Thực hiện phép tính a. \(A = \left[ { – \sqrt {2,25} + 4\sqrt {{{\left( { – 2,15} \right)}^2}} – {{\left( {3\sqrt {\frac{7}{6}} } \right)}^2}} \right].\sqrt {1\frac{9}{{16}}} \) b. \(H = \frac{5}{3} – 0,6\sqrt {\frac{{50}}{8}} + \frac{3}{2}{\left( {\frac{{\sqrt 5 }}{3}} \right)^2}\) c. \(A = \left[ {\sqrt {64} + 2.\sqrt {{{\left( { – 3} \right)}^2}} – 7.\sqrt {1,69} + 3.\sqrt {\frac{{25}}{{16}}} } \right]:{\left( {5\sqrt {\frac{2}{3}} } \right)^2}\) d. \(A = 1,68 + \left[ {\frac{4}{5} – 1,2\left( {\frac{5}{2} – 1\frac{3}{4}} \right)} \right]:\left[ {{{\left( {\frac{2}{3}} \right)}^2} + \frac{1}{9}} \right]\)

Thực hiện phép tính

a. \(A = \left[ { – \sqrt {2,25} + 4\sqrt {{{\left( { – 2,15} \right)}^2}} – {{\left( {3\sqrt {\frac{7}{6}} } \right)}^2}} \right].\sqrt {1\frac{9}{{16}}} \)

b. \(H = \frac{5}{3} – 0,6\sqrt {\frac{{50}}{8}} + \frac{3}{2}{\left( {\frac{{\sqrt 5 }}{3}} \right)^2}\)

c. \(A = \left[ {\sqrt {64} + 2.\sqrt {{{\left( { – 3} \right)}^2}} – 7.\sqrt {1,69} + 3.\sqrt {\frac{{25}}{{16}}} } \right]:{\left( {5\sqrt {\frac{2}{3}} } \right)^2}\)

d. \(A = 1,68 + \left[ {\frac{4}{5} – 1,2\left( {\frac{5}{2} – 1\frac{3}{4}} \right)} \right]:\left[ {{{\left( {\frac{2}{3}} \right)}^2} + \frac{1}{9}} \right]\)

A. a) \( – 4,25\)

b) \(1\)

c) \(\frac{{519}}{{1000}}\)

d) \(\frac{3}{2}\)

B. a) \( – 4,05\)

b) \(2\)

c) \(\frac{{195}}{{1000}}\)

d) \(\frac{2}{3\)

C. a) \( – 5,25\)

b) \(3\)

c) \(\frac{{915}}{{999}}\)

d) \(\frac{4}{5}\)

D. a) \( – 4,85\)

b) \(4\)

c) \(\frac{{196}}{{1000}}\)

d) \(\frac{3}{4}\)

Hướng dẫn

Chọn đáp án là: A

Phương pháp giải:

+) Ta tính giá trị của biểu thức dưới dấu căn

+) Sau đó thực hiện phép tính theo thức tự thực hiện: nhân chia trước, cộng trừ sau; trong ngoặc trước và ngoài ngoặc sau.

\(\begin{array}{l}{\rm{a)}}A = \left[ { – \sqrt {2,25} + 4\sqrt {{{\left( { – 2,15} \right)}^2}} – {{\left( {3\sqrt {\frac{7}{6}} } \right)}^2}} \right].\sqrt {1\frac{9}{{16}}} \\A = \left[ { – 1,5 + 4.2,15 – 9.\frac{7}{6}} \right].\sqrt {\frac{{25}}{{16}}} \\A = \left[ { – 1,5 + 8,6 – \frac{{21}}{2}} \right].\frac{5}{4}\\A = \left[ {7,1 – 10,5} \right].1,25\\A = – 3,4.1,25\\A = – 4,25\end{array}\)

\(\begin{array}{l}{\rm{b)}}H = \frac{5}{3} – 0,6\sqrt {\frac{{50}}{8}} + \frac{3}{2}{\left( {\frac{{\sqrt 5 }}{3}} \right)^2}\\H = \frac{5}{3} – \frac{6}{{10}}.\sqrt {\frac{{25}}{4}} + \frac{3}{2}.\frac{{{{\left( {\sqrt 5 } \right)}^2}}}{{{3^2}}}\\H = \frac{5}{3} – \frac{3}{5}.\sqrt {\frac{{{5^2}}}{{{2^2}}}} + \frac{3}{2}.\frac{5}{9}\\H = \frac{5}{3} – \frac{3}{5}.\sqrt {{{\left( {\frac{5}{2}} \right)}^2}} + \frac{5}{6}\\H = \frac{5}{3} – \frac{3}{5}.\frac{5}{2} + \frac{5}{6} = \frac{5}{3} – \frac{3}{2} + \frac{5}{6}\\H = \frac{{10 – 9 + 5}}{6} = \frac{6}{6} = 1\end{array}\)

\(\begin{array}{l}{\rm{c)}}A = \left[ {\sqrt {64} + 2.\sqrt {{{\left( { – 3} \right)}^2}} – 7.\sqrt {1,69} + 3.\sqrt {\frac{{25}}{{16}}} } \right]:{\left( {5\sqrt {\frac{2}{3}} } \right)^2}\\A = \left[ {\sqrt {{8^2}} + 2.3 – 7.\sqrt {{{(1,3)}^2}} + 3.\sqrt {{{\left( {\frac{5}{4}} \right)}^2}} } \right]:\left( {25.\frac{2}{3}} \right)\\A = \left[ {8 + 6 – 7.1,3 + 3.\frac{5}{4}} \right]:\frac{{50}}{3}\\A = \left[ {8 + 6 – 9,1 + 3,75} \right]:\frac{{50}}{3}\\A = 8,65:\frac{{50}}{3} = \frac{{173}}{{20}}.\frac{3}{{50}} = \frac{{519}}{{1000}}\end{array}\)

\(\begin{array}{l}{\rm{d)}}A = 1,68 + \left[ {\frac{4}{5} – 1,2\left( {\frac{5}{2} – 1\frac{3}{4}} \right)} \right]:\left[ {{{\left( {\frac{2}{3}} \right)}^2} + \frac{1}{9}} \right]\\A = \frac{{42}}{{25}} + \left[ {\frac{4}{5} – \frac{6}{5}\left( {\frac{5}{2} – \frac{7}{4}} \right)} \right]:\left[ {\frac{4}{9} + \frac{1}{9}} \right]\\A = \frac{{42}}{{25}} + \left[ {\frac{4}{5} – \frac{6}{5}.\frac{3}{4}} \right]:\frac{5}{9}\\A = \frac{{42}}{{25}} + \left[ {\frac{4}{5} – \frac{9}{{10}}} \right]:\frac{5}{9}\\A = \frac{{42}}{{25}} + \frac{{ – 1}}{{10}}:\frac{5}{9} = \frac{{42}}{{25}} + \frac{{ – 9}}{{50}}\\A = \frac{{84}}{{50}} + \frac{{ – 9}}{{50}} = \frac{{75}}{{50}} = \frac{3}{2}\end{array}\)

Chọn A