Question

 

Bài kiểm tra học kì I

Bài 1 : Thực hiện phép tính :

a, \(\frac{-5}{9}\left(\frac{3}{10}-\frac{2}{5}\right)\)

b, \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{9}{25}}+1^{2016}\)

c, \(2^8:2^5+3^2\cdot2-12\)

d, \(3^x+\sqrt{\frac{16}{81}}-\sqrt{9}+\frac{\sqrt{81}}{3}=9\frac{4}{9}\) (Tìm \(x\))

Answer 1

\(-\frac{5}{9}\left(\frac{3}{10}-\frac{2}{5}\right)=-\frac{5}{9}\left(\frac{3}{10}-\frac{4}{10}\right)=-\frac{5}{9}.\frac{-1}{10}=\frac{1}{18}\)

\(\frac{1}{2}\sqrt{64}-\sqrt{\frac{9}{25}}+1^{2016}=\frac{1}{2}.8-\frac{3}{5}+1=4+\frac{2}{5}=\frac{22}{5}\)

\(2^8:2^5+3^2.2-12=2^3+9.2-12=8+18-12=8+6=14\)

Answer 2

\(3^x+\sqrt{\frac{16}{81}}-\sqrt{9}+\frac{\sqrt{81}}{3}=9\frac{4}{9}\)

\(3^x+\frac{4}{9}-3+\frac{9}{3}=9\frac{4}{9}\)

\(3^x+\frac{4}{9}-3+3=9\frac{4}{9}\)

\(3^x+\frac{4}{9}=9+\frac{4}{9}\)

\(\Rightarrow3^x=9+\frac{4}{9}-\frac{4}{9}\)

\(3^x=9\)

\(3^x=3^2\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

Answer 3

a, \(\frac{-5}{9}\left(\frac{3}{10}-\frac{2}{5}\right)\)

\(=\frac{-5}{9}\left(\frac{3}{10}-\frac{4}{10}\right)\)

\(=\frac{-5}{9}.\frac{-1}{10}\)

\(=\frac{5}{90}\)

\(=\frac{1}{18}\)

Answer 4

b, \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{9}{25}}+1^{2016}\)

\(=\frac{1}{2}.8-\frac{3}{5}+1\)

\(=4-\frac{8}{5}\)

\(=\frac{20}{5}-\frac{8}{5}\)

\(=\frac{12}{5}\)

Answer 5

c, \(2^8:2^5+3^2.2-12\)

\(=2^3+9.2-12\)

\(=8+18-12\)

\(=26-12\)

\(=14\)

Answer 6

d, \(3^x+\sqrt{\frac{16}{81}}-\sqrt{9}+\frac{\sqrt{81}}{3}=9\frac{4}{9}\)

\(3^x+\frac{4}{9}-3+3=9\frac{4}{9}\)

\(3^x+\frac{4}{9}-3=9\frac{4}{9}-3\)

\(3^x+\frac{4}{9}=9\frac{4}{9}-3+3\)

\(3^x+\frac{4}{9}=9\frac{4}{9}\)

\(3^x=9\frac{4}{9}-\frac{4}{9}\)

\(3^x=9\)

\(3^x=3^2\)

\(\Rightarrow x=2\)