Chương III : Phân số

cau 1

de a dat gia tri lon nhat suy ra5a-17/4a-23 lon nhat

suy ra 4a-23 phai nho nhat khac 0 va la so nguyen duong

suy ra 4a-23=1

suy ra 4a=1+23=24

suy ra a=24 chia 4=6

vay de a nho nhat thi a=6

đề bài đúng đấy bạn,hông sai đâu,có cần mk giải lun cho hông???

\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)

\(\Rightarrow\dfrac{1}{3}\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)

\(\Rightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

\(\Rightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

\(\Rightarrow\dfrac{x+3-5}{5\left(x+3\right)}=\dfrac{303}{1540}\)

\(\Rightarrow\dfrac{x-2}{5\left(x+3\right)}=\dfrac{303}{1540}\)

\(\Rightarrow1540\left(x-2\right)=303.5\left(x+3\right)\)

\(\Rightarrow1540x-3080=1515x+4545\)

\(\Rightarrow25x=7625\)

\(\Rightarrow x=305\)

Vậy x = 305

\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+.....+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{1}{3}\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+.....+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\\ \Rightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+....+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\\ \Rightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\\ \Rightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{1540}:\dfrac{1}{3}\\ \Rightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\\ \Rightarrow\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\\ \Rightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\\ \Rightarrow x+3=308\\ \Rightarrow x+=308-3=5\\ \\ \)

\(E=\left[\dfrac{1919}{2121}-\dfrac{54}{99}+\left(\dfrac{2456}{3799}\right)^5\right].\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

E \(=\left[\dfrac{1919}{2121}-\dfrac{54}{99}+\left(\dfrac{2456}{3799}\right)^5\right].0\)

E = 0

Vậy E = 0

=0

\(E=\left[\dfrac{1919}{2121}-\dfrac{54}{99}+\left(\dfrac{2456}{3799}\right)^5\right].\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(\Rightarrow E=\left[\dfrac{1919}{2121}-\dfrac{54}{99}+\left(\dfrac{2456}{3799}\right)^5\right].0\)

\(\Rightarrow E=0\)

a) a)A=7/10.11+7/11.12+7/12.13+...+7/69.70

=7.1/10.11.1+...+7.1/69.70.1

=7/1.1/10.11+...+7/1.1/69.70

=7/1.(1/10.11+...+1/69.70)

=7/1.(1/10-1/11+...+1/69-1/70)

=7/1.(1/10-1/70)

=7/1.(70/70-1/70)

=7/1.69/70

=69/10

bài b làm tương tự như vậy đó

Theo đề, ta có:

a : b = \(\dfrac{3}{2}\)

a = \(\dfrac{3}{2}\)b (1)

a - b = 8 (2)

Thế (1) vào (2)

\(\dfrac{3}{2}\)b - b = 8

\(\dfrac{1}{2}\)b = 8

b = 8 :\(\dfrac{1}{2}\)

b = 16

a = 24

Theo đề bài ta có:

a : b =\(\dfrac{3}{2}\) và a - b = 8

=> a= \(\dfrac{3}{2}\) .b

Vs a = \(\dfrac{3}{2}\).b

=> \(\dfrac{3}{2}\) .b - b.1 = 8

=> b.(\(\dfrac{3}{2}\) - 1) =8

=> b. \(\dfrac{1}{2}\) = 8

=> b = 8 : \(\dfrac{1}{2}\)

=> b = 8 . \(\dfrac{2}{1}\)

=> b = 16

Vs b =16

=> a = \(\dfrac{3}{2}\). 16

=> a = \(\dfrac{3.16}{2}\)

=> a = 24

Vậy a = 24

b = 16

Help me!!!

Bài này giải ra dài lắm;

Gợi ý : với câu a) cm 1<A<2

với câ u b) 0<B<1

với câu c) áp dụng bài toán của ông gao í; cách tỉnh tổng từ 1->100 trong sách GK 6 có nhé

Mong bạn giải ra

\(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)

\(=\dfrac{-1}{4.5}+\dfrac{-1}{5.6}+\dfrac{-1}{6.7}+\dfrac{-1}{7.8}+\dfrac{-1}{8.9}+\dfrac{-1}{9.10}\)

\(=\dfrac{-1}{4}-\dfrac{-1}{5}+\dfrac{-1}{5}-\dfrac{-1}{6}+\dfrac{-1}{6}-\dfrac{-1}{7}+\dfrac{-1}{7}-\dfrac{-1}{8}+\dfrac{-1}{8}-\dfrac{-1}{9}+\dfrac{-1}{9}-\dfrac{-1}{10}\)

\(=\dfrac{-1}{4}-\dfrac{-1}{10}\)

\(=\dfrac{-3}{20}\)

\(\dfrac{1}{2}\cdot\left(\dfrac{2}{9}+\dfrac{3}{7}-\dfrac{5}{27}\right)\\ =\dfrac{1}{2}\left(\dfrac{3}{7}+\dfrac{1}{27}\right)\\ =\dfrac{1}{2}\cdot\dfrac{88}{189}\\ =\dfrac{44}{189}\)