10+..=50
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
50 + 50 + 50 + 50 x 10 x 10 + 10
= (50 + 50) + 50 x 10 x 10 + 10
= 100 + 500 x 10 + 10
= 100 + 5000 + 10
= 5110
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
50 x 2 + 8 x 50
= (2 + 8) x 50
= 10 x 50
= 500
10 + 10 + 10 + 50 + 50 + 10 + 10
= (10 x 5) + (50 + 50)
= 50 + 100
= 150
50 x 2 + 8 x 50
= 50 x ( 2 + 8 )
= 50 x 10
= 500
10 + 10 + 10 + 50 + 50 + 10 + 10
= 10 x 5 + 50 x 5
= 50 + 250
= 300
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{10^{50}-3}{10^{50}+1}\)<\(\dfrac{10^{50}+1}{10^{50}+1}\)<\(\dfrac{10^{50}+1}{10^{50}-3}\)
=>\(\dfrac{10^{50}-3}{10^{50}+1}\)<\(\dfrac{10^{50}+1}{10^{50}-3}\)
vậy (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
C1:A = \(\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=\frac{10^{50}-1}{10^{50}-1}+\frac{3}{10^{50}-1}\)
= \(1+\frac{3}{10^{50}-1}\)
B = \(\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=\frac{10^{50}-3}{10^{50}-3}+\frac{3}{10^{50}-3}\)
= \(1+\frac{3}{10^{50}-3}\)
Vì \(\frac{3}{10^{50}-1}< \frac{3}{10^{50}-3}\)=) \(1+\frac{3}{10^{50}-1}< 1+\frac{3}{10^{50}-3}\)=) \(A< B\)
C2: Áp dụng tính chất : Nếu \(\frac{a}{b}>1\)=) \(\frac{a}{b}>\frac{a+m}{b+m}\)
Vì B > 1 =) B > \(\frac{10^{50}+2}{10^{50}-3+2}=\frac{10^{50}+2}{10^{50}-1}=A\)
(=) B > A
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=1+\frac{3}{10^{50}-1}\)
\(B=\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=1+\frac{3}{10^{50}-3}\)
Vì \(10^{50}-1>10^{50}-3\Rightarrow\frac{3}{10^{50}-1}< \frac{3}{10^{50}-3}\)(2 phân số có cùng tử số, mẫu số của phân số nào lớn hơn thì phân
số đó nhỏ hơn)
\(\Rightarrow1+\frac{3}{10^{50}-1}< 1+\frac{3}{10^{50}-3}\Rightarrow A< B\)
\(A=\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=1+\frac{3}{10^{50}-1}.\)
\(B=\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=1+\frac{3}{10^{50}-3}.\)
Do 1050-1 > 1050-3 ; => \(1+\frac{3}{10^{50}-3}>1+\frac{3}{10^{50}-1}\)
=> B > A
![](https://rs.olm.vn/images/avt/0.png?1311)
60 + 50 + 50 + 40 + 10 + 10 + 10 + 10
= ( 60 + 40 ) + ( 50 + 50 ) + 10 + 10 + 10 + 10
= 100 + 100 + 10 . 4
= 100 + 100 + 40
= 200 + 40
= 240
60 + 50 + 50 + 40 + 10 + 10 + 10 + 10
= ( 60 + 40 ) + ( 50 + 50 ) + 10 + 10 + 10 + 10
= 100 + 100 + 10 . 4
= 100 + 100 + 40
= 200 + 40
= 240
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{10^{50}+1}{10^{50}-3}=\frac{\left(10^{50}-3\right)+4}{10^{50}-3}=1+\frac{4}{10^{50}-3}\)
\(\frac{10^{50}+3}{10^{50}-1}=\frac{\left(10^{50}-1\right)+4}{10^{50}-1}=1+\frac{4}{10^{50}-1}\)
Ta so sánh \(\frac{4}{10^{50}-3}với\frac{4}{10^{50}-1}\) . Ta có \(\frac{4}{10^{50}-3}\) > \(\frac{4}{10^{50}-1}\) => 1050+1/1050-3 > 1050+3/1050-1
Ta có :
\(\frac{10^{50}+1}{10^{50}-3}=\frac{10^{50}-3+4}{10^{50}-3}=1+\frac{4}{10^{50}-3}\)
\(\frac{10^{50}+3}{10^{50}-1}=\frac{10^{50}-1+4}{10^{50}-1}=1+\frac{4}{10^{50}-1}\)
Do \(\frac{4}{10^{50}-3}>\frac{4}{10^{50}-1}\)
\(\Rightarrow1+\frac{4}{10^{50}-3}>1+\frac{4}{10^{50}-1}\)
\(\Rightarrow\frac{10^{50}+1}{10^{50}-3}>\frac{10^{50}+3}{10^{50}-1}\)
Chúc bạn học tốt !!!
Gọi số cần tìm là x
10 + x= 50
=> x= 50-10
=> x= 40
Tick nhé
x= 40 nha loan tick cho tớ