x + 1 = \dfrac{8}{{3{x^2} + 6x + 5}}\\
\to – 2 – 4x = \dfrac{8}{{3{x^2} + 6x + 5}}\\
\to – 1 – 2x = \dfrac{4}{{3{x^2} + 6x + 5}}\\
\to 2x + 1 = \dfrac{{ – 4}}{{3{x^2} + 6x + 5}}\\
\to 6{x^3} + 12{x^2} + 10x + 3{x^2} + 6x + 5 = – 4\\
\to 6{x^3} + 16{x^2} + 16x + 9 = 0\\
\to x = – 1,579665221\left( {TM} \right)
\end{array}\)
\(KL:\left[ \begin{array}{l}
x = \dfrac{{ – 3 + \sqrt 6 }}{3}\\
x = – 1,579665221
\end{array} \right.\)
|2x+3|>0
|2x-1|>0
=>|2x+3|+|2x-1|>0
=>8/[3(x+1)^2+2]>0
=>3(x+1)^2+2 <8
=>3(x+1)^2<6
=>(x+1)^2<2
=>(x+1)<căn bậc 2 của 2
>x<2.41421356…
=>x<2