by Giải các bài toán sau: Tìm x:
a) \(x – \frac{1}{2} = \frac{3}{2}\)
b) \(\left| {x + \frac{3}{4}} \right| – \frac{1}{2} = \sqrt {\frac{9}{4}} \)
c) \(4\frac{2}{3}.x – {\left( {\frac{{ – 2}}{3}} \right)^2} = {\left( {2,3 – 5,7} \right)^0}\)
A. \(a)x = 2;\,\,b)\,\left[ \begin{array}{l}x = \frac{5}{4}\\x = \frac{{ – 11}}{4}\end{array} \right.;\,\,\,c)\,x = \frac{{15}}{{43}}\,\)
B. \(a)x = 2;\,\,b)\,\left[ \begin{array}{l}x = \frac{5}{4}\\x = \frac{{ – 11}}{4}\end{array} \right.;\,\,\,c)\,x = \frac{{13}}{{43}}\,\)
C. \(a)x = 3;\,\,b)\,\left[ \begin{array}{l}x = \frac{5}{4}\\x = \frac{{ – 11}}{4}\end{array} \right.;\,\,\,c)\,x = \frac{{13}}{{43}}\,\)
D. \(a)x = 2;\,\,b)\,\left[ \begin{array}{l}x = \frac{7}{4}\\x = \frac{{ – 11}}{4}\end{array} \right.;\,\,\,c)\,x = \frac{{13}}{{43}}\,\)
Hướng dẫn
Chọn đáp án là: B
Phương pháp giải:
a) Chuyển vế đổi dấu, quy đồng.
b) Áp dụng \(\left| A \right| = \left\{ \begin{array}{l}A\,\,khi\,\,A \ge 0\\ – A\,\,khi\,\,A < 0\end{array} \right.\)
c) Áp dụng \({\left( {\frac{a}{b}} \right)^n} = \frac{{{a^n}}}{{{b^n}}}\,\,;\,\,{a^0} = 1\)
a) \(x – \frac{1}{2} = \frac{3}{2} \Leftrightarrow x = \frac{3}{2} + \frac{1}{2} = 2\)
b) \(\left| {x + \frac{3}{4}} \right| – \frac{1}{2} = \sqrt {\frac{9}{4}} \Leftrightarrow \left| {x + \frac{3}{4}} \right| = \frac{3}{2} + \frac{1}{2} = 2\) (1)
TH1: \(x + \frac{3}{4} \ge 0 \Leftrightarrow x \ge – \frac{3}{4}\)
(1) \( \Rightarrow x + \frac{3}{4} = 2 \Leftrightarrow x = 2 – \frac{3}{4} = \frac{5}{4}\,\,(N)\)
TH2: \(x + \frac{3}{4} < 0 \Leftrightarrow x < – \frac{3}{4}\)
(1) \( \Rightarrow – x – \frac{3}{4} = 2 \Leftrightarrow x = – 2 – \frac{3}{4} = – \frac{{11}}{4}\,\,(N)\)
Vậy \(x = \frac{5}{4}\) hoặc \(x = – \frac{{11}}{4}\)
c) \(4\frac{2}{3}.x – {\left( {\frac{{ – 2}}{3}} \right)^2} = {\left( {2,3 – 5,7} \right)^0} \Leftrightarrow \frac{{14}}{3}x – \frac{4}{9} = 1 \Leftrightarrow \frac{{14}}{3}x = 1 + \frac{4}{9} = \frac{{13}}{9}\)
\( \Leftrightarrow x = \frac{{13}}{9}.\frac{3}{{14}} = \frac{{13}}{{42}}\)
Chọn đáp án B
