\(\dfrac{10}{14}-\dfrac{17}{18}=\)
Tính hợp lý:
\(a.\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+\dfrac{-8}{13}\)
\(b.\dfrac{-3}{8}\cdot\dfrac{1}{6}+\dfrac{3}{-8}\cdot\dfrac{5}{6}+\dfrac{-10}{16}\)
\(c.\dfrac{-4}{11}\cdot\dfrac{5}{15}\cdot\dfrac{11}{-4}\)
a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)
=1-1-1
=-1
b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)
c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)
a) \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(-\dfrac{5}{13}+-\dfrac{8}{13}\right)+\left(-\dfrac{18}{35}+-\dfrac{17}{35}\right)=1+-1+-1=-1\)
b) \(=-\dfrac{3}{8}\cdot\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}-\dfrac{10}{16}=-1\)
c) \(=\left(-\dfrac{4}{11}\cdot-\dfrac{11}{4}\right)\cdot\dfrac{5}{15}=1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)
a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)
b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)
c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)
So sánh:
a) \(\dfrac{-11}{12};\dfrac{17}{-18}\)
b) \(\dfrac{-14}{-21};\dfrac{-60}{-72}\)
c) \(\dfrac{2135}{13790};\dfrac{4}{3}\)
d) \(\dfrac{2022}{2021};\dfrac{10}{9}\)
e) \(\dfrac{35}{36};\dfrac{16}{17}\)
f) -1,3; -1,2
giúp mình với, mình cảm ơn
\(a)\dfrac{-11}{12}và\dfrac{17}{-18}\) \(\Leftrightarrow\dfrac{-11}{12}và\dfrac{-17}{18}\) \(\Leftrightarrow\dfrac{-33}{36}và\dfrac{-34}{36}\)
Ta thấy rằng : \(-33>-34\Rightarrow\dfrac{-33}{36}>\dfrac{-34}{36}\)
Hay : \(\dfrac{-11}{12}>\dfrac{17}{-18}\)
\(b)\dfrac{-14}{-21}và\dfrac{-60}{-72}\)
Ta có : \(\dfrac{-14}{-21}\text{=}\dfrac{-14:-7}{-21:-7}\text{=}\dfrac{2}{3}\text{=}\dfrac{4}{6}\)
\(\dfrac{-60}{-72}\text{=}\dfrac{-60:-12}{-72:-12}=\dfrac{5}{6}\)
Do đó : \(\dfrac{-14}{-21}< \dfrac{-60}{-72}\)
\(c)\dfrac{2135}{13790}và\dfrac{4}{3}\)
Xét phân số : \(\dfrac{2135}{13790}\) ta thấy rằng : \(tử< mẫu\left(2135< 13790\right)\)
\(\Rightarrow\dfrac{2135}{13790}< 1\)
Xét phân số : \(\dfrac{4}{3}có\) : \(tử>mẫu\left(4>3\right)\)
\(\Rightarrow\dfrac{4}{3}>1\)
Do đó : \(\dfrac{2135}{13790}< \dfrac{4}{3}\)
\(d)\dfrac{2022}{2021}và\dfrac{10}{9}\)
Ta thấy rằng : \(\dfrac{2022}{2021}-\dfrac{1}{2021}\text{=}1\)
\(\dfrac{10}{9}-\dfrac{1}{9}\text{=}1\)
Mà : \(\dfrac{1}{9}>\dfrac{1}{2021}\)
\(\Rightarrow\dfrac{2022}{2021}< \dfrac{10}{9}\)
\(e)\dfrac{35}{36}và\dfrac{16}{17}\)
Ta có : \(\dfrac{35}{36}+\dfrac{1}{36}\text{=}1\)
\(\dfrac{16}{17}+\dfrac{1}{17}\text{=}1\)
Mà : \(\dfrac{1}{36}< \dfrac{1}{17}\)
\(\Rightarrow\dfrac{35}{36}>\dfrac{16}{17}\)
\(f)-1,3< -1,2\)
a) Ta có:
\(-\dfrac{11}{12}=\dfrac{1}{12}-1\)
\(-\dfrac{17}{18}=\dfrac{1}{18}-1\)
Mà: \(\dfrac{1}{12}>\dfrac{1}{18}\)
Hay: \(\dfrac{1}{12}-1>\dfrac{1}{18}-1\Rightarrow-\dfrac{11}{12}>-\dfrac{17}{18}\)
b) Ta có:
\(\dfrac{-14}{-21}=\dfrac{2}{3}=\dfrac{4}{6}\)
\(\dfrac{-60}{-72}=\dfrac{5}{6}\)
Mà: \(5>4\Rightarrow\dfrac{-60}{-72}>\dfrac{-14}{-21}\)
c) Ta có:
\(\dfrac{2135}{13790}=\dfrac{61}{394}< 1\) (tử nhỏ hơn mẫu)
\(\dfrac{4}{3}>1\) (tử lớn hơn mẫu)
Ta có: \(\dfrac{61}{394}< \dfrac{4}{3}\Rightarrow\dfrac{2135}{13790}< \dfrac{4}{3}\)
d) Ta có:
\(\dfrac{2022}{2021}=\dfrac{1}{2021}+1\)
\(\dfrac{10}{9}=\dfrac{1}{9}+1\)
Ta thấy: \(\dfrac{1}{2021}< \dfrac{1}{9}\Rightarrow\dfrac{1}{2021}+1< \dfrac{1}{9}+1\)
Hay \(\dfrac{2022}{2021}< \dfrac{10}{9}\)
e) Ta có:
\(\dfrac{35}{36}=1-\dfrac{1}{36}\)
\(\dfrac{16}{17}=1-\dfrac{1}{17}\)
Ta có: \(\dfrac{1}{36}< \dfrac{1}{17}\Rightarrow1-\dfrac{1}{36}>1-\dfrac{1}{17}\)
Hay \(\dfrac{35}{36}>\dfrac{16}{17}\)
f) Ta có: \(1,3>1,2\)
\(\Rightarrow-1,3< -1,2\)
Tính một cách hợp lí:
\(\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
= 11/125 - [(17/18 - 4/9) + (5/7 - 17/14)]
= 11/125 - (1/2 - 1/2)
= 11/125
\(\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}=\dfrac{11}{125}-\left(\dfrac{17}{18}-\dfrac{4}{9}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)=\dfrac{11}{125}-\dfrac{17-4.2}{18}+\dfrac{17-5.2}{14}=\dfrac{11}{125}-\dfrac{9}{18}+\dfrac{7}{14}=\dfrac{11}{125}-\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{11}{125}\)
Tính hợp lí
1) \((\)\(\dfrac{7}{-18}\)+\(\dfrac{-5}{12}\)) -\(\dfrac{13}{-18}\)
2) \(\dfrac{-13}{17}\)+(\(\dfrac{13}{-21}\)+\(\dfrac{-4}{17}\))
3) ( \(\dfrac{13}{-10}\)-\(\dfrac{-4}{13}\))+\(\dfrac{11}{-10}\)
4) \(\dfrac{13}{17}\)\(\times\)\(\dfrac{4}{-5}\)+\(\dfrac{13}{17}\)\(\times\)\(\dfrac{-3}{4}\)
5) \(\dfrac{-5}{12}\)\(\times\)\(\dfrac{7}{-17}\)\(\times\)\(\dfrac{9}{-20}\)
6) \(\dfrac{5}{9}\)\(\times\)\(\dfrac{11}{23}\)+ \(\dfrac{11}{23}\)\(\times\)\(\dfrac{17}{9}\)\(-\)\(\dfrac{13}{9}\)\(\times\)\(\dfrac{11}{23}\)
\(1.\dfrac{-7}{18}+\dfrac{-5}{12}-\dfrac{-13}{18}\text{=}\left(\dfrac{-7}{18}-\dfrac{-13}{18}\right)+\dfrac{-5}{12}\text{=}\dfrac{1}{3}+\dfrac{-5}{12}\text{=}\dfrac{-1}{12}\)
\(2.\dfrac{-13}{17}+\dfrac{-13}{21}+\dfrac{-4}{17}\text{=}\left(\dfrac{-13}{17}+\dfrac{-4}{17}\right)+\dfrac{-13}{21}\text{=}-1+\dfrac{-13}{21}\text{=}\dfrac{-34}{21}\)
\(3.\dfrac{-13}{10}-\dfrac{-4}{13}+\dfrac{-11}{10}\text{=}\dfrac{-12}{5}-\dfrac{-4}{13}\text{=}\dfrac{-136}{65}\)
\(4.\dfrac{13}{17}\times\left(\dfrac{-4}{5}+\dfrac{-3}{4}\right)\text{=}\dfrac{13}{17}\times\dfrac{-31}{20}\text{=}\dfrac{-403}{340}\)
\(5.\left(\dfrac{-5}{12}\times\dfrac{-9}{20}\right)\times\dfrac{-7}{17}\text{=}\dfrac{3}{16}\times\dfrac{-7}{17}\text{=}\dfrac{-21}{272}\)
\(6.\dfrac{11}{23}\times\left(\dfrac{5}{9}+\dfrac{17}{9}-\dfrac{13}{9}\right)\text{=}\dfrac{11}{23}\times1\text{=}\dfrac{11}{23}\)
\(\dfrac{3}{15}.\dfrac{5}{9}:\dfrac{-18}{17}.\dfrac{14}{17}\)
\(\dfrac{12}{7}.\dfrac{7}{4}+\dfrac{35}{11}:\dfrac{245}{121}\)
giúp mình với
a: \(=\dfrac{1}{3}\cdot\dfrac{1}{3}\cdot\dfrac{-17}{18}\cdot\dfrac{14}{17}=\dfrac{-14}{18\cdot9}=\dfrac{-14}{162}=\dfrac{-7}{81}\)
b: \(=\dfrac{12}{4}+\dfrac{35}{11}\cdot\dfrac{121}{245}=3+\dfrac{11}{7}=\dfrac{32}{7}\)
\(\dfrac{3}{15}.\dfrac{5}{9}:\dfrac{-18}{17}.\dfrac{14}{17}\)
\(\dfrac{12}{7}.\dfrac{7}{4}+\dfrac{35}{11}:\dfrac{245}{121}\)
giúp mình với
a: \(=\dfrac{15}{135}\cdot\dfrac{-17}{18}\cdot\dfrac{14}{17}=\dfrac{1}{9}\cdot\dfrac{-7}{9}=\dfrac{-7}{81}\)
b: \(=3+\dfrac{35}{11}\cdot\dfrac{121}{245}=3+\dfrac{11}{7}=\dfrac{32}{7}\)
So sánh:
a/ \(A=\dfrac{17^{18}+1}{17^{19}+1};B=\dfrac{17^{17}+1}{17^{18}+1}\)
b/ \(A=\dfrac{10^8-2}{10^8+2};B=\dfrac{10^8}{10^8+4}\)
c/ \(A=\dfrac{20^{10}+1}{20^{10}-1};B=\dfrac{20^{10}-1}{20^{10}-3}\)
GIÚP MÌNH VỚI
Giải:
a) A=1718+1/1719+1
17A=1719+17/1719+1
17A=1719+1+16/1719+1
17A=1+16/1719+1
Tương tự:
B=1717+1/1718+1
17B=1718+17/1718+1
17B=1718+1+16/1718+1
17B=1+16/1718+1
Vì 16/1719+1<16/1718+1 nên 17A<17B
⇒A<B
b) A=108-2/108+2
A=108+2-4/108+2
A=1+-4/108+2
Tương tự:
B=108/108+4
B=108+4-4/108+1
B=1+-4/108+1
Vì -4/108+2>-4/108+1 nên A>B
c)A=2010+1/2010-1
A=2010-1+2/2010-1
A=1+2/2010-1
Tương tự:
B=2010-1/2010-3
B=2010-3+2/2010-3
B=1+2/2010-3
Vì 2/2010-3>2/2010-1 nên B>A
⇒A<B
Chúc bạn học tốt!
a, \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)
b, \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)
Cảm ơn khi đã giúp mình
a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)
\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)
\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)
\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)
mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)
nên x-17=0
hay x=17
Vậy: x=17
b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)
\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)
\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)
\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)
mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)
nên x+20=0
hay x=-20
Vậy: x=-20
Bài 1.Tính
a)\(\dfrac{4}{28}\)+\(\dfrac{12}{36}\) b)\(\dfrac{-12}{18}\)+\(\dfrac{15}{-21}\) c)\(\dfrac{14}{28}\)-\(\dfrac{-16}{32}\)-\(\dfrac{17}{51}\)
\(a,\dfrac{1}{7}+\dfrac{1}{3}=\dfrac{3}{21}+\dfrac{7}{21}=\dfrac{10}{21}\\ b,\dfrac{-2}{3}+\dfrac{-5}{7}=\dfrac{-14+\left(-15\right)}{21}=\dfrac{-29}{21}\\ c,\dfrac{1}{2}-\dfrac{-1}{2}-\dfrac{1}{3}=\dfrac{3+3-2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
\(a.\dfrac{4}{28}+\dfrac{12}{36}=\dfrac{1}{7}+\dfrac{1}{3}=\dfrac{3}{21}+\dfrac{7}{21}=\dfrac{10}{21}\\ b.\dfrac{-12}{18}+\dfrac{-15}{21}=\dfrac{-2}{3}+\dfrac{-5}{7}=\dfrac{-14}{21}+\dfrac{-15}{21}=\dfrac{-29}{21}\\ c.\dfrac{14}{28}+\dfrac{16}{32}-\dfrac{17}{51}=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{17}{51}=1-\dfrac{17}{51}=\dfrac{2}{3}\)
\(\dfrac{4}{28}+\dfrac{12}{36}=\dfrac{144}{1008}+\dfrac{336}{1008}=\\ \dfrac{480}{1008}=\dfrac{10}{21}\)
Câu b và c em làm tương tự.