Bài 1 trang 24 sách giáo khoa Toán 7 Chân trời sáng tạo tập 1
Bỏ dấu ngoặc rồi tính:
a)(left( {frac{{ – 3}}{7}} right) + left( {frac{5}{6} – frac{4}{7}} right);)
b)(frac{3}{5} – left( {frac{2}{3} + frac{1}{5}} right);)
c)(left[ {left( {frac{{ – 1}}{3} + 1} right) – left( {frac{2}{3} – frac{1}{5}} right)} right];)
d)(1frac{1}{3} + left( {frac{2}{3} – frac{3}{4}} right) – left( {0,8 + 1frac{1}{5}} right)).
Lời giải:
a)
(begin{array}{l}left( {frac{{ – 3}}{7}} right) + left( {frac{5}{6} – frac{4}{7}} right) = left( {frac{{ – 3}}{7}} right) + frac{5}{6} – frac{4}{7} = left[ {left( {frac{{ – 3}}{7}} right) – frac{4}{7}} right] + frac{5}{6} =frac{-7}{7}+frac{5}{6}= – 1 + frac{5}{6} = frac{{ – 1}}{6}end{array})
b)
(begin{array}{l}frac{3}{5} – left( {frac{2}{3} + frac{1}{5}} right) = frac{3}{5} – frac{2}{3} – frac{1}{5} = (frac{3}{5} – frac{1}{5}) – frac{2}{3} = frac{2}{5} – frac{2}{3} = frac{6}{{15}} – frac{{10}}{{15}} = frac{{ – 4}}{{15}}end{array})
c)
(begin{array}{l}left[ {left( {frac{{ – 1}}{3}} right) + 1} right] – left( {frac{2}{3} – frac{1}{5}} right) = left( {frac{{ – 1}}{3}} right) + 1 – frac{2}{3} + frac{1}{5} = left( {frac{{ – 1}}{3} – frac{2}{3}} right) + 1 + frac{1}{5} = frac{-3}{3}+1+frac{1}{5}= – 1 + 1 + frac{1}{5} = frac{1}{5}end{array})
d)
(begin{array}{l}1frac{1}{3} + left( {frac{2}{3} – frac{3}{4}} right) – left( {0,8 + 1frac{1}{5}} right) = 1 + frac{1}{3} + frac{2}{3} – frac{3}{4} – left( {frac{4}{5} + 1 + frac{1}{5}} right)=1+frac{3}{3}-frac{3}{4}-(frac{5}{5}+1) = 1 + 1 – frac{3}{4} – (1+1) = – frac{3}{4}end{array}).
Bài 2 trang 25 sách giáo khoa Toán 7 Chân trời sáng tạo tập 1
Tính:
a) (left( {frac{3}{4}:1frac{1}{2}} right) – left( {frac{5}{6}:frac{1}{3}} right))
b) (left[ {left( {frac{{ – 1}}{5}} right):frac{1}{{10}}} right] – frac{5}{7}.left( {frac{2}{3} – frac{1}{5}} right))
c) (left( { – 0,4} right) + 2frac{2}{5}.{left[ {left( {frac{{ – 2}}{3}} right) + frac{1}{2}} right]^2})
d)(left{ {left[ {{{left( {frac{1}{{25}} – 0,6} right)}^2}:frac{{49}}{{125}}} right].frac{5}{6}} right} – left[ {left( {frac{{ – 1}}{3}} right) + frac{1}{2}} right])
Lời giải:
a)
(begin{array}{l}left( {frac{3}{4}:1frac{1}{2}} right) – left( {frac{5}{6}:frac{1}{3}} right) = left( {frac{3}{4}:frac{3}{2}} right) – left( {frac{5}{6}.3} right) = left( {frac{3}{4}.frac{2}{3}} right) – frac{5}{2} = frac{1}{2} – frac{5}{2} = frac{-4}{2}= – 2.end{array})
b)
(begin{array}{l}left[ {left( {frac{{ – 1}}{5}} right):frac{1}{{10}}} right] – frac{5}{7}.left( {frac{2}{3} – frac{1}{5}} right) = left( {frac{{ – 1}}{5}} right).10 – frac{5}{7}.left( {frac{{10}}{{15}} – frac{3}{{15}}} right) = – 2 – frac{5}{7}.frac{7}{{15}} = – 2 – frac{1}{3} = frac{{ – 6}}{3} – frac{1}{3} = frac{{ – 7}}{3}end{array})
c)
(begin{array}{l}left( { – 0,4} right) + 2frac{2}{5}.{left[ {left( {frac{{ – 2}}{3}} right) + frac{1}{2}} right]^2} = left( { – frac{2}{5}} right) + frac{{12}}{5}.{left[ {left( {frac{{ – 4}}{6}} right) + frac{3}{6}} right]^2} = left( { – frac{2}{5}} right) + frac{{12}}{5}.{left( {frac{{ – 1}}{6}} right)^2} = left( { – frac{2}{5}} right) + frac{{12}}{5}.frac{1}{{36}} = left( { – frac{2}{5}} right) + frac{1}{{15}} = left( { – frac{6}{{15}}} right) + frac{1}{{15}} = frac{{ – 5}}{{15}} = frac{{ – 1}}{3}end{array})
d)
(begin{array}{l}left{ {left[ {{{left( {frac{1}{{25}} – 0,6} right)}^2}:frac{{49}}{{125}}} right].frac{5}{6}} right} – left[ {left( {frac{{ – 1}}{3}} right) + frac{1}{2}} right] = left{ {left[ {{{left( {frac{1}{{25}} – frac{3}{5}} right)}^2}.frac{{125}}{{49}}} right].frac{5}{6}} right} – left[ {left( {frac{{ – 2}}{6}} right) + frac{3}{6}} right] = left{ {left[ {{{left( {frac{{ 1}}{{25}}-frac{15}{25}} right)}^2}.frac{{125}}{{49}}} right].frac{5}{6}} right} – frac{1}{6} = left{ {left[ {{{left( {frac{{ – 14}}{{25}}} right)}^2}.frac{{125}}{{49}}} right].frac{5}{6}} right} – frac{1}{6} = left{ {frac{{196}}{{{{25}^2}}}.frac{{25.5}}{{49}}.frac{5}{6}} right} – frac{1}{6} = left( {frac{{4.49.25.5.5}}{{{{25}^2}.49.6}}} right) – frac{1}{6} = frac{4}{6} – frac{1}{6} = frac{3}{6} = frac{1}{2}end{array})
Bài 3 trang 25 sách giáo khoa Toán 7 Chân trời sáng tạo tập 1
Cho biểu thức: (A = left( {2 + frac{1}{3} – frac{2}{5}} right) – left( {7 – frac{3}{5} – frac{4}{3}} right) – left( {frac{1}{5} + frac{5}{3} – 4} right).)
Hãy tính giá trị của A theo hai cách:
a) Tính giá trị của từng biểu thức trong dấu ngoặc trước.
b) Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp.
Lời giải:
a)
(begin{array}{l}A = left( {2 + frac{1}{3} – frac{2}{5}} right) – left( {7 – frac{3}{5} – frac{4}{3}} right) – left( {frac{1}{5} + frac{5}{3} – 4} right).A = left( {frac{{30}}{{15}} + frac{5}{{15}} – frac{6}{{15}}} right) – left( {frac{{105}}{{15}} – frac{9}{{15}} – frac{{20}}{{15}}} right) – left( {frac{3}{{15}} + frac{{25}}{{15}} – frac{{60}}{{15}}} right)A = frac{{29}}{{15}} – frac{{76}}{{15}} – left( {frac{{ – 32}}{{15}}} right)A = frac{{29}}{{15}} – frac{{76}}{{15}} + frac{{32}}{{15}}A = frac{{ – 15}}{{15}}A = – 1end{array})
b)
(begin{array}{l}A = left( {2 + frac{1}{3} – frac{2}{5}} right) – left( {7 – frac{3}{5} – frac{4}{3}} right) – left( {frac{1}{5} + frac{5}{3} – 4} right)A = 2 + frac{1}{3} – frac{2}{5} – 7 + frac{3}{5} + frac{4}{3} – frac{1}{5} – frac{5}{3} + 4A = left( {2 – 7 + 4} right) + left( {frac{1}{3} + frac{4}{3} – frac{5}{3}} right) + left( { – frac{2}{5} + frac{3}{5} – frac{1}{5}} right)A = – 1 + 0 + 0 = – 1end{array})
Bài 4 trang 25 sách giáo khoa Toán 7 Chân trời sáng tạo tập 1
Tìm x, biết:
a)(x + frac{3}{5} = frac{2}{3};)
b)(frac{3}{7} – x = frac{2}{5};)
c)(frac{4}{9} – frac{2}{3}x = frac{1}{3};)
d)(frac{3}{{10}}x – 1frac{1}{2} = left( {frac{{ – 2}}{7}} right):frac{5}{{14}})
Lời giải:
a)
(begin{array}{l}x + frac{3}{5} = frac{2}{3}x = frac{2}{3} – frac{3}{5}x = frac{{10}}{{15}} – frac{9}{{15}}x = frac{1}{{15}}end{array})
Vậy (x = frac{1}{{15}}).
b)
(begin{array}{l}frac{3}{7} – x = frac{2}{5}x = frac{3}{7} – frac{2}{5}x = frac{{15}}{{35}} – frac{{14}}{{35}}x = frac{1}{{35}}end{array})
Vậy (x = frac{1}{{35}}).
c)
(begin{array}{l}frac{4}{9} – frac{2}{3}x = frac{1}{3}frac{2}{3}x = frac{4}{9} – frac{1}{3}frac{2}{3}x = frac{4}{9} – frac{3}{9}frac{2}{3}x = frac{1}{9}x = frac{1}{9}:frac{2}{3}x = frac{1}{9}.frac{3}{2}x = frac{1}{6}end{array})
Vậy (x = frac{1}{6}).
d)
(begin{array}{l}frac{3}{{10}}x – 1frac{1}{2} = left( {frac{{ – 2}}{7}} right):frac{5}{{14}}frac{3}{{10}}x – frac{3}{2} = left( {frac{{ – 2}}{7}} right).frac{{14}}{5}frac{3}{{10}}x – frac{3}{2} = frac{{ – 4}}{5}frac{3}{{10}}x = frac{{ – 4}}{5} + frac{3}{2}frac{3}{{10}}x = frac{{ – 8}}{{10}} + frac{{15}}{{10}}frac{3}{{10}}x = frac{7}{{10}}x = frac{7}{{10}}:frac{3}{{10}}x = frac{7}{3}end{array})
Vậy (x = frac{7}{3}).
Bài 5 trang 25 sách giáo khoa Toán 7 Chân trời sáng tạo tập 1
Tìm x, biết:
a)(frac{2}{9}:x + frac{5}{6} = 0,5;)
b)(frac{3}{4} – left( {x – frac{2}{3}} right) = 1frac{1}{3};)
c)(1frac{1}{4}:left( {x – frac{2}{3}} right) = 0,75;)
d)(left( { – frac{5}{6}x + frac{5}{4}} right):frac{3}{2} = frac{4}{3}).
Lời giải:
a)
(begin{array}{l}frac{2}{9}:x + frac{5}{6} = 0,5frac{2}{9}:x = frac{1}{2} – frac{5}{6}frac{2}{9}:x = frac{3}{6} – frac{5}{6}frac{2}{9}:x = frac{{ – 2}}{6}x = frac{2}{9}:frac{{ – 2}}{6}x = frac{2}{9}.frac{{ – 6}}{2}x = frac{{ – 2}}{3}end{array})
Vậy (x = frac{{ – 2}}{3}).
b)
(begin{array}{l}frac{3}{4} – left( {x – frac{2}{3}} right) = 1frac{1}{3}x – frac{2}{3} = frac{3}{4} – 1frac{1}{3}x – frac{2}{3} = frac{3}{4} – frac{4}{3}x – frac{2}{3} = frac{9}{{12}} – frac{{16}}{{12}}x – frac{2}{3} = frac{{ – 7}}{{12}}x = frac{{ – 7}}{{12}} + frac{2}{3}x = frac{{ – 7}}{{12}} + frac{8}{{12}}x = frac{1}{12}end{array})
Vậy(x = frac{1}{12}).
c)
(begin{array}{l}1frac{1}{4}:left( {x – frac{2}{3}} right) = 0,75frac{5}{4}:left( {x – frac{2}{3}} right) = frac{3}{4}x – frac{2}{3} = frac{5}{4}:frac{3}{4}x – frac{2}{3} = frac{5}{4}.frac{4}{3}x – frac{2}{3} = frac{5}{3}x = frac{5}{3} + frac{2}{3}x = frac{7}{3}end{array})
Vậy (x = frac{7}{3}).
d)
(begin{array}{l}left( { – frac{5}{6}x + frac{5}{4}} right):frac{3}{2} = frac{4}{3} – frac{5}{6}x + frac{5}{4} = frac{4}{3}.frac{3}{2} – frac{5}{6}x + frac{5}{4} = 2 – frac{5}{6}x = 2 – frac{5}{4} – frac{5}{6}x = frac{8}{4} – frac{5}{4} – frac{5}{6}x = frac{3}{4}x = frac{3}{4}:left( { – frac{5}{6}} right)x = frac{3}{4}.frac{{ – 6}}{5}x = frac{{ – 9}}{{10}}end{array})
Vậy (x = frac{{ – 9}}{{10}}).
Bài 6 trang 25 sách giáo khoa Toán 7 Chân trời sáng tạo tập 1
Tính nhanh:
a)(frac{{13}}{{23}}.frac{7}{{11}} + frac{{10}}{{23}}.frac{7}{{11}};)
b) (frac{5}{9}.frac{{23}}{{11}} – frac{1}{{11}}.frac{5}{9} + frac{5}{9})
c)(left[ {left( { – frac{4}{9}} right) + frac{3}{5}} right]:frac{{13}}{{17}} + left( {frac{2}{5} – frac{5}{9}} right):frac{{13}}{{17}};)
d) (frac{3}{{16}}:left( {frac{3}{{22}} – frac{3}{{11}}} right) + frac{3}{{16}}:left( {frac{1}{{10}} – frac{2}{5}} right))
Lời giải:
a)
(begin{array}{l}frac{{13}}{{23}}.frac{7}{{11}} + frac{{10}}{{23}}.frac{7}{{11}} = frac{7}{{11}}left( {frac{{13}}{{23}} + frac{{10}}{{23}}} right) = frac{7}{{11}}.1 = frac{7}{{11}}end{array})
b)
(begin{array}{l}frac{5}{9}.frac{{23}}{{11}} – frac{1}{{11}}.frac{5}{9} + frac{5}{9} = frac{5}{9}.left( {frac{{23}}{{11}} – frac{1}{{11}} + 1} right) = frac{5}{9}.left( {2 + 1} right) = frac{5}{9}.3 = frac{5}{3}end{array})
c)
(begin{array}{l}left[ {left( { – frac{4}{9} + frac{3}{5}} right):frac{{13}}{{17}}} right] + left( {frac{2}{5} – frac{5}{9}} right):frac{{13}}{{17}} = left( { – frac{4}{9} + frac{3}{5}} right).frac{{17}}{{13}} + left( {frac{2}{5} – frac{5}{9}} right).frac{{17}}{{13}} = frac{{17}}{{13}}.left( { – frac{4}{9} + frac{3}{5} + frac{2}{5} – frac{5}{9}} right) = frac{{17}}{{13}}.left[ {left( { – frac{4}{9} – frac{5}{9}} right) + left( {frac{3}{5} + frac{2}{5}} right)} right] = frac{{17}}{{13}}.left( { – 1 + 1} right) = frac{{17}}{{13}}.0 = 0end{array})
d)
(begin{array}{l}frac{3}{{16}}:left( {frac{3}{{22}} – frac{3}{{11}}} right) + frac{3}{{16}}:left( {frac{1}{{10}} – frac{2}{5}} right) = frac{3}{{16}}:left( {frac{3}{{22}} – frac{6}{{22}}} right) + frac{3}{{16}}:left( {frac{1}{{10}} – frac{4}{{10}}} right) = frac{3}{{16}}:frac{{ – 3}}{{22}} + frac{3}{{16}}:frac{{ – 3}}{{10}} = frac{3}{{16}}.frac{{ – 22}}{3} + frac{3}{{16}}.frac{{ – 10}}{3} = frac{3}{{16}}.left( {frac{{ – 22}}{3} + frac{{ – 10}}{3}} right) = frac{3}{{16}}.frac{{ – 32}}{3} = – 2end{array})
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