Giải Bài 10 trang 39 sách bài tập toán 7 tập 1 - Cánh diều

Đề bài

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) \(2 + \dfrac{1}{6} - x = 10.\sqrt {0,01} - \sqrt {\dfrac{{25}}{{36}}} \).

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 \(x = 2\).