by Tìm \(x\) , biết
a. \(\frac{2}{3} + \frac{5}{3}x = \frac{5}{7}\)
b. \(\sqrt {1,69} .\left( {2\sqrt x + \sqrt {\frac{{81}}{{121}}} } \right) = \frac{{13}}{{10}}\)
c. \(\frac{{x – 5}}{4} = \frac{{1 – 2x}}{7}\)
d. \(\left| {\frac{3}{5}\sqrt x – \frac{1}{{20}}} \right| – \frac{3}{4} = \frac{1}{5}\)
A. a) \(x = \frac{1}{{5}}\)
b) \(x = \frac{{13 }}{5}\)
c) \(x = \frac{1}{{12}}\)
d) \(x = \frac{{29}}{5}\)
B. a) \(x = \frac{1}{{53}}\)
b) \(x = \frac{{5 }}{13}\)
c) \(x = \frac{1}{{21}}\)
d) \(x = \frac{{2}}{59}\)
C. a) \(x = \frac{1}{{35}}\)
b) \(x = \frac{{13 }}{5}\)
c) \(x = \frac{1}{{121}}\)
d) \(x = \frac{{25}}{9}\)
D. a) \(x = \frac{1}{{3}}\)
b) \(x = \frac{{35 }}{5}\)
c) \(x = \frac{12}{{121}}\)
d) \(x = \frac{{5}}{9}\)
Hướng dẫn
Chọn đáp án là: C
Phương pháp giải:
Ta áp dụng thứ tự thực hiện phép tính để tìm \(x\). Đối với bài toán tìm \(x\) có chứa dấu giá trị tuyệt đối ta áp dụng quy tắc phá dấu giá trị tuyệt đối: \(\left| x \right| = \left\{ \begin{array}{l}
x\,\,\,\,\,\,khi\,\,x \ge 0\\
– x\,\,\,\,khi\,\,\,x < 0
\end{array} \right.\( sau đó tìm \(x\).
a. \(\frac{2}{3} + \frac{5}{3}x = \frac{5}{7}\)
\(\begin{array}{l}\frac{5}{3}x = \frac{5}{7} – \frac{2}{3}\\\frac{5}{3}x = \frac{1}{{21}}\\x = \frac{1}{{21}}:\frac{5}{3}\\x = \frac{1}{{35}}\end{array}\)
Vậy \(x = \frac{1}{{35}}\)
b. \(\frac{{x – 5}}{4} = \frac{{1 – 2x}}{7}\)
\(\begin{array}{l}7.\left( {x – 5} \right) = 4.\left( {1 – 2x} \right)\\7x – 35 = 4 – 8x\\7x + 8x = 35 + 4\\15x = 39\\x = \frac{{39}}{{15}} = \frac{{13}}{5}\end{array}\)
Vậy \(x = \frac{{13 }}{5}\)
c. \(\sqrt {1,69} .\left( {2\sqrt x + \sqrt {\frac{{81}}{{121}}} } \right) = \frac{{13}}{{10}}\)
\(\begin{array}{l}\sqrt {1,69} .\left( {2\sqrt x + \sqrt {\frac{{81}}{{121}}} } \right) = \frac{{13}}{{10}}\\1,3.\left( {2\sqrt x + \frac{9}{{11}}} \right) = 1,3\\2\sqrt x + \frac{9}{{11}} = 1,3:1,3\\2\sqrt x + \frac{9}{{11}} = 1\\2\sqrt x = 1 – \frac{9}{{11}}\\2\sqrt x = \frac{2}{{11}}\\\sqrt x = \frac{2}{{11}}:2\\\sqrt x = \frac{1}{{11}}\\x = \frac{1}{{121}}\end{array}\)
Vậy \(x = \frac{1}{{121}}\)
d. \(\left| {\frac{3}{5}\sqrt x – \frac{1}{{20}}} \right| – \frac{3}{4} = \frac{1}{5}\)
\(\begin{array}{l}\left| {\frac{3}{5}\sqrt x – \frac{1}{{20}}} \right| = \frac{1}{5} + \frac{3}{4}\\\left| {\frac{3}{5}\sqrt x – \frac{1}{{20}}} \right| = \frac{{19}}{{20}}\end{array}\)
Trường hợp 1: \(\frac{3}{5}\sqrt x – \frac{1}{{20}} = \frac{{19}}{{20}}\)
\(\begin{array}{l}\frac{3}{5}\sqrt x = \frac{{19}}{{20}} + \frac{1}{{20}} = 1\\\sqrt x = 1:\frac{3}{5} = \frac{5}{3}\\x = \frac{{25}}{9}\end{array}\)
Trường hợp 2: \(\frac{3}{5}\sqrt x – \frac{1}{{20}} = \frac{{ – 19}}{{20}}\)
\(\begin{array}{l}\frac{3}{5}\sqrt x = \frac{{ – 19}}{{20}} + \frac{1}{{20}} = \frac{{ – 18}}{{20}} = – \frac{9}{{10}}\\\sqrt x = \frac{{ – 9}}{{10}}:\frac{3}{5}\end{array}\)
\(\sqrt x = – \frac{3}{2} < 0\) (vô lý)
Vậy \(x = \frac{{25}}{9}\)
Chọn C
