Ta có

 \(C=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}...+\frac{1}{17.18}>A=\frac{1}{2.3}+\frac{1}{5.4}+...+\frac{1}{18.19}\)

\(C< =>\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{18-17}{17.18}\)\(>A\)

\(C< =>\frac{1}{2}-\frac{1}{18}\)\(>A\)

\(C< =>\frac{4}{9}\)\(>A\left(1\right)\)

Lại có  \(C=\frac{4}{9}< \frac{9}{19}=B\left(2\right)\)

Từ (1),(2) => B>A

a. 24 cm2                   b . 6cm2                                 c . ai = 1/2 ic

\(x^2+6x+11\)

\(=\left(x^2+6x+9\right)+2\)

\(=\left(x+3\right)^2+2\)\(>0\)

Vậy pt vô nghiệm

\(x^2+6x+11=\left(x^2+2.x.3+3^2\right)+2=\left(x+3\right)^2+2\)

Ta có: \(\left(x+3\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+3\right)^2+2\ge2>0\forall x\)

\(\Rightarrow\)đa thức \(x^2+6x+11\) vô nghiệm

                                          đpcm

Talk => Told

Plant => Planted

@_@ ^^

Talk -> Told

Plant -> Planted

Code : Breacker

đề sai 1/x(x + 1) phải là 2/x(x + 1)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1010}{1012}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1010}{1012}\)

\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1010}{1012}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1010}{1012}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1010}{1012}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{505}{1012}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{1012}\)

\(\Rightarrow x+1=1012\)

\(\Rightarrow x=1011\)

(x - 1) (x + 1) - (x + 1) (x + 2) = - 19

<=> x² - 1 - (x² + 3x + 2) = - 19

<=> x² - 1 - x² - 3x - 2 = -19

<=> - 3x = -16

<=> x = 16/3

vậy x = 16/3

(2x - 1) .3 - (x + 3). (-2) = -x +1

<=> (6x - 3) - (-2x - 6 ) = 1 - x

<=> 6x - 3 + 2x + 6 = 1 - x

<=> 8x + 3 = 1 - x

<=> 9x = - 2

<=> x = - 2/9

vậy x = - 2/9