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B = 1/6 + 1/12 + 1/20 + ... + 1/90

B = 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/9.10

B = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10

B = 1/2 - 1/10

B = 5/10 - 1/10

B = 4/10 = 2/5

Ủng hộ mk nha ♡_♡☆_☆

[2x-2=0=>x=1

x-1=0=>x=1

x+1=0=>x=-1

5=0=>x=5

a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b) Ta có: \(B=\left(\dfrac{x-2}{2x-2}+\dfrac{3}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(1-\dfrac{x-3}{x+1}\right)\)

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

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

\(=\dfrac{x^2-1-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{-2}\)

\(=\dfrac{-2x+2}{2\left(x-1\right)}\cdot\dfrac{-1}{2}\)

\(=\dfrac{-2\left(x-1\right)}{2\left(x-1\right)}\cdot\dfrac{-1}{2}\)

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

Vậy: Khi x=2005 thì \(B=\dfrac{1}{2}\)

a/

Để biểu thức được xác định

\(=>\left\{{}\begin{matrix}2x-2\ne0\\2x+2\ne0\\x+1\ne0\end{matrix}\right.\)

\(\odot2x-2\ne0\)

\(2x\ne2\)

\(x\ne1\)

\(\odot2x+2\ne0\)

\(2x\ne-2\)

\(x\ne-1\)

\(\odot x+1\ne0\)

\(x\ne-1\)

Vậy điều kiện xác định của bt là: \(x\ne-1;x\ne\pm2\)

Hình như là thế này 

A,2.308+7,062

=9,370

2,308+9,91

=12,218

B,1,002+25,068+4,03

=(1,002+25,068)+4,03

=26,070               +4,03

=30,100

B=42-y/y-15=27-(y-15)/y-15=27/(y-15)-1

để B có giá trị nhỏ nhất =>27/y-15 - 1 có GTNN=>27/y-15 có GTNN

=>y-15=-1 => y=14

=> B có GTNN = -28 <=>y=14

\(\dfrac{2^{50}.6^{10}.9^9}{12}=\dfrac{2^{50}.3^{10}.2^{10}.3^{27}}{12}=\dfrac{2^{60}.3^{37}}{3.2^2}=2^{48}.3^{36}\)

\(B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)

\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(B=\frac{1}{2}-\frac{1}{10}\)

\(B=\frac{2}{5}\)