Giúp em với ạ

a, Để A nhận giá trị dương thì \(A>0\)hay \(x-1>0\Leftrightarrow x>1\)

b, \(B=2\sqrt{2^2.5}-3\sqrt{3^2.5}+4\sqrt{4^2.5}\)

\(=4\sqrt{5}-9\sqrt{5}+16\sqrt{5}=\left(4-9+16\right)\sqrt{5}=11\sqrt{5}\)

( theo công thức \(A\sqrt{B}=\sqrt{A^2B}\))

c, Với \(a\ge0;a\ne1\)

\(C=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)

\(=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(\sqrt{a}+1\right)^2}=1\)

+) \(M=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2019\cdot2020}\)

\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2019}-\frac{1}{2010}\)

\(M=1-\frac{1}{2010}=\frac{2009}{2010}\)

Vậy M=\(\frac{2009}{2010}\)

+) Đặt A=\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{50}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\cdot\cdot\cdot\frac{49}{50}\)

\(A=\frac{1\cdot2\cdot\cdot\cdot\cdot49}{2\cdot3\cdot\cdot\cdot\cdot50}=\frac{1}{50}\)

Ta có: \(M=1\cdot2+3\cdot4+\cdots+57\cdot58\)

\(=1\left(1+1\right)+3\left(3+1\right)+\cdots+57\left(57+1\right)\)

\(=\left(1^2+3^2+\cdots+57^2\right)+\left(1+3+\cdots+57\right)\)

Đặt \(A=1^2+3^2+\cdots+57^2\)

\(=1^2+2^2+3^2+4^2+\cdots+57^2+58^2-\left(2^2+4^2+\cdots+58^2\right)\)

\(=\frac{58\left(58+1\right)\left(2\cdot58+1\right)}{6}-2^2\left(1^2+2^2+\cdots+29^2\right)\)

\(=\frac{58\cdot59\cdot117}{6}-4\cdot\frac{29\left(29+1\right)\left(2\cdot29+1\right)}{6}\)

\(=29\cdot59\cdot39-4\cdot\frac{29\cdot30\cdot59}{6}=29\cdot59\cdot39-4\cdot29\cdot5\cdot59\)

=32509

Đặt B=1+3+...+57

Số số hạng của dãy số là:

(57-1):2+1=56:2+1=28+1=29(số)

Tổng của dãy số là: \(B=\left(57+1\right)\cdot\frac{29}{2}=58\cdot\frac{29}{2}=29\cdot29=841\)

M=A+B

=32509+841

=33350

\(N=1^2+3^2+\cdots+57^2\)

=>N=A

=32509

M-N

=33350-32509

=841

\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\cdot\frac{5^2}{5\cdot6}=\frac{1^2}{1\cdot6}=\frac{1}{6}\)

lan sau nho ghi de cho dung nha bn

\(\frac{1.1.2.2.3.3.4.4.5.5}{1.2.2.3.3.4.4.5.5.6}\)=\(\frac{\left(1.2.3.4.5\right).\left(1.2.3.4.5\right)}{\left(1.2.3.4.5\right)\left(2.3.4.5.6\right)}=\frac{1}{6}\)

Trước tiên, chúng ta cần có lý thuyết về biến đổi phân số.

\(\dfrac{b-a}{a\cdot b}=\dfrac{1}{a}-\dfrac{1}{b}\)

Ta có:

\(S=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\)

\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\)

\(S=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+...-\dfrac{1}{2018}\)

\(S=1-\dfrac{1}{2018}\)

\(S=\dfrac{2017}{2018}\)

=1/1.2+1/2.3+1/3.4+...1/2017.2018

=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2017-1/2018

=1-1/2018

=2018/2018-1/2018

=2017/2018

hình như là 3² chứ k f 3³

\(B=\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\)

\(B=\frac{\left(1\cdot1\right)\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)}{\left(1\cdot2\right)\left(2\cdot3\right)\left(3\cdot4\right)\left(4\cdot5\right)}\)

\(B=\frac{\left(1\cdot2\cdot3\cdot4\right)\left(1\cdot2\cdot3\cdot4\right)}{\left(1\cdot2\cdot3\cdot4\right)\left(2\cdot3\cdot4\cdot5\right)}\)

\(=\frac{1}{5}\)

\(B=\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\)

\(B=\frac{1^2\cdot2^2\cdot3^2\cdot4^2}{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5}\)

\(B=\frac{1^2\cdot2^2\cdot3^2\cdot4^2}{1^2\cdot2^2\cdot3^2\cdot4^2\cdot5}=\frac{1}{5}\)

\(\frac{9}{2}:\left(2.5-\frac{15}{4}\right)+\left(\frac{-1}{2}\right)^2\)

\(=\frac{9}{2}:\left(10-\frac{15}{4}\right)+\frac{1}{4}\)

\(=\frac{9}{2}:\left(\frac{40}{4}-\frac{15}{4}\right)+\frac{1}{4}\)

\(=\frac{9}{2}:\frac{25}{4}+\frac{1}{4}\)

\(=\frac{9}{2}.\frac{4}{25}+\frac{1}{4}\)

\(=\frac{18}{25}+\frac{1}{4}\)

\(=\frac{97}{100}\)

\(\frac{9}{2}\div\left(2\times5-\frac{15}{4}\right)+\left(\frac{-1}{2}\right)^2\)

\(=\frac{9}{2}\div\left(10-\frac{15}{4}\right)+\left(\frac{-1}{2}\right)^2\)

\(=\frac{9}{2}\div\left(\frac{40-15}{4}\right)+\frac{1}{4}\)

\(=\frac{9}{2}\div\frac{25}{4}+\frac{1}{4}\)

\(=\frac{9}{2}\times\frac{4}{25}+\frac{1}{4}\)

\(=\frac{18}{25}+\frac{1}{4}\)

\(=\frac{72+25}{100}=\frac{97}{100}\)