Giải:

Đặt \(A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}\)

Ta có:

\(A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}\)

\(\Rightarrow A=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)

Nhận xét:

\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}< \dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}=\dfrac{1}{3}\)

\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}< \dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{1}{4}\)

\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{1}{5}\)

\(\Rightarrow A< \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}< \dfrac{48}{60}=\dfrac{4}{5}\)

\(\Rightarrow A< \dfrac{4}{5}\left(1\right)\)

Lại có:

\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}>\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{1}{4}\)

\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}>\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{1}{5}\)

\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{1}{6}\)

\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{37}{60}>\dfrac{36}{60}=\dfrac{3}{5}\)

\(\Rightarrow A>\dfrac{3}{5}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\)

\(\Rightarrow\dfrac{3}{5}< A< \dfrac{4}{5}\)

Vậy \(\dfrac{3}{5}< \dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}< \dfrac{4}{5}\) (Đpcm)

Đặt

Ta có:

Nhận xét:

Lại có:

Từ và

Vậy

bài 1 :

\(a,\dfrac{2}{7}+\dfrac{1}{3}=\dfrac{6}{21}+\dfrac{7}{21}=\dfrac{13}{21}\)

\(b,\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{9}{15}-\dfrac{5}{15}=\dfrac{4}{15}\)

\(c,\dfrac{13}{4}:5=\dfrac{13}{4}:\dfrac{5}{1}=\dfrac{13}{4}x\dfrac{1}{5}=\dfrac{13}{20}\)

\(d,\dfrac{6}{23}x\dfrac{1}{18}=\dfrac{1}{69}\)

bài 2 :

\(a,x+\dfrac{1}{3}=\dfrac{5}{12}\)

   \(x=\dfrac{5}{12}-\dfrac{1}{3}\)

  \(x=\dfrac{1}{12}\)

\(b,x:\dfrac{7}{4}=\dfrac{2}{5}\)

   \(x=\dfrac{2}{5}x\dfrac{7}{4}\)

   \(x=\dfrac{7}{10}\)

bài 3 :

đổi : 2 dm 1cm = 21cm

chiều cao hình bình hành là;

       21 x\(\dfrac{3}{7}=\)9(cm)

diện tích hình bình hành là;

       21 x 9 =189 (cm²)

                 đáp số : 189 cm²

bài 4 :

\(\dfrac{2}{3}x\dfrac{2}{10}+\dfrac{2}{3}x\dfrac{5}{10}x\dfrac{3}{3}\)

\(\dfrac{2}{3}x\left(\dfrac{2}{10}+\dfrac{5}{10}\right)x\dfrac{2}{3}\)

=\(\dfrac{2}{3}x1x\dfrac{2}{3}\)

\(=\dfrac{2}{3}x\dfrac{2}{3}\)

=\(\dfrac{4}{9}\)

Bài 1)

a) \(\dfrac{6}{21}+\dfrac{7}{21}=\dfrac{13}{21}\)

b) \(\dfrac{9}{15}-\dfrac{5}{15}=\dfrac{4}{15}\)

c) \(\dfrac{13}{4}x\dfrac{1}{5}=\dfrac{13}{20}\)

d) \(\dfrac{6}{414}=\dfrac{1}{69}\)

Bài 2)

a) \(x=\dfrac{5}{12}-\dfrac{1}{3}\)

\(x=\dfrac{1}{12}\)

b) \(x=\dfrac{2}{5}x\dfrac{7}{4}\)

\(x=\dfrac{7}{10}\)

Bài 3)

2dm 1cm = 21 cm

Chiều cao tấm bìa la

\(21x\dfrac{3}{7}=9\left(cm\right)\)

Diện tích tấm bìa là

\(21x9=189\left(cm2\right)\)

Bài 4)

\(\dfrac{2}{3}x\dfrac{2}{10}+\dfrac{2}{3}x\dfrac{5}{10}x\dfrac{2}{3}=\dfrac{2}{3}x\left(\dfrac{2}{10}+\dfrac{5}{10}\right)x\dfrac{2}{3}=\dfrac{2}{3}x\dfrac{7}{10}x\dfrac{2}{3}=\dfrac{14}{45}\)

a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)

b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)

c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)

a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)

\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)

\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)

b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)

\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)

\(=\dfrac{-5}{3}+\dfrac{42}{29}\)

\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)

c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)

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

\(=-1-1-1+4\)

=1

a) Ta có: =−2590+6490−8190=−2590+6490−8190

(−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)

=−53+4229=−53+4229

1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1

bài này có trong sách Nâng cao và Phát triển bạn nhé

a \(\dfrac{6}{15}+\dfrac{11}{15}=\dfrac{17}{15}\)

b \(\dfrac{22}{77}-\dfrac{14}{77}=\dfrac{8}{77}\)

c \(\dfrac{11}{13}\times\dfrac{26}{31}=11\times\dfrac{2}{31}=\dfrac{22}{31}\)

d \(\dfrac{1}{2}\times3\times\dfrac{2}{5}=\dfrac{3}{5}\)

a, 2/5 + 11/15 = 6/15 + 11/15 = 17/15

b,2/7 - 2/11 = 22/77 - 14/77 = 8/77

c, 11/13 x 26/31= 11 x 2/31 =22/31

d, 1/2 : 1/3 x 2/5

=  ( 1/2 x 2/5 ) : 1/3

=     1/5        : 1/3

=         3/5