Đề cương ôn tập cuối kì 1 Toán 7 Kết nối tri thức cấu trúc mới (Tự luận) có đáp án - Phần 1

a) \(\frac{1}{9} - 0,3 \cdot \frac{5}{9} + \frac{1}{3} + 5\) 

\( = \frac{1}{9} - \frac{3}{{10}} \cdot \frac{5}{9} + \frac{1}{3} + 5\)

\( = \frac{1}{9} - \frac{1}{6} + \frac{1}{3} + 5\)

c) \(\left( {{2^2}:\frac{4}{3} - \frac{1}{2}} \right) \cdot \frac{6}{5} - 17\)

\( = \left( {4\, \cdot \,\frac{3}{4} - \frac{1}{2}} \right) \cdot \frac{6}{5} - 17\)

\( = \left( {3 - \frac{1}{2}} \right) \cdot \frac{6}{5} - 17\)

e) \(\left| {\frac{{ - 1}}{2}} \right| + {\left( {\frac{{ - 1}}{3}} \right)^2}:\left| { - 2} \right| - {\left( {\frac{{ - 2}}{3}} \right)^0}\)

\( = \frac{1}{2} + \frac{1}{9}:2 - 1\)\( = \frac{1}{2} + \frac{1}{{18}} - 1\)

f) \(\frac{{\sqrt {{3^2}} + \sqrt {{{39}^2}} }}{{\sqrt {{{91}^2}} - \sqrt {{{( - 7)}^2}} }}\)

\( = \frac{{\left| 3 \right| + \left| {39} \right|}}{{\left| {91} \right| - \left| { - 7} \right|}}\)

g) \(10\,.\,\sqrt {0,01} \, \cdot \,\sqrt {\frac{{16}}{9}} + 3\sqrt {49} - \frac{1}{6}\, \cdot \,\sqrt 4 \)

\( = 10 \cdot 0,1 \cdot \frac{4}{3} + 3.7 - \frac{1}{6}.2\)

\( = \frac{4}{3} + 21 - \frac{1}{3}\)

\( = \left( {\frac{4}{3} - \frac{1}{3}} \right) + 21\)

\[ = 1 + 21\]\[ = 22\].

h) \(0,8:\left\{ {0,2 - 7\left[ {\frac{1}{6} + \left( {\frac{5}{{21}} - \frac{5}{{14}}} \right)} \right]} \right\}\)

\( = 0,8:\left\{ {0,2 - 7\left[ {\frac{1}{6} + \left( { - \frac{5}{{42}}} \right)} \right]} \right\}\)

\[ = 0,8:\left\{ {0,2 - 7\left[ {\frac{1}{{21}}} \right]} \right\}\]

\[ = \frac{4}{5}:\left\{ {\frac{1}{5} - \frac{1}{3}} \right\}\]