b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)

Đặt \(x=15k;y=20k;z=24k\)

Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)

lk

a, \(\dfrac{x}{7}-\dfrac{1}{2}=\dfrac{y}{y+1}\Leftrightarrow\dfrac{2x-7}{14}=\dfrac{y}{y+1}\Rightarrow\left(2x-7\right)\left(y+1\right)=14y\)

\(\Leftrightarrow2xy+2x-7y-7=14y\Leftrightarrow2xy+2x-21y-7=0\)

\(\Leftrightarrow2x\left(y+1\right)-21\left(y+1\right)+14=0\Leftrightarrow\left(2x-21\right)\left(y+1\right)=-14\)

\(\Rightarrow2x-21;y+1\inƯ\left(-14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

Bài 4:

\(\dfrac{3}{4}+y:\dfrac{2}{5}=\dfrac{37}{16}\)

\(\Rightarrow y:\dfrac{2}{5}=\dfrac{37}{16}-\dfrac{3}{4}\)

\(\Rightarrow y:\dfrac{2}{5}=\dfrac{25}{16}\)

\(\Rightarrow y=\dfrac{2}{5}\cdot\dfrac{25}{16}\)

\(\Rightarrow y=\dfrac{5}{8}\)

________________

\(456+y:87=23987\)

\(\Rightarrow y:87=23987-456\)

\(\Rightarrow y:87=23531\)

\(\Rightarrow y=23531\cdot87\)

\(\Rightarrow y=2047197\)

a)\(\dfrac{4}{5}\times\dfrac{5}{8}:\dfrac{4}{5}\)

\(=\left(\dfrac{4}{5}:\dfrac{4}{5}\right)\times\dfrac{5}{8}\)

\(=1\times\dfrac{5}{8}=\dfrac{5}{8}\)

b)\(\dfrac{5}{6}+\left(\dfrac{1}{2}:\dfrac{3}{2}+\dfrac{4}{5}\right)\)

\(=\dfrac{5}{6}+\left(\dfrac{1}{3}+\dfrac{4}{5}\right)\)

\(=\dfrac{5}{6}+\dfrac{17}{15}\)

\(=\dfrac{59}{30}\)

Bài 2:

a) \(\dfrac{3}{4}+y:\dfrac{2}{5}=\dfrac{37}{16}\)

\(y:\dfrac{2}{5}=\dfrac{37}{16}-\dfrac{3}{4}\)

\(y:\dfrac{2}{5}=\dfrac{25}{16}\)

\(y=\dfrac{25}{16}\times\dfrac{2}{5}\)

\(y=\dfrac{5}{8}\)

b)\(456+y:87=23987\)

\(y:87=23987-456\)

\(y:87=23531\)

\(y=23531\times87\)

\(y=2047197\)

a) 4/5 x 5/8 : 4/5

= 5/8

b) 5/6 + ( 1/2 : 3/2 + 4/5)

= 5/6 + (1/3 + 4/5)

= 5/6 + 17/15

= 59/30

B4:

3/4 + y : 2/5 = 37/16

y : 2/5 = 25/16

y = 5/8.

456 + y : 87 = 23987

y : 87 = 23531

y = 2047197.

Bài 3 :

\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)

\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)

\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)

\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)

.....

\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)

\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)

Bạn xem lại đề 2, phần mẫu của N

@Nguyễn Đức Trí: Đề bài nó như vậy mà

a, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{299.300}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{299}-\dfrac{1}{300}\)

\(=1-\dfrac{1}{300}=\dfrac{299}{300}\)

Vậy \(A=\dfrac{299}{300}\)

b, \(B=\dfrac{10^2}{16.26}+\dfrac{10^2}{26.36}+...+\dfrac{10^2}{86.96}\)

\(=10\left(\dfrac{10}{16.26}+\dfrac{10}{26.36}+...+\dfrac{10}{86.96}\right)\)

\(=10\left(\dfrac{1}{16}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{36}+...+\dfrac{1}{86}-\dfrac{1}{96}\right)\)

\(=10\left(\dfrac{1}{16}-\dfrac{1}{96}\right)\)

\(=10.\dfrac{5}{96}=\dfrac{25}{48}\)

Vậy...

a,\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{299.300}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{299}-\dfrac{1}{300}\)

(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với mọi \(a\in N\)*)

\(A=\dfrac{1}{1}-\dfrac{1}{300}=\dfrac{299}{300}\)

a) \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3.4}+....+\dfrac{1}{299\cdot300}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{299}-\dfrac{1}{300}\)

\(=1-\dfrac{1}{300}=\dfrac{299}{300}\)

b) \(B=\dfrac{10^2}{16\cdot26}+\dfrac{10^2}{26\cdot36}+\dfrac{10^2}{36\cdot46}+...+\dfrac{10^2}{86.96}\)

\(\dfrac{B}{10}=\dfrac{10}{16\cdot26}+\dfrac{10}{26\cdot36}+\dfrac{10}{36\cdot46}+...+\dfrac{10}{86\cdot96}\)

\(\dfrac{B}{10}=\dfrac{1}{16}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{36}+\dfrac{1}{36}-\dfrac{1}{46}+...+\dfrac{1}{86}-\dfrac{1}{96}\)

\(\dfrac{B}{10}=\dfrac{1}{16}-\dfrac{1}{96}=\dfrac{5}{96}\)

\(\Rightarrow B=\dfrac{5}{96}\cdot10=\dfrac{25}{48}\)

a) \(\dfrac{1}{2}:y\times\dfrac{3}{5}=\dfrac{4}{3}+\dfrac{3}{4}\)

\(\dfrac{1}{2}:y\times\dfrac{3}{5}=\dfrac{25}{12}\)

\(\dfrac{1}{2}:y=\dfrac{25}{12}:\dfrac{3}{5}\)

\(\dfrac{1}{2}:y=\dfrac{125}{36}\)

\(y=\dfrac{1}{2}:\dfrac{125}{36}\)

\(y=\dfrac{18}{125}\)

b) \(\dfrac{4}{3}-\dfrac{1}{2}\times y=1\)

\(\dfrac{1}{2}\times y=\dfrac{4}{3}-1\)

\(\dfrac{1}{2}\times y=\dfrac{1}{3}\)

\(y=\dfrac{1}{3}:\dfrac{1}{2}\)

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

c) \(\dfrac{1}{4}+y:\dfrac{1}{3}=\dfrac{5}{6}\)

\(y:\dfrac{1}{3}=\dfrac{5}{6}-\dfrac{1}{4}\)

\(y:\dfrac{1}{3}=\dfrac{7}{12}\)

\(y=\dfrac{7}{12}\cdot\dfrac{1}{3}\)

\(y=\dfrac{7}{36}\)

Bài 4: 

a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)

\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)

\(1,25-x=\dfrac{11}{12}\)

\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)

\(x=\dfrac{1}{3}\)

b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)

\(x-\dfrac{7}{6}=\dfrac{13}{12}\)

\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)

\(x=\dfrac{27}{12}=\dfrac{9}{4}\)

c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)

\(4-\left(2x+1\right)=\dfrac{8}{3}\)

\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)

\(2x+1=\dfrac{20}{3}\)

\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)

\(2x=\dfrac{17}{3}\)

\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)

Bài 15:

a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)

\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)

\(=>x=\left(\dfrac{-2}{3}\right)^8\)

b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)

\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)

\(=>x=\left(\dfrac{4}{9}\right)^9\)

c) \(\left(x+4\right)^3=-125\)

\(\left(x+4\right)^3=\left(-5\right)^3\)

\(=>x+4=-5\)

\(x=-5-4\)

\(=>x=-9\)

d) \(\left(10-5x\right)^3=64\)

\(\left(10-5x\right)^3=4^3\)

\(=>10-5x=4\)

\(5x=10-4\)

\(5x=6\)

\(=>x=\dfrac{6}{5}\)

e) \(\left(4x+5\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)

Bài 16:

a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)

\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)

\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)

b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)

\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)

\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)

c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)

\(=\dfrac{7}{4}.\dfrac{-12}{5}\)

\(=\dfrac{-21}{5}\)

\(#Wendy.Dang\)

Uh, chừa sau k dám học muộn nx

Đăng từng bài `1` thôi cậu ơii

\(B=\frac{1}{16}+\frac{6}{16.26}+\frac{6}{26.36}+....+\frac{6}{2006.2016}\)

\(=\frac{6}{6.16}+\frac{6}{16.26}+\frac{6}{26.36}+....+\frac{6}{2006.2016}\)

\(=\frac{6}{10}\left(\frac{1}{6}-\frac{1}{16}+\frac{1}{16}-\frac{1}{26}+...+\frac{1}{2006}-\frac{1}{2016}\right)\)

\(=\frac{3}{5}\left(\frac{1}{6}-\frac{1}{2016}\right)\)

\(=\frac{67}{672}\)

\(B=\frac{1}{16}+\frac{6}{16\cdot26}+\frac{6}{26\cdot36}+...+\frac{6}{2006\cdot2016}\)

\(B=\frac{1}{16}+6\left(\frac{1}{16\cdot26}+\frac{1}{26\cdot36}+...+\frac{1}{2006\cdot2016}\right)\)

\(B=\frac{1}{16}+6\left[\frac{1}{10}\left(\frac{10}{16\cdot26}+\frac{10}{26\cdot36}+...+\frac{10}{2006\cdot1016}\right)\right]\)

\(B=\frac{1}{16}+6\left[\frac{1}{10}\left(\frac{1}{16}-\frac{1}{26}+\frac{1}{26}-\frac{1}{36}+...+\frac{1}{2006}-\frac{1}{2016}\right)\right]\)

\(B=\frac{1}{16}+6\left[\frac{1}{10}\left(\frac{1}{16}-\frac{1}{2016}\right)\right]\)

\(B=\frac{1}{16}+6\cdot\left[\frac{1}{10}\cdot\frac{125}{2016}\right]\)

\(B=\frac{1}{16}+6\cdot\frac{26}{4032}\)

\(B=\frac{1}{16}+\frac{25}{672}\)

\(B=\frac{57}{672}\)

2:

a: =>2/3:x=1,4-2,4=-1

=>x=-2/3

b: =>x/5=25/30-19/30=6/30=1/5

=>x=1

3:

Số học sinh giỏi là 40*1/4=10 bạn

Số học sinh khá là 30*3/5=18 bạn

Số học sinh TB là 30-18=12 bạn