mk biết làm

1) \(D=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+....+\frac{10}{1400}\)

\(D=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+.....+\frac{5}{700}\)

\(D=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+......+\frac{5}{25.28}\)

\(D=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+.....+\frac{3}{25.28}\right)\)

\(D=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{25}-\frac{1}{28}\right)\)

\(D=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}.\frac{6}{28}=\frac{5}{14}\)

\(E=\frac{1}{1+2}+\frac{1}{1+2+3}+.......+\frac{1}{1+2+3+....+24}\)

Ta có: \(1+2=\)\(\frac{2.\left(2+1\right)}{2}=3\);\(1+2+3=\frac{3.\left(3+1\right)}{2}=6\);\(1+2+3+...+24=\frac{24.\left(24+1\right)}{2}=300\)

\(E=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{300}\)

=>\(\frac{1}{2}E=\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{600}=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{24.25}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{24}-\frac{1}{25}=\frac{1}{2}-\frac{1}{25}=\frac{23}{50}\)

=>\(E=\frac{46}{50}\)

Vậy \(\frac{D}{E}=\frac{5}{14}:\frac{46}{50}=\frac{250}{644}=\frac{125}{322}\)

2) Theo t/c dãy tỉ số=nhau:

\(\frac{a+b}{a+c}=\frac{a-b}{a-c}=\frac{a+b-\left(a-b\right)}{a+c-\left(a-c\right)}=\frac{a+b-a+b}{a+c-a+c}=\frac{2b}{2c}=1\)

=>b=c

do đó \(A=\frac{10b^2+9bc+c^2}{2b^2+bc+2c^2}=\frac{10b^2+9b^2+b^2}{2b^2+b^2+2b^2}=\frac{\left(10+9+1\right).b^2}{\left(2+1+2\right).b^2}=4\)

Câu hỏi của Nguyễn Thị Huế ở ngay bên dưới đó bạn!!!

K cho mik nhé Gumm

Đâu có
 

\(-\frac{54}{55}\) và \(-\frac{55}{56}\)

Ta có: \(-\frac{54}{55}>-\frac{54}{56}>-\frac{55}{56}\)

\(\Rightarrow-\frac{54}{55}>-\frac{55}{56}\)

\(\frac{-54}{55}< \frac{-55}{56}\)

cách 1: so sánh phần bù

\(A=\frac{-54}{55}=1-\frac{-54}{55}=\frac{-1}{55}.\)\(;\)\(B=\frac{-55}{56}=1-\frac{-55}{56}=\frac{-1}{56}\)

ta có -1=-1; 55>56 nên A>B

\(\hept{\begin{cases}x\left(x-y\right)=\frac{3}{10}\\y\left(x-y\right)=\frac{-3}{50}\end{cases}}\)\(\Rightarrow x\left(x-y\right)-y\left(x-y\right)=\frac{3}{10}-\frac{-3}{50}\)

\(\Rightarrow\left(x-y\right)^2=\frac{3}{10}+\frac{3}{50}=\frac{9}{25}\)

\(\Rightarrow\orbr{\begin{cases}x-y=\frac{3}{5}\\x-y=\frac{-3}{5}\end{cases}}\)

Đến đây bn xét từng trường hợp r` thay vào đề bài là ra

\(\hept{\begin{cases}x\left(x-y\right)=\frac{3}{10}\\y\left(x-y\right)=\frac{-3}{50}\end{cases}}\Rightarrow x\left(x-y\right)-y\left(x-y\right)=\frac{3}{10}-\frac{-3}{50}\)

\(\Rightarrow\left(x-y\right)^2=\frac{3}{10}+\frac{3}{50}=\frac{9}{25}\)

\(\orbr{\begin{cases}x-y=\frac{3}{5}\\x-y=\frac{-3}{5}\end{cases}}\)

Đến đây bạn xét từng trường hợp rồi thay vào đề là ta ra

\(D=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)

\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{5}{3}.\dfrac{6}{28}=\dfrac{5}{14}\)

\(E=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{24.25}=2\left(\dfrac{1}{2}-\dfrac{1}{25}\right)=\dfrac{2.23}{50}=\dfrac{23}{25}.\)

\(\dfrac{D}{E}=\dfrac{5}{24}.\dfrac{25}{23}=\dfrac{125}{552}.\)

sai ròi kìa

3.

Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Leftrightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}\) và \(a+2b-3c=-20\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\)

+) \(\dfrac{a}{2}=5\Rightarrow a=5.2=10\)

+) \(\dfrac{2b}{6}=5\Rightarrow2b=5.6=30\Rightarrow b=30:2=15\)

+) \(\dfrac{3c}{12}=5\Rightarrow3c=5.12=60\Rightarrow c=60:3=20\)

Vậy ...

3.

ta có:\(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)=\(\dfrac{c}{4}\)=>\(\dfrac{a}{2}\)=\(\dfrac{2b}{6}\)=\(\dfrac{3c}{12}\) và a+2b-3c=-20

áp dụng tính chất của dãy tỉ số bằng nhau ta có

\(\dfrac{a}{2}\)=\(\dfrac{2b}{6}\)=\(\dfrac{3c}{12}\)=\(\dfrac{a+2b-3c}{2+6-12}\)\(\dfrac{-20}{-4}\)=5

vì\(\dfrac{a}{2}\)=5=>a=2.5=10

\(\dfrac{2b}{6}\)=5=>2b=5.6=30=>b=30:2=15

\(\dfrac{3c}{12}\)=5=>3c=5.12=60=>c=60:3=20

vậy a=10,b=15,c=20

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