Sửa N=\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{100}{101}\)

Ta có : \(\frac{1}{2}< \frac{2}{3}\); \(\frac{3}{4}< \frac{4}{5}\); \(\frac{5}{6}< \frac{6}{7}\); ... ; \(\frac{99}{100}< \frac{100}{101}\)

\(\Rightarrow\)\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)hay M < N

b) M .N = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}.\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}=\frac{1.2.3.4.5.6...99.100}{2.3.4.5.6.7...100.101}=\frac{1}{101}\)

c) vì M < N nên M. M < M . N = \(\frac{1}{101}\)\(< \frac{1}{100}\)

\(\Rightarrow M< \frac{1}{10}\)

Nhỏ hơn 

\(A=\frac{x+2}{x-5}\)là số nguyên dương

     \(\rightarrow x+2⋮x-5\)

Ta có:\(x-5⋮x-5\)

\(\rightarrow[\left(x+2\right)-\left(x-5\right)]⋮x-5\)

\(\rightarrow\left(x+2-x+5\right)⋮x-5\)

\(\rightarrow7⋮x-5\)\(\rightarrow x-5\inƯ\left(7\right)=\left\{1;7\right\}\)

\(x-5=1\rightarrow x=6\)

\(x-5=7\rightarrow x=12\)

Vậy x=6 hoặc x=12

thì tử chia hết cho mẫu

Thiếu \(|a|\)=\(b^2\left(b-c\right)\)

+, Nếu a=0 => b=0 hoặc b-c=0 => b=c hoặc b=c  ( đều vô lí ) => a khác 0

+, Nếu b = 0 => a = 0 ( vô lí ) => b khác 0

=>c=0

=> |a| = b^2.b = b^3

=> b^3 >= 0

=> b là số nguyên dương

=> a là số nguyên âm

Vậy a là số nguyên âm , b là số nguyên dương và c = 0

bài 1

1750 san phẩm

bài 2

tỉ lệ\(\frac{1}{3000000}\)

40°C  = ?  °F

104 độ Pha-ren-hét

chúc hok tốt

104 độ nha

\(A=\frac{17^{18}+1}{17^{19}+1}\)

\(17A=\frac{17^{19}+17}{17^{19}+1}=\frac{\left(17^{19}+1\right)+16}{17^{19}+1}=1+\frac{16}{17^{19}+1}\)

\(B=\frac{17^{17}+1}{17^{18}+1}\)

\(17B=\frac{17^{18}+17}{17^{18}+1}=\frac{\left(17^{18}+1\right)+16}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)

\(\text{Vì}\)\(1+\frac{16}{17^{19}+1}< 1+\frac{16}{17^{18}+1}\)

\(\Leftrightarrow17A< 17B\)

\(\Leftrightarrow A< B\)

Trả lời

\(17A=\frac{\left(17^{18}+1\right)17}{17^{19}+1}=\frac{17^{19}+17}{17^{19}+1}=\frac{17^{19}+1+16}{17^{19}+1}=\frac{17^{19}+1}{17^{19}+1}+\frac{16}{17^{19}+1}=1+\frac{16}{17^{19}+1}\)

\(17B=\frac{\left(17^{17}+1\right)17}{17^{18}+1}=\frac{17^{18}+17}{17^{18}+1}=\frac{17^{18}+1+16}{17^{18}+1}=\frac{17^{18}+1}{17^{18}+1}+\frac{16}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)

Vì \(17^{19}+1>17^{18}+1\)

\(\Rightarrow\frac{16}{17^{18}+1}>\frac{16}{17^{19}+1}\)

\(\Rightarrow1+\frac{16}{17^{18}+1}>1+\frac{16}{17^{19}+1}\)

\(\Rightarrow B>A\)