1)

\(A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+...+\dfrac{500}{5^{500}}\\ 5A=1+\dfrac{2}{5}+\dfrac{3}{5^2}+...+\dfrac{500}{5^{49}}\\ 5A-A=\left(1+\dfrac{2}{5}+\dfrac{3}{5^2}+...+\dfrac{500}{5^{49}}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+...+\dfrac{500}{5^{500}}\right)\\ 4A=1-\dfrac{500}{5^{500}}\\ A=\left(1-\dfrac{500}{5^{500}}\right):4\\ A=1:4-\dfrac{500}{5^{500}}:4\\ A=\dfrac{1}{4}-\dfrac{500}{5^{500}\cdot4}< \dfrac{1}{4}< \dfrac{5}{16}\)

Vậy \(A< \dfrac{5}{16}\)

2,

\(M=\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\) =\(\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)

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

e: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x+5}{2}=\dfrac{y-2}{3}=\dfrac{x-y+5+2}{2-3}=\dfrac{10+7}{-1}=-17\)

=>x+5=-34; y-2=-51

=>x=-39; y=-49

g: Áp dụng tính chất của DTSBN, ta được

\(\dfrac{a-1}{2}=\dfrac{b+3}{4}=\dfrac{c-5}{6}=\dfrac{5a-3b-4c-5-9+20}{5\cdot2-3\cdot4-6\cdot4}=\dfrac{-253}{13}\)

=>a-1=-506/13; b+3=-1012/13; c-5=-1518/13

=>a=-493/13; b=-1051/13; c=-1453/13

Lời giải:
e. Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{x+5}{2}=\frac{y-2}{3}=\frac{x+5-(y-2)}{2-3}=\frac{(x-y)+5+2}{2-3}=\frac{10+5+2}{-1}=-17$

Suy ra:

$x+5=2(-17)=-34\Rightarrow x=-39$

$y-2=3(-17)=-51\Rightarrow y=-49$

f. Đề thiếu. Bạn xem lại

h. Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{a-1}{2}=\frac{b+3}{4}=\frac{c-5}{6}$

$=\frac{5a-5}{10}=\frac{3b+9}{12}=\frac{4c-20}{24}$

$=\frac{5a-5-(3b+9)-(4c-20)}{10-12-24}$

$=\frac{5a-3b-4c-5-9+20}{-26}=\frac{500-5-9+20}{-26}=\frac{-253}{13}$

Suy ra:
$a-1=2.\frac{-253}{13}\Rightarrow a=\frac{-493}{13}$

$b+3=4.\frac{-253}{13}\Rightarrow b=\frac{-1051}{13}$

$c-5=6.\frac{-253}{13}\Rightarrow c=\frac{-1453}{13}$

\(A=\dfrac{\left(3+\dfrac{2}{15}+\dfrac{1}{5}\right):\dfrac{5}{2}}{\left(5+\dfrac{3}{7}-2-\dfrac{1}{4}\right):\left(4+\dfrac{43}{56}\right)}\)

\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{2}{5}}{\dfrac{89}{28}:\dfrac{267}{56}}=\dfrac{4}{3}:\dfrac{2}{3}=2\)

\(B=\dfrac{\dfrac{6}{5}:\left(\dfrac{6}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{8}{25}+\dfrac{2}{25}}=\dfrac{\dfrac{6}{5}:\dfrac{3}{2}}{\dfrac{2}{5}}=2\)

Do đó: A=B

a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)

b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)

c) \(=-\dfrac{80}{9}\)

d) \(=-\dfrac{1}{6}\)

a.=\(\dfrac{157}{8}:\dfrac{7}{12}-\dfrac{61}{4}:\dfrac{7}{12}=\dfrac{471}{14}-\dfrac{183}{7}=\dfrac{15}{2}\)

b.=\(\dfrac{2}{15}-\dfrac{2}{3}+\dfrac{1}{5}=-\dfrac{1}{3}\)

c.\(\left(\dfrac{10}{3}+2.5\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}=\dfrac{35}{6}:\left(-\dfrac{31}{30}\right)-\dfrac{11}{31}=-\dfrac{175}{31}-\dfrac{11}{31}=-6\)

d.\(\left[6+\dfrac{1}{8}-\dfrac{1}{2}\right]:\dfrac{3}{12}=\dfrac{45}{8}:\dfrac{3}{12}=\dfrac{45}{2}\)

a: \(A=\dfrac{19}{9}+\dfrac{4}{11}+\dfrac{2}{3}=\dfrac{209}{99}+\dfrac{44}{99}+\dfrac{66}{99}=\dfrac{319}{99}\)

b: \(B=\dfrac{-50}{60}+\dfrac{-35}{60}+\dfrac{12}{60}=\dfrac{-73}{60}\)

c: \(C=\dfrac{-27}{36}+\dfrac{132}{36}+\dfrac{10}{36}=\dfrac{115}{36}\)

d: \(D=\dfrac{-19}{3}+\dfrac{2}{3}-\dfrac{4}{5}=\dfrac{-17}{3}-\dfrac{4}{5}=\dfrac{-85-12}{15}=-\dfrac{97}{15}\)

a) `1/3 - 1/4 : 2/5 = 1/3 - 5/8 = -7/24`

b) `6/7-(5/6+1/3)-(2/3+1/7) = 6/7-5/6-1/3-2/3-1/7`

`=(6/7-1/7)-(1/3+2/3)-5/6`

`=5/7-1-5/6`

`=-47/42`

c) `-5/9 . 2/5 + 4 5/9 + 5/9 . (-3/5)`

`= -5/9 . 2/5 + 4 + 5/9 + (-5/9) . 3/5`

`=-5/9 . (2/5 + 3/5-1) + 4`

`=-5/9 . 0 +4`

`=4`

d) 3 1/2 - (5 4/7 - 1 1/2) : 0,75`

`=7/2 - (39/7 - 3/2) : 3/4`

`= 7/2 - 57/14 : 3/4`

`=7/2 - 38/7`

`=-27/14`

De tke nma khonq bt lamm

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