Toán Tìm GTTĐ : a, |x+1|+|x+2|=x b, | |2x-1| +3 | = 3-x 17/10/2021 By Ximena Tìm GTTĐ : a, |x+1|+|x+2|=x b, | |2x-1| +3 | = 3-x
Đáp án: a. Phương trình vô nghiệm Giải thích các bước giải: \(\begin{array}{l}a.DK:x \ge 0\\{\left( {x + 1} \right)^2} + 2\left( {x + 1} \right)\left( {x + 2} \right) + {\left( {x + 2} \right)^2} = {x^2}\\ \to {x^2} + 2x + 1 + 2{x^2} + 6x + 4 + {x^2} + 4x + 4 = {x^2}\\ \to 3{x^2} + 12x + 9 = 0\\ \to {x^2} + 4x + 3 = 0\\ \to {x^2} + x + 3x + 3 = 0\\ \to x\left( {x + 1} \right) + 3\left( {x + 1} \right) = 0\\ \to \left[ \begin{array}{l}x + 1 = 0\\x + 3 = 0\end{array} \right.\\ \to \left[ \begin{array}{l}x = – 1\left( l \right)\\x = – 3\left( l \right)\end{array} \right.\end{array}\) ⇒ Phương trình vô nghiệm \(\begin{array}{l}b.\left| {\left| {2x – 1} \right| + 3} \right| = 3 – x\\ \to \left[ \begin{array}{l}\left| {2x – 1} \right| + 3 = 3 – x\\\left| {2x – 1} \right| + 3 = – 3 + x\end{array} \right.\\ \to \left[ \begin{array}{l}\left| {2x – 1} \right| = – x\\\left| {2x – 1} \right| = x – 6\end{array} \right.\\ \to \left[ \begin{array}{l}2x – 1 = x\left( {DK:x < \frac{1}{2}} \right)\\2x – 1 = – x\left( {DK:x \ge \frac{1}{2}} \right)\\2x – 1 = x – 6\left( {DK:x \ge \frac{1}{2}} \right)\\2x – 1 = – x + 6\left( {DK:x < \frac{1}{2}} \right)\end{array} \right.\\ \to \left[ \begin{array}{l}x = 1\\3x = 1\\x = – 5\\3x = 7\end{array} \right.\\ \to \left[ \begin{array}{l}x = 1\left( l \right)\\x = \frac{1}{3}\left( l \right)\\x = – 5\left( l \right)\\x = \frac{7}{3}\left( l \right)\end{array} \right.\end{array}\) ⇒ Phương trình vô nghiệm Trả lời
Đáp án:
a. Phương trình vô nghiệm
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0\\
{\left( {x + 1} \right)^2} + 2\left( {x + 1} \right)\left( {x + 2} \right) + {\left( {x + 2} \right)^2} = {x^2}\\
\to {x^2} + 2x + 1 + 2{x^2} + 6x + 4 + {x^2} + 4x + 4 = {x^2}\\
\to 3{x^2} + 12x + 9 = 0\\
\to {x^2} + 4x + 3 = 0\\
\to {x^2} + x + 3x + 3 = 0\\
\to x\left( {x + 1} \right) + 3\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x + 1 = 0\\
x + 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – 1\left( l \right)\\
x = – 3\left( l \right)
\end{array} \right.
\end{array}\)
⇒ Phương trình vô nghiệm
\(\begin{array}{l}
b.\left| {\left| {2x – 1} \right| + 3} \right| = 3 – x\\
\to \left[ \begin{array}{l}
\left| {2x – 1} \right| + 3 = 3 – x\\
\left| {2x – 1} \right| + 3 = – 3 + x
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left| {2x – 1} \right| = – x\\
\left| {2x – 1} \right| = x – 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x – 1 = x\left( {DK:x < \frac{1}{2}} \right)\\
2x – 1 = – x\left( {DK:x \ge \frac{1}{2}} \right)\\
2x – 1 = x – 6\left( {DK:x \ge \frac{1}{2}} \right)\\
2x – 1 = – x + 6\left( {DK:x < \frac{1}{2}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
3x = 1\\
x = – 5\\
3x = 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = \frac{1}{3}\left( l \right)\\
x = – 5\left( l \right)\\
x = \frac{7}{3}\left( l \right)
\end{array} \right.
\end{array}\)
⇒ Phương trình vô nghiệm