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 \(x = 2\).