Bài 1:
1, \(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{45}\)
\(\dfrac{1}{2}A=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{45}\right)\)
\(\dfrac{1}{2}A=\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\)
\(\dfrac{1}{2}A=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+..+\dfrac{1}{9\cdot10}\)
\(\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{10}\)
\(A=1-\dfrac{1}{5}\)
\(A=\dfrac{4}{5}\)
2, \(x+10=y\left(x-1\right)\)
\(\Rightarrow x-1+11=y\left(x-1\right)\)
\(\Rightarrow\left(x-1\right)-y\left(x-1\right)=-11\)
\(\Rightarrow\left(x-1\right)\left(1-y\right)=-11\)
\(\Rightarrow\left(x-1\right)\left(y-1\right)=11\)
Ta có bảng:
| x - 1 | 1 | -1 | 11 | -11 |
| y - 1 | 11 | -11 | 1 | -1 |
| x | 2 | 0 | 12 | -10 |
| y | 12 | -10 | 2 | 0 |
Mà x và y là các số tự nhiên
Nên: (x;y)={(2;12);(12;2)}
Bài 4:
a: Xét ΔMOB và ΔCOD có
\(\widehat{MOB}=\widehat{COD}\)(hai góc đối đỉnh)
\(\widehat{OMB}=\widehat{OCD}\)(hai góc so le trong, MB//CD)
Do đó: ΔMOB~ΔCOD
=>\(\dfrac{MO}{CO}=\dfrac{OB}{OD}=\dfrac{MB}{DC}=\dfrac{1}{2}\)
MO/CO=1/2
=>\(\dfrac{S_{MBO}}{S_{BCO}}=\dfrac{1}{2}\)
b: Ta có: \(\dfrac{S_{MBO}}{S_{BCO}}=\dfrac{1}{2}\)
mà \(S_{MBO}+S_{BCO}=S_{MBC}=9\left(cm^2\right)\)
nên \(S_{MBO}=\dfrac{1}{3}\cdot9=3\left(cm^2\right)\)
