1: =5/8*17/13*7/17*8/3*6

=5/3*7/13*6

=5*7*2/13=70/13

2: =-7/9(24/19-1)

=-7/9*5/19

=-35/171

3: =-3/8(7/6-9/6)

=-3/8*(-2/6)

=3/8*1/3=1/8

\(=\dfrac{17\left(375-126-150\right)}{33\cdot\left(99+3\right):2}=\dfrac{17\cdot99}{33\cdot102:2}=\dfrac{17\cdot3\cdot33}{33\cdot51}=1\)

\(\frac{51.125-51.42-17.150}{3+6+9+...+99}\)=\(\frac{51.125-51.42-17.3.50}{1683}\)=\(\frac{51.125-51.42-71.50}{1683}\)=\(\frac{51\left(125-42-50\right)}{1683}=\frac{51.33}{51.33}=1\)

a. = 1

b. = 2/15

c. = 1

d. = 1

a,=17/17=1

b,=7-4-1/15=2/15

c,=9+11+31/51=51/51=1

d,9-1-3/5=5/5=1

a) 11/17 + 6/17 = 1

b) 7/15 - 4/15 - 1/15 = 2/15

c) 9/51 + 11/51 + 31/51 = 1

d) 9/5 - 1/5 - 3/5 = 1

# Học tốt # k mình nha mn

\(\dfrac{33}{38}:\dfrac{11}{19}=\dfrac{33\times19}{38\times11}=\dfrac{3\times1}{2\times1}=\dfrac{3}{2}\)

a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)

\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)

\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)

b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)

\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)

\(=\dfrac{-5}{3}+\dfrac{42}{29}\)

\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)

c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)

\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)

\(=-1-1-1+4\)

=1

a) Ta có: =−2590+6490−8190=−2590+6490−8190

(−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)

=−53+4229=−53+4229

1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1

\(A=17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)

\(=11-\dfrac{15}{17}=\dfrac{172}{17}\)

\(B=\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{12}=36\dfrac{363}{533}-36\dfrac{6}{12}=\dfrac{193}{1066}\)

\(C=27\dfrac{51}{59}-\left(7\dfrac{51}{59}-\dfrac{1}{3}\right)=27\dfrac{51}{59}-7\dfrac{51}{59}+\dfrac{1}{3}=20+\dfrac{1}{3}=\dfrac{61}{3}\)

\(A=17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)

\(=\left(17\dfrac{2}{31}-6\dfrac{2}{31}\right)-\dfrac{15}{17}=11-\dfrac{15}{17}=\dfrac{172}{17}\)

\(B=\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{12}=36\dfrac{363}{533}-36\dfrac{1}{2}=\dfrac{193}{1066}\) (Casio :>)

\(C=27\dfrac{51}{59}-\left(7\dfrac{51}{59}-\dfrac{1}{3}\right)=27\dfrac{51}{59}-7\dfrac{51}{59}+\dfrac{1}{3}\)

\(=20+\dfrac{1}{3}=\dfrac{61}{3}\)

Casio hân hạnh tài trợ chương trình này :))

Làm chi tiết : Đáp số: \(\dfrac{-3}{40}\)

Ta có: \(D=\left(2\dfrac{2}{15}\cdot\dfrac{9}{17}\cdot\dfrac{3}{32}\right):\left(-\dfrac{3}{17}\right)\)

\(=\dfrac{32}{15}\cdot\dfrac{9}{17}\cdot\dfrac{3}{32}\cdot\dfrac{-17}{3}\)

\(=\dfrac{-9}{15}=-\dfrac{3}{5}\)