a.\(\frac{17.5-17}{-17}=\frac{17.4}{-17}=-4\)

b.\(\frac{49.8}{49}=8\)

a) \(\frac{17.5-17}{3-20}\)= \(\frac{85-17}{3-20}\)= \(\frac{68}{-17}\)=-4

b) \(\frac{49+7.49}{49}\)= \(\frac{49+343}{49}\)= \(\frac{392}{49}\)= 8

CHÚC BẠN HỌC TỐT

a)rút gọn 3 vs 15

14vs21

=1.5

b)=8

a)\(\frac{3.21}{14.15}=\frac{1.3}{2.5}=\frac{3}{10}\)

b)\(\frac{49+7.49}{49}=\frac{49.\left(1+7\right)}{49}=\frac{8}{1}=8\)

a)\(\frac{3.21}{14.15}=\frac{1.3}{2.5}=\frac{3}{10}\)

b)\(\frac{49+7.49}{49}=\frac{49.\left(1+7\right)}{49}=1.8=8\)

\(\left(\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}\right)-\left(\frac{-5}{4}+\frac{6}{11}-\frac{48}{49}\right)=\left(\frac{-1}{4}-\frac{16}{11}\right)-\left(-\frac{31}{44}-\frac{48}{49}\right)=-\frac{1}{4}-\frac{16}{11}+\frac{31}{44}+\frac{48}{49}=-\frac{1}{49}\)

BÀI 1

\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}.\)

bài 2

a)           \(\frac{1}{2}-\frac{1}{3}+\frac{1}{12}=\frac{6}{12}-\frac{4}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)

b)          \(\frac{9^9.27^4}{3^8.81^5}=\frac{\left(3^2\right)^9.\left(3^3\right)^4}{3^8.\left(3^4\right)^5}=\frac{3^{18}.3^{12}}{3^8.3^{20}}=\frac{3^{30}}{3^{28}}=3^2=9\)

Study well 

Bài 1: \(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}\)

Bài 2: 

a)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{12}=\frac{6}{12}-\frac{4}{12}+\frac{1}{12}=\frac{6-4+1}{12}=\frac{1}{4}\)

b)\(\frac{9^9.27^4}{3^8.81^5}=\frac{9^9.3^{12}}{3^8.9^{10}}=\frac{3^4}{9}=\frac{3^4}{3^2}=3^2=9\)

Bài 1 :

\(\left(\frac{1}{8}+\frac{1}{8.15}+\frac{1}{15.22}+...+\frac{1}{43.50}\right)\frac{4-3-5-7-...-49}{217}\)

\(=\frac{1}{7}\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+\frac{1}{15}-\frac{1}{22}+...+\frac{1}{43}-\frac{1}{50}\right).\frac{5-\left(1+3+5+7+...+49\right)}{217}\)

\(=\frac{1}{7}\left(1-\frac{1}{50}\right).\frac{5-\left(12.50\right)+25}{217}\)

\(=\frac{1}{7}.\frac{49}{50}.\frac{5-625}{217}\)

\(=\frac{-2}{5}\)

Bài 2 :

\(B=\frac{x^2+17}{x^2+7}=\frac{\left(x^2+7\right)+10}{x^2+7}=1+\frac{10}{x^2+7}\)

Ta có : \(x^2\ge0\). Dấu '' = '' xảy ra khi :

\(x=0\Rightarrow x^2+7\ge7\)( 2 vế dương )

\(\Rightarrow\frac{10}{x^2+7}\le\frac{10}{7}\)

\(\Rightarrow1+\frac{10}{x^2+7}\le1+\frac{10}{7}\)

\(\Rightarrow B\le\frac{17}{7}\)

Dấu '' = '' xảy ra < = > x = 0

Vậy Max \(B=\frac{17}{7}\Leftrightarrow x=0\)