1* 10 ^50

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

50x2+8x50

=50x(8+2)

=50x10

500

10+10+10+50+50+10

=10x5+50+50

=50+50+50

=50x3

=150

Bằng 50 nhé! : )

10 + 10 + 10 + 10 + 10 - 50 + 100 - 50 = 50

\(10+10+10+10+10-50+100-50=50\)

=5110

k mk đi mk k bn rùi 

đó n,nha !!!!!!!!!!!

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

=5110

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)

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

60 + 50 + 50 + 40 + 10 + 10 + 10 + 10 = 240

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

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

\(A=\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=\)\(\frac{10^{50}-1}{10^{50}-1}+\frac{3}{10^{50}-1}\)\(=1+\frac{3}{10^{50}-1}\)

\(B=\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}\)\(=\frac{10^{50}-3}{10^{50}-3}+\frac{3}{10^{50}-3}\)\(=1+\frac{3}{10^{50}-3}\)

Ta có: 1<3 suy ra \(10^{50}-1>10^{50}-3\)

Suy ra\(\frac{3}{10^{50}-1}< \frac{3}{10^{50}-3}\)

Suy ra \(1+\frac{3}{10^{50}-1}< 1+\frac{3}{10^{50}-3}\)

Suy ra A<B

\(\frac{10^{50}+1}{10^{50}-3}=\frac{\left(10^{50}-3\right)+4}{10^{50}-3}=1+\frac{4}{10^{50}-3}\)

\(\frac{10^{50}+3}{10^{50}-1}=\frac{\left(10^{50}-1\right)+4}{10^{50}-1}=1+\frac{4}{10^{50}-1}\)

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 !!! 

10 mũ 50 +1:10 mũ 50-3 = 10 mũ 50 -1