Tìm x biết:
a) \(x + 2.\sqrt {16} = – 3.\sqrt {49} \);
b) \(2x – \sqrt {1,69} = \sqrt {1,21} \);
c) \(5.\left( {\sqrt {\dfrac{1}{{25}}} – x} \right) – \sqrt {\dfrac{1}{{81}}} = – \dfrac{1}{9}\);
d) \(2 + \dfrac{1}{6} – x = 10.\sqrt {0,01} – \sqrt {\dfrac{{25}}{{36}}} \).
Phương pháp giải:
Tìm giá trị của x dựa vào các biểu thức đã cho.
Lời giải chi tiết:
a)
\(\begin{array}{l}x + 2.\sqrt {16} = – 3.\sqrt {49} \\{\rm{ }}x + 2.4 = – 3.7\\{\rm{ }}x + 8 = – 21\\{\rm{ }}x = – 21 – 8\\{\rm{ }}x = – 29\end{array}\)
Vậy \(x = – 29\).
b)
\(\begin{array}{l}2x – \sqrt {1,69} = \sqrt {1,21} \\{\rm{ }}2x – 1,3 = 1,1\\{\rm{ }}2x = 1,1 + 1,3\\{\rm{ }}2x = 2,4\\{\rm{ }}x = 2,4:2\\{\rm{ }}x = 1,2\end{array}\)
Vậy \(x = 1,2\).
c)
\(\begin{array}{l}5.\left( {\sqrt {\dfrac{1}{{25}}} – x} \right) – \sqrt {\dfrac{1}{{81}}} = – \dfrac{1}{9}\\{\rm{ }}5.\left( {\dfrac{1}{5} – x} \right) – \dfrac{1}{9} = – \dfrac{1}{9}\\{\rm{ }}5.\left( {\dfrac{1}{5} – x} \right) = – \dfrac{1}{9} + \dfrac{1}{9}\\{\rm{ }}5.\left( {\dfrac{1}{5} – x} \right) = 0\\{\rm{ }}\dfrac{1}{5} – x = 0:5\\{\rm{ }}\dfrac{1}{5} – x = 0\\{\rm{ }}x = \dfrac{1}{5} – 0\\{\rm{ }}x = \dfrac{1}{5}\end{array}\)
Vậy \(x = \dfrac{1}{5}\).
d)
\(\begin{array}{l}2 + \dfrac{1}{6} – x = 10.\sqrt {0,01} – \sqrt {\dfrac{{25}}{{36}}} \\{\rm{ }}\dfrac{{13}}{6} – x = 10{\rm{ }}.{\rm{ }}0,1 – \dfrac{5}{6}\\{\rm{ }}\dfrac{{13}}{6} – x = \dfrac{1}{6}\\{\rm{ }}x = \dfrac{{13}}{6} – \dfrac{1}{6}\\{\rm{ }}x = \dfrac{{12}}{6}\\{\rm{ }}x = 2\end{array}\)
Vậy \(x = 2\).