Giúp em với ạ |A|=|B| ⇔ A ²= B ² ⇔ A=B hoặc A=-B Giải phương trình sau: |x ²-x|+|2x-4|=3 29/08/2021 Bởi Mackenzie Giúp em với ạ |A|=|B| ⇔ A ²= B ² ⇔ A=B hoặc A=-B Giải phương trình sau: |x ²-x|+|2x-4|=3
Giải thích các bước giải: +) Nếu x>2 thì $\begin{array}{l} \left\{ \begin{array}{l} {x^2} – x > 0\\ 2x – 4 > 0 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} \left| {{x^2} – x} \right| = {x^2} – x\\ \left| {2x – 4} \right| = 2x – 4 \end{array} \right. \end{array}$ Khi đó pt trở thành: $\begin{array}{l} {x^2} – x + 2x – 4 = 3\\ \Leftrightarrow {x^2} + x – 7 = 0\\ \Leftrightarrow {(x + \frac{1}{2})^2} = \frac{{29}}{4}\\ \Leftrightarrow \left[ \begin{array}{l} x + \frac{1}{2} = \frac{{\sqrt {29} }}{2}\\ x + \frac{1}{2} = \frac{{ – \sqrt {29} }}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{{\sqrt {29} – 1}}{2}\\ x = \frac{{ – \sqrt {29} – 1}}{2} \end{array} \right.\\ \Leftrightarrow x = \frac{{\sqrt {29} – 1}}{2} \end{array}$ +) Nếu 1≤x≤2 thì $\begin{array}{l} \left\{ \begin{array}{l} {x^2} – x ≥ 0\\ 2x – 4 ≤ 0 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} \left| {{x^2} – x} \right| = {x^2} – x\\ \left| {2x – 4} \right| = -2x + 4 \end{array} \right. \end{array}$ Khi đó pt trở thành: $\begin{array}{l} {x^2} – x – 2x + 4 = 3\\ \Leftrightarrow {x^2} – 3x + 1 = 0\\ \Leftrightarrow {(x – \frac{3}{2})^2} = \frac{5}{4}\\ \Leftrightarrow \left[ \begin{array}{l} x – \frac{3}{2} = \frac{{\sqrt 5 }}{2}\\ x – \frac{3}{2} = \frac{{ – \sqrt 5 }}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{{\sqrt 5 + 3}}{2}\\ x = \frac{{ – \sqrt 5 + 3}}{2} \end{array} \right. \end{array}$ (không thoả mãn) +) Nếu 0≤x≤1 thì $\begin{array}{l} \left\{ \begin{array}{l} {x^2} – x ≤ 0\\ 2x – 4 ≤ 0 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} \left| {{x^2} – x} \right| = -{x^2} + x\\ \left| {2x – 4} \right| = -2x + 4 \end{array} \right. \end{array}$ Khi đó pt trở thành: $\begin{array}{l} – {x^2} + x – 2x + 4 = 3\\ \Leftrightarrow x + x – 1 = 0\\ \Leftrightarrow {(x + \frac{1}{2})^2} = \frac{5}{4}\\ \Leftrightarrow \left[ \begin{array}{l} x + \frac{1}{2} = \frac{{\sqrt 5 }}{2}\\ x + \frac{1}{2} = \frac{{ – \sqrt 5 }}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{{\sqrt 5 – 1}}{2}\\ x = \frac{{ – \sqrt 5 – 1}}{2} \end{array} \right.\\ \Leftrightarrow x = \frac{{\sqrt 5 – 1}}{2} \end{array}$ +) Nếu x<0 thì $\begin{array}{l} \left\{ \begin{array}{l} {x^2} – x > 0\\ 2x – 4 < 0 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} \left| {{x^2} – x} \right| = {x^2} – x\\ \left| {2x – 4} \right| = 2x – 4 \end{array} \right. \end{array}$ Khi đó pt trở thành: $\begin{array}{l} {x^2} – x – 2x + 4 = 3\\ \Leftrightarrow x – 3x + 1 = 0\\ \Leftrightarrow {(x – \frac{3}{2})^2} = \frac{5}{4}\\ \Leftrightarrow \left[ \begin{array}{l} x – \frac{3}{2} = \frac{{\sqrt 5 }}{2}\\ x – \frac{3}{2} = \frac{{ – \sqrt 5 }}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{{\sqrt 5 + 3}}{2}\\ x = \frac{{ – \sqrt 5 + 3}}{2} \end{array} \right. \end{array}$ (không thoả mãn) Bình luận
Giải thích các bước giải:
+) Nếu x>2 thì
$\begin{array}{l} \left\{ \begin{array}{l} {x^2} – x > 0\\ 2x – 4 > 0 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} \left| {{x^2} – x} \right| = {x^2} – x\\ \left| {2x – 4} \right| = 2x – 4 \end{array} \right. \end{array}$
Khi đó pt trở thành:
$\begin{array}{l} {x^2} – x + 2x – 4 = 3\\ \Leftrightarrow {x^2} + x – 7 = 0\\ \Leftrightarrow {(x + \frac{1}{2})^2} = \frac{{29}}{4}\\ \Leftrightarrow \left[ \begin{array}{l} x + \frac{1}{2} = \frac{{\sqrt {29} }}{2}\\ x + \frac{1}{2} = \frac{{ – \sqrt {29} }}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{{\sqrt {29} – 1}}{2}\\ x = \frac{{ – \sqrt {29} – 1}}{2} \end{array} \right.\\ \Leftrightarrow x = \frac{{\sqrt {29} – 1}}{2} \end{array}$
+) Nếu 1≤x≤2 thì
$\begin{array}{l} \left\{ \begin{array}{l} {x^2} – x ≥ 0\\ 2x – 4 ≤ 0 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} \left| {{x^2} – x} \right| = {x^2} – x\\ \left| {2x – 4} \right| = -2x + 4 \end{array} \right. \end{array}$
Khi đó pt trở thành:
$\begin{array}{l} {x^2} – x – 2x + 4 = 3\\ \Leftrightarrow {x^2} – 3x + 1 = 0\\ \Leftrightarrow {(x – \frac{3}{2})^2} = \frac{5}{4}\\ \Leftrightarrow \left[ \begin{array}{l} x – \frac{3}{2} = \frac{{\sqrt 5 }}{2}\\ x – \frac{3}{2} = \frac{{ – \sqrt 5 }}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{{\sqrt 5 + 3}}{2}\\ x = \frac{{ – \sqrt 5 + 3}}{2} \end{array} \right. \end{array}$
(không thoả mãn)
+) Nếu 0≤x≤1 thì
$\begin{array}{l} \left\{ \begin{array}{l} {x^2} – x ≤ 0\\ 2x – 4 ≤ 0 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} \left| {{x^2} – x} \right| = -{x^2} + x\\ \left| {2x – 4} \right| = -2x + 4 \end{array} \right. \end{array}$
Khi đó pt trở thành:
$\begin{array}{l} – {x^2} + x – 2x + 4 = 3\\ \Leftrightarrow x + x – 1 = 0\\ \Leftrightarrow {(x + \frac{1}{2})^2} = \frac{5}{4}\\ \Leftrightarrow \left[ \begin{array}{l} x + \frac{1}{2} = \frac{{\sqrt 5 }}{2}\\ x + \frac{1}{2} = \frac{{ – \sqrt 5 }}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{{\sqrt 5 – 1}}{2}\\ x = \frac{{ – \sqrt 5 – 1}}{2} \end{array} \right.\\ \Leftrightarrow x = \frac{{\sqrt 5 – 1}}{2} \end{array}$
+) Nếu x<0 thì
$\begin{array}{l} \left\{ \begin{array}{l} {x^2} – x > 0\\ 2x – 4 < 0 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} \left| {{x^2} – x} \right| = {x^2} – x\\ \left| {2x – 4} \right| = 2x – 4 \end{array} \right. \end{array}$
Khi đó pt trở thành:
$\begin{array}{l} {x^2} – x – 2x + 4 = 3\\ \Leftrightarrow x – 3x + 1 = 0\\ \Leftrightarrow {(x – \frac{3}{2})^2} = \frac{5}{4}\\ \Leftrightarrow \left[ \begin{array}{l} x – \frac{3}{2} = \frac{{\sqrt 5 }}{2}\\ x – \frac{3}{2} = \frac{{ – \sqrt 5 }}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{{\sqrt 5 + 3}}{2}\\ x = \frac{{ – \sqrt 5 + 3}}{2} \end{array} \right. \end{array}$
(không thoả mãn)