1.100809941\(:>\)

\(\frac{9991}{9992}+\frac{1}{9}-\frac{2}{19984}-\frac{5}{495}\)

\(=0,99989991993+0,11111111111-0,00010008006-0,0101010101\)

\(=1,10080994088\)

Rút gọn phân số rồi tính như thường :)

\(-\frac{9991}{9992}+\frac{1}{99}-\frac{2}{19984}-\frac{5}{495}\)

\(=-\frac{9991}{9992}+\frac{1}{99}-\frac{1}{9992}-\frac{1}{99}\)

\(=\left(-\frac{9991}{9992}-\frac{1}{9992}\right)+\left(\frac{1}{99}-\frac{1}{99}\right)\)

\(=-1+0\)

\(=-1\)

1)
S1=(1+999)+(2+998)+...+(501+499)+500
S1=1000.(999-1)+500
S1=998 000 + 500
S1=998 500
 

:/ câu 1 quá ez mik bik làm rồi còn các câu sau nó thì mik chịu

Bài 1: 

a: =(172,56-72,56)-(35,32-5,32)+4,37-5,37

=100-30-1

=69

b: =7,2(111+890+999)

=7,2x2000

=14400

a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)

\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )

Vậy x = 1

b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)

=> x + 100 = 0

=> x           = -100

c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)

\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)

=> x - 100 = 0

=> x           = 100

Chúc bạn học tốt

có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai

\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)

\(\Rightarrow3M-M=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{99}}\right)\)

\(\Rightarrow2M=1-\frac{1}{3^{99}}< 1\Rightarrow M< \frac{1}{2}\left(đpcm\right)\)

đpcm là j bạn

mk cx ko biết nhưng thường mk làm là x - 1

còn 1 - x thì có một trường hợp là x = 0

`A=3/4+8/9+.............+9999/10000`

`=1-1/4+1-1/9+,,,,,,,,,,+1-1/10000`

`=99-(1/4+1/9+.........+1/10000)<99-0=99`

`=>A<99`

Thanks

x =0 và y=5