\(a,ĐKXĐ:x-1\ge0\Leftrightarrow x\ge1\)

Đặt \(\hept{\begin{cases}\sqrt[3]{2-x}=a\\\sqrt{x-1}=b\left(b\ge0\right)\end{cases}\Rightarrow}a^3+b^2=2-x+x-1=1\)

Lại có: \(a=1-b\)

Thay vào được

\(\left(1-b\right)^3+b^2=1\)

\(\Leftrightarrow1-3b+3b^2-b^3+b^2-1=0\)

\(\Leftrightarrow-b^3+4b^2-3b=0\)

\(\Leftrightarrow b^3-4b^2+3b=0\)

\(\Leftrightarrow b\left(b^2-4b+3\right)=0\)

\(\Leftrightarrow b\left(b-1\right)\left(b-3\right)=0\)

\(\Leftrightarrow b=0\left(h\right)b=1\left(h\right)b=3\)(T/m ĐK b>0)

*Với b = 0

\(\Leftrightarrow\sqrt{x-1}=0\)

\(\Leftrightarrow x=1\left(TmĐKXĐ\right)\)

*Với b = 1

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(TmĐKXĐ\right)\)

*Với b = 3

\(\Leftrightarrow\sqrt{x-1}=3\)

\(\Leftrightarrow x-1=9\)

\(\Leftrightarrow x=10\)

Vậy \(S\in\left\{1;2;10\right\}\)

em chỉ bt bài 2 nha!

\(A=\left(1-\frac{2}{2\cdot3}\right)\left(1-\frac{2}{3\cdot4}\right)...\left(1-\frac{2}{2020\cdot2021}\right)\)

\(\frac{2}{3}\cdot\frac{5}{6}\cdot\frac{9}{10}\cdot...\cdot\frac{2020\cdot2021-2}{2020\cdot2021}\left(1\right)\)

Mặt khác:\(2020\cdot2021-2=2020\left(2022-1\right)+2020-2022\)

\(=2020\cdot2022-2022\)

\(=2022\left(2020-1\right)=2019\cdot2022\left(2\right)\)

Từ (1),(2) ta có:

\(A=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot...\cdot\frac{2022\cdot2019}{2020\cdot2021}\)

\(=\frac{\left(4\cdot5\cdot6\cdot...\cdot2022\right)\left(1\cdot2\cdot3\cdot...\cdot2019\right)}{\left(2\cdot3\cdot4\cdot...\cdot2020\right)\left(3\cdot4\cdot5\cdot...\cdot2021\right)}\)

\(=\frac{2021\cdot2022}{2\cdot3}\cdot\frac{1\cdot2}{2020\cdot2021}=\frac{2022}{3\cdot2020}=\frac{2022}{6060}\)

2/3

4

\(\dfrac{13}{15}-\dfrac{1}{5}=\dfrac{13}{15}=\dfrac{3}{15}=\dfrac{10}{15}=\dfrac{2}{3}\)

\(\dfrac{7}{8}\times\dfrac{2}{7}=\dfrac{14}{56}=\dfrac{1}{4}\)

\(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\)

\(=\frac{1\times2\times3}{2\times3\times4}\)

\(=\frac{1}{4}\)

1/2 x 2/3 x 3/4 = 1/3 x 3/4 = 1/4

hok tốt

1/2 . 2/3 . 3/4

= 1/3 . 3/4

= 1/4

S = 1\(\dfrac{1}{3}\).1\(\dfrac{1}{8}\).1\(\dfrac{1}{15}\).1\(\dfrac{1}{24}\).1\(\dfrac{1}{35}\)....

S = \(\dfrac{4}{3}\).\(\dfrac{9}{8}\).\(\dfrac{16}{15}\).\(\dfrac{25}{24}\).\(\dfrac{36}{35}\)....

S = \(\dfrac{2^2}{1.3}\).\(\dfrac{3^2}{2.4}\).\(\dfrac{4^2}{3.5}\).\(\dfrac{5^2}{4.6}\).\(\dfrac{6^2}{5.7}\)...

Phân số thứ 100 của dãy số trên là: \(\dfrac{101^2}{100.102}\)

Tích của 100 số đầu tiên của dãy trên là:

S = \(\dfrac{2^2}{1.3}\).\(\dfrac{3^2}{2.4}\).\(\dfrac{4^2}{3.5}\).\(\dfrac{5^2}{4.6}\).\(\dfrac{6^2}{5.7}\)....\(\dfrac{101^2}{100.102}\)

S = \(\dfrac{\left(1.2.3...100.101\right)\times\left(2.3.4.5...101\right)}{\left(1.2.3.4...100\right)\times\left(3.4.5....101.102\right)}\)

S = \(\dfrac{101.2}{1.102}\)

S = \(\dfrac{101}{51}\)

51xS = \(\dfrac{101}{51}\) x 51 = 101

S = 1\(\dfrac{1}{3}\).1\(\dfrac{1}{8}\).1\(\dfrac{1}{15}\).1\(\dfrac{1}{24}\).1\(\dfrac{1}{35}\)....

S = \(\dfrac{4}{3}\).\(\dfrac{9}{8}\).\(\dfrac{16}{15}\).\(\dfrac{25}{24}\).\(\dfrac{36}{35}\)....

S = \(\dfrac{2^2}{1.3}\).\(\dfrac{3^2}{2.4}\).\(\dfrac{4^2}{3.5}\).\(\dfrac{5^2}{4.6}\).\(\dfrac{6^2}{5.7}\)...

Phân số thứ 100 của dãy số trên là: \(\dfrac{101^2}{100.102}\)

Tích của 100 số đầu tiên của dãy trên là:

S = \(\dfrac{2^2}{1.3}\).\(\dfrac{3^2}{2.4}\).\(\dfrac{4^2}{3.5}\).\(\dfrac{5^2}{4.6}\).\(\dfrac{6^2}{5.7}\)....\(\dfrac{101^2}{100.102}\)

S = \(\dfrac{\left(1.2.3...100.101\right)\times\left(2.3.4.5...101\right)}{\left(1.2.3.4...100\right)\times\left(3.4.5....101.102\right)}\)

S = \(\dfrac{101.2}{1.102}\)

S = \(\dfrac{101}{51}\)

51xS = \(\dfrac{101}{51}\) x 51 = 101

6/7 mình làm rồi 

trên violympic

5/14 đúng chắc luôn

1/2+1/6+1/12+1/20+1/30+1/42