Bài tập cuối chương 1

a)

\(\dfrac{2}{5}+\dfrac{3}{5}:\left(-\dfrac{3}{2}\right)+\dfrac{1}{2}\)

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

\(=\dfrac{2}{5}+\dfrac{-2}{5}+\dfrac{1}{2}\)

\(=0+\dfrac{1}{2}\)

\(=\dfrac{1}{2}\)

b)

\(2\dfrac{1}{3}+\left(-\dfrac{1}{3}\right)^2-\dfrac{3}{2}\)

\(=\dfrac{7}{3}+\dfrac{1}{9}-\dfrac{3}{2}\)

\(=\dfrac{21}{9}+\dfrac{1}{9}-\dfrac{3}{2}\)

\(=\dfrac{22}{9}-\dfrac{3}{2}\)

\(=\dfrac{44}{18}-\dfrac{27}{18}\)

\(=\dfrac{17}{18}\)

c)

\(\left(\dfrac{7}{8}-0,25\right):\left(\dfrac{5}{6}-0,75\right)^2\)

\(=\left(\dfrac{7}{8}-\dfrac{1}{4}\right):\left(\dfrac{5}{6}-\dfrac{3}{4}\right)^2\)

\(=\left(\dfrac{7}{8}-\dfrac{2}{8}\right):\left(\dfrac{20}{24}-\dfrac{18}{24}\right)^2\)

\(=\dfrac{5}{8}:\left(\dfrac{1}{12}\right)^2\)

\(=\dfrac{5}{8}:\dfrac{1}{144}\)

\(=\dfrac{5}{8}.\dfrac{144}{1}\)

\(=90\)

d)

\(\left(-0,75\right)-\left[\left(-2\right)+\dfrac{3}{2}\right]:1,5+\left(-\dfrac{5}{4}\right)\)

\(=\dfrac{-3}{4}-\left[\left(-2\right)+\dfrac{3}{2}\right]:\dfrac{3}{2}+\dfrac{-5}{4}\)

\(=\dfrac{-3}{4}-\dfrac{-1}{2}:\dfrac{3}{2}+\dfrac{-5}{4}\)

\(=\dfrac{-3}{4}-\dfrac{-1}{2}.\dfrac{2}{3}+\dfrac{-5}{4}\)

\(=\dfrac{-3}{4}-\dfrac{-1}{3}+\dfrac{-5}{4}\)

\(=\dfrac{-9}{12}-\dfrac{-4}{12}+\dfrac{-5}{4}\)

\(=\dfrac{-5}{12}+\dfrac{-5}{4}\)

\(=\dfrac{-5}{12}+\dfrac{-15}{12}\)

\(=\dfrac{-5}{3}\)

a)

\(\dfrac{5}{23}+\dfrac{7}{17}+0,25-\dfrac{5}{23}+\dfrac{10}{17}\)

\(=\dfrac{5}{23}+\dfrac{7}{17}+\dfrac{1}{4}-\dfrac{5}{23}+\dfrac{10}{17}\)

\(=\left(\dfrac{5}{23}-\dfrac{5}{23}\right)+\left(\dfrac{7}{17}+\dfrac{10}{17}\right)+\dfrac{1}{4}\)

\(=0+1+\dfrac{1}{4}\)

\(=1+\dfrac{1}{4}\)

\(=\dfrac{5}{4}\)

b)

\(\dfrac{3}{7}.2\dfrac{2}{3}-\dfrac{3}{7}.1\dfrac{1}{2}\)

\(=\dfrac{3}{7}.\dfrac{8}{3}-\dfrac{3}{7}.\dfrac{3}{2}\)

\(=\dfrac{3}{7}.\left(\dfrac{8}{3}-\dfrac{3}{2}\right)\)

\(=\dfrac{3}{7}.\left(\dfrac{16}{6}-\dfrac{9}{6}\right)\)

\(=\dfrac{3}{7}.\dfrac{7}{6}\)

\(=\dfrac{1}{2}\)

c)

\(13\dfrac{1}{4}:\left(-\dfrac{4}{7}\right)-17\dfrac{1}{4}:\left(-\dfrac{4}{7}\right)\)

\(=\dfrac{53}{4}:\left(-\dfrac{4}{7}\right)-\dfrac{69}{4}:\left(-\dfrac{4}{7}\right)\)

\(=\dfrac{53}{4}.\dfrac{-7}{4}-\dfrac{69}{4}.\dfrac{-7}{4}\)

\(=\dfrac{-7}{4}.\left(\dfrac{53}{4}-\dfrac{69}{4}\right)\)

\(=\dfrac{-7}{4}.\left(-4\right)\)

\(=7\)

d)

\(\dfrac{100}{123}:\left(\dfrac{3}{4}+\dfrac{7}{12}\right)+\dfrac{23}{123}:\left(\dfrac{9}{5}-\dfrac{7}{15}\right)\)

\(=\dfrac{100}{123}:\left(\dfrac{9}{12}+\dfrac{7}{12}\right)+\dfrac{23}{123}:\left(\dfrac{27}{15}-\dfrac{7}{15}\right)\)

\(=\dfrac{100}{123}:\dfrac{4}{3}+\dfrac{23}{123}:\dfrac{4}{3}\)

\(=\dfrac{100}{123}.\dfrac{3}{4}+\dfrac{23}{123}.\dfrac{3}{4}\)

\(=\dfrac{3}{4}.\left(\dfrac{100}{123}+\dfrac{23}{123}\right)\)

\(=\dfrac{3}{4}.1\)

\(=\dfrac{3}{4}\)

\(a.\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{5^{16}.\left(3^3\right)^7}{\left(5^3\right)^5.\left(3^2\right)^{11}}=\dfrac{5^{16}.3^{21}}{5^{15}.3^{22}}=\dfrac{5}{3}\\ b.\left(0,2\right)^2.5-\dfrac{2^{13}.7^3}{4^6.9^5}\\ =0,04.5-\dfrac{2^{13}.7^3}{\left(2^2\right)^6.\left(3^2\right)^5}\\ =0,2-\dfrac{2^{13}.343}{2^{12}.3^{10}}\\ =\dfrac{1}{5}-\dfrac{2.343}{59049}\\ =\dfrac{1}{5}-\dfrac{686}{59049}\\ =0,1883825298\)

\(c.\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\\ =\dfrac{5^6+2^2.\left(5^2\right)^3+2^3.\left(5^3\right)^2}{26.5^6}\\ =\dfrac{5^6+2^2.5^6+2^3.5^6}{26.5^6}\\ =\dfrac{5^6.\left(1+2^2+2^3\right)}{26.5^6}\\ =\dfrac{5^6.13}{26.5^6}=\dfrac{13}{26}=\dfrac{1}{2}\)

a)

\(\begin{array}{l}A = \left[ {\left( { - 0,5} \right) - \frac{3}{5}} \right]:\left( { - 3} \right) + \frac{1}{3} - \left( { - \frac{1}{6}} \right):\left( { - 2} \right)\\ = \left( {\frac{{ - 5}}{{10}} - \frac{6}{{10}}} \right).\frac{{ - 1}}{3} + \frac{1}{3} - \frac{-1}{6}.\frac{{ - 1}}{2}\\ = \frac{{ - 11}}{{10}}.\frac{{ - 1}}{3} + \frac{1}{3} - \frac{1}{12}\\ = \frac{{11}}{{30}} + \frac{1}{3} + \frac{{ - 1}}{{12}}\\ = \frac{{22}}{{60}} + \frac{{20}}{{60}} - \frac{{ 5}}{{60}}\\ = \frac{{37}}{{60}}\end{array}\)

b)

\(\begin{array}{l}B = \left( {\frac{2}{{25}} - 0,036} \right):\frac{{11}}{{50}} - \left[ {\left( {3\frac{1}{4} - 2\frac{4}{9}} \right)} \right].\frac{9}{{29}}\\ = \left( {\frac{2}{{25}} - \frac{{36}}{{1000}}} \right).\frac{{50}}{{11}} - \left[ {\left( {\frac{{13}}{4} - \frac{{22}}{9}} \right)} \right].\frac{9}{{29}}\\ = \left( {\frac{{10}}{{125}} - \frac{9}{{250}}} \right).\frac{{50}}{{11}} - \left[ {\left( {\frac{{117}}{{36}} - \frac{{88}}{{36}}} \right)} \right].\frac{9}{{29}}\\ = \frac{{ 11}}{{250}}.\frac{{50}}{{11}} - \frac{{29}}{{36}}.\frac{9}{{29}}\\ = \frac{{ 1}}{{5}} - \frac{1}{4}\\ = \frac{{ 4}}{{20}} - \frac{{5}}{{20}}\\ = \frac{{ - 1}}{{20}}\end{array}\)

a)

\(-\dfrac{3}{5}.x=\dfrac{12}{25}\)

\(x=\dfrac{12}{25}:\left(-\dfrac{3}{5}\right)\)

\(x=\dfrac{12}{25}.\dfrac{-5}{3}\)

\(x=\dfrac{-4}{5}\)

b)

\(\dfrac{3}{5}x-\dfrac{3}{4}=-1\dfrac{1}{2}\)

\(\dfrac{3}{5}x-\dfrac{3}{4}=\dfrac{-3}{2}\)

\(\dfrac{3}{5}x=\dfrac{-3}{2}+\dfrac{3}{4}\)

\(\dfrac{3}{5}x=\dfrac{-6}{4}+\dfrac{3}{4}\)

\(\dfrac{3}{5}x=\dfrac{-3}{4}\)

\(x=\dfrac{-3}{4}:\dfrac{3}{5}\)

\(x=\dfrac{-3}{4}.\dfrac{5}{3}\)

\(x=\dfrac{-5}{4}\)

c)

\(\dfrac{2}{5}+\dfrac{3}{5}:x=0,5\)

\(\dfrac{2}{5}+\dfrac{3}{5}:x=\dfrac{1}{2}\)

\(\dfrac{3}{5}:x=\dfrac{1}{2}-\dfrac{2}{5}\)

\(\dfrac{3}{5}:x=\dfrac{5}{10}-\dfrac{4}{10}\)

\(\dfrac{3}{5}:x=\dfrac{1}{10}\)

\(x=\dfrac{3}{5}:\dfrac{1}{10}\)

\(x=\dfrac{3}{5}.\dfrac{10}{1}\)

\(x=6\)

d)

\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)=1\dfrac{2}{3}\)

\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)=\dfrac{5}{3}\)

\(x-\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{5}{3}\)

\(x-\dfrac{1}{2}=\dfrac{9}{12}-\dfrac{20}{12}\)

\(x-\dfrac{1}{2}=\dfrac{-11}{12}\)

\(x=\dfrac{-11}{12}+\dfrac{1}{2}\)

\(x=\dfrac{-11}{12}+\dfrac{6}{12}\)

\(x=\dfrac{-5}{12}\)

e)

\(2\dfrac{2}{15}:\left(\dfrac{1}{3}-5x\right)=-2\dfrac{2}{5}\)

\(\dfrac{32}{15}:\left(\dfrac{1}{3}-5x\right)=\dfrac{-12}{5}\)

\(\dfrac{1}{3}-5x=\dfrac{32}{15}:\dfrac{-12}{5}\)

\(\dfrac{1}{3}-5x=\dfrac{32}{15}.\dfrac{-5}{12}\)

\(\dfrac{1}{3}-5x=\dfrac{-8}{9}\)

\(5x=\dfrac{1}{3}-\dfrac{-8}{9}\)

\(5x=\dfrac{3}{9}+\dfrac{8}{9}\)

\(5x=\dfrac{11}{9}\)

\(x=\dfrac{11}{9}:5\)

\(x=\dfrac{11}{45}\)

g)

\(x^2+\dfrac{1}{9}=\dfrac{5}{3}:3\)

\(x^2+\dfrac{1}{9}=\dfrac{5}{9}\)

\(x^2=\dfrac{5}{9}-\dfrac{1}{9}\)

\(x^2=\dfrac{4}{9}\)

\(x^2=\left(\dfrac{2}{3}\right)^2\) hoặc \(x^2=\left(-\dfrac{2}{3}\right)^2\)

Nên \(\Rightarrow x=\dfrac{2}{3}\) hoặc \(x=-\dfrac{2}{3}\)

a) Diện tích hình thang ABCD là :

\(\dfrac{\left(\dfrac{11}{3}+\dfrac{17}{2}\right)\times3}{2}=\dfrac{73}{4}\left(m^2\right)\)

b) Theo đầu bài:

\(\Rightarrow\) Diện tích hình thoi MNPQ bằng diện tích hình thang ABCD ở câu a nên diện tích hình thoi là \(\dfrac{73}{4}m^2\)

Độ dài NQ là:

\(2\times\dfrac{73}{4}:\dfrac{35}{4}=\dfrac{146}{35}\left(m\right)\)

Đổi \(-3\dfrac{3}{4}=-\dfrac{15}{4}\)

Số hữu tỉ a là:

\(\left(-\dfrac{15}{4}.-\dfrac{1}{4}-\dfrac{3}{4}\right):\dfrac{1}{2}=\dfrac{3}{8}\)

\(a,35,6^oF\)=\(\dfrac{5}{9}.\left(35,6-32\right)=2^oC\)

\(22,64^oF=\dfrac{5}{9}.\left(22,64-32\right)=-5,2^oC\)

Độ chênh lệch nhiệt độ từ 5 giờ chiều đến 10 giờ tối là:

\(2-\left(-5,2\right)=7,2\left(^oC\right)\)

Vậy ...

Số tiền lãi mẹ của Minh nhận đc :

\(321600000-300000000=21600000\left(đồng\right)\)

Lãi xuất là:

\(21600000:300000000\times100\%=7,2\%\)

Món hàng thứ nhất sau khi giảm có giá là:

\(125\,\,000.\left( {100 - 30} \right):100 = 87\,\,500\)(đồng)

Món hàng thứ hai sau khi giảm có giá là:

\(300\,000.\left( {100 - 15} \right):100 = 255\,\,000\)(đồng)

Giá tiền món hàng thứ ba khi đã giảm là:

692 500 – 87 500 – 255 000 = 350 000 (đồng)

Giá tiền món hàng thứ ba khi chưa giảm là:

\(350\,\,000:(100-40).100 ≈ 583333,333...\) (đồng)

Ta làm tròn số tiền lên cho hợp với đơn vị tiền tệ.

Vậy số tiền mà bác Thu cần trả cho món hàng thứ ba khoảng 584 000 (đồng).