\(S=\dfrac{3}{2}+\dfrac{7}{6}+\dfrac{13}{12}+...+\dfrac{9901}{9900}\)

\(=1+\dfrac{1}{2}+1+\dfrac{1}{6}+...+1+\dfrac{1}{9900}\)

\(=\left(1+1+1+...+1\right)+\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\)

\(=99+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=99+\left(1-\dfrac{1}{100}\right)=100-\dfrac{1}{100}=\dfrac{10000-1}{100}=\dfrac{9999}{100}\)

S = ( 1+\(\dfrac{1}{2}\) ) + ( 1 + \(\dfrac{1}{6}\) ) + .... + ( 1 + \(\dfrac{1}{9900}\) )

   = 9900 + ( \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ..... + \(\dfrac{1}{99.100}\) )

   = 9900 + ( 1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + ..... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\) )

   = 9900 + 1 - \(\dfrac{1}{100}\)

   = 9901 - \(\dfrac{1}{100}\)

A = {20; 30; 40; 50; 60; 70}

A = {x ∈ N|12 < x ≤ 70 và x ⋮ 10}

A = { 20, 30, 40, 50, 60, 70 }

A = { x ϵ N; x ⋮ 10 }

50-(20+40)

=50-60=-10

\(30+\left(31+69\right)-210\)

\(=30+100-210\)

\(=30-110=-80\)

$50-(20+40)=50-60=-10$

---

$30+(31+69)-210$

$=30+100-210$

$=130-210=-80$

Bài 7:

\(\dfrac{x-2}{5}=\dfrac{-2}{2y+1}\)

=>\(\left(x-2\right)\left(2y+1\right)=5\cdot\left(-2\right)=-10\)

mà 2y+1 lẻ

nên \(\left(x-2;2y+1\right)\in\left\{\left(10;-1\right);\left(-10;1\right);\left(2;-5\right);\left(-2;5\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(12;-1\right);\left(-8;0\right);\left(4;-3\right);\left(0;2\right)\right\}\)

Bài 6:

\(\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{70}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{101}{770}\)

=>\(\dfrac{2}{40}+\dfrac{2}{88}+\dfrac{2}{140}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{101}{770}\)

=>\(\dfrac{2}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{770}\)

=>\(\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{770}\)

=>\(\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{770}\)

=>\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{770}:\dfrac{2}{3}=\dfrac{101}{770}\cdot\dfrac{3}{2}=\dfrac{303}{1540}\)

=>\(\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}=\dfrac{1}{308}\)

=>x+3=308

=>x=305

Bài 8:

a: \(\left(2x-1\right)^2+4>=4\forall x\)

=>\(B=\dfrac{20}{\left(2x-1\right)^2+4}< =\dfrac{20}{4}=5\forall x\)

Dấu '=' xảy ra khi 2x-1=0

=>\(x=\dfrac{1}{2}\)

b: \(x^2+1>=1\forall x\)

=>\(\left(x^2+1\right)^2>=1^2=1\forall x\)

=>\(\left(x^2+1\right)^2+5>=1+5=6\forall x\)

=>\(C=\dfrac{10}{\left(x^2+1\right)^2+5}< =\dfrac{10}{6}=\dfrac{5}{3}\forall x\)

Dấu '=' xảy ra khi x=0

a: KHi xét nghiệm viêm gan thì có 2 kết quả có thể xảy ra: Dương tính, Âm tính

b: Xác suất thực nghiệm là:

\(\dfrac{26}{230}=\dfrac{13}{115}\)

$2,5\times20,21\times5\times40\times0,2$

$=(2,5\times40)\times(5\times0,2)\times20,21$

$=100\times1\times20,21$

$=100\times20,21=2021$

\(2,5\cdot20,21\cdot5\cdot40\cdot0,2\\=\left(2,5\cdot40\right)\cdot20,21\cdot\left(5\cdot0,2\right)\\ =\left(2,5\cdot4\cdot10\right)\cdot20,21\cdot1\\ =100\cdot20,21\\ =2021\)

\(1-\left(4\dfrac{2}{5}+x-7\dfrac{2}{3}\right):15\dfrac{1}{3}=0\\ 1-\left(\dfrac{22}{5}+x-\dfrac{23}{3}\right):\dfrac{46}{3}=0\\ 1-\left(\dfrac{-49}{15}+x\right):\dfrac{46}{3}=0\\ \left(\dfrac{-49}{15}+x\right):\dfrac{46}{3}=1\\ -\dfrac{49}{15}+x=\dfrac{46}{3}\\ x=\dfrac{46}{3}+\dfrac{49}{15}\\ x=\dfrac{279}{15}=\dfrac{93}{5}\)

$1-\left(4.\frac25+x-\frac{7.2}{3}\right):15.\frac13=0$

$\Rightarrow \left(\frac85-\frac{14}{3}+x\right):15:3=1$

$\Rightarrow \left(-\frac{46}{15}+x\right):15=3$

$\Rightarrow -\frac{46}{15}+x=3.15$

$\Rightarrow -\frac{46}{15}+x=45$

$\Rightarrow x=45-\left(-\frac{46}{15}\right)=\frac{721}{15}$

Giúp với, mình cần rất gấp, pls

\(56=2^3\cdot7;140=2^2\cdot5\cdot7\)

=>\(ƯCLN\left(56;140\right)=2^2\cdot7=28\)

Ta có: 

\(56=7\cdot8=7\cdot2^3\)

\(140=4\cdot35=2^2\cdot5\cdot7\)

\(=>ƯCLN\left(56;140\right)=7\cdot2^2=28\)

\(x+x:2+x:3=11\)

=>\(x+\dfrac{1}{2}x+\dfrac{1}{3}x=11\)

=>\(x\left(\dfrac{1}{2}+\dfrac{1}{3}+1\right)=11\)

=>\(x\cdot\dfrac{11}{6}=11\)

=>\(x=11:\dfrac{11}{6}=6\)