Sử dụng tính chất bắc cầu để so sánh các phân số sau:
a) \(\dfrac{1997}{1996}và\dfrac{1996}{1997}\)
b) \(\dfrac{3}{5}và\dfrac{15}{13}\)
Rút gọn phân số:
a) \(\dfrac{2929-101}{2.1149+404}\)
b) \(\dfrac{6.9-2.17}{63.3-119}\)
c) \(\dfrac{3.13-13.18}{15.40-80}\)
d) \(\dfrac{-1997-1996+1}{\left(-1995\right).\left(-1997\right)+1996}\)
e) \(\dfrac{2.3+4.6+14.21}{3.5+6.10+21.35}\)
g) \(\dfrac{\left(-5\right)^3.40.4^3}{135.\left(-3\right)^{14}.\left(-100\right)^0}\)
h) \(\dfrac{18.34+\left(-18\right).124}{-36.17+9.\left(52\right)}\)
j) \(\dfrac{9.11+32.9}{23.15+12.23}\)
k) \(\dfrac{12.13+24.26+36.39}{24.26+48.52+72.78}\)
b) \(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}\)
\(=\dfrac{2\left(3\cdot9-17\right)}{7\cdot\left(3\cdot9-17\right)}\)
\(=\dfrac{2}{7}\)
So sánh
\(\dfrac{1}{3993}\) và \(\dfrac{1997^2-1996^2}{1997^2+1996^2}\)
Ta có :
\(\dfrac{1997^2-1996^2}{1997^2+1996^2}=\dfrac{1.\left(1997+1996\right)}{1997^2+1996^2}=\dfrac{3993}{1997^2+1996^2}\)
Lại có : \(\dfrac{1}{3993}=\dfrac{3993}{3993^2}\)
Do \(3993^2=\left(1997+1996\right)^2>1997^2+1996^2\)
\(\Rightarrow\dfrac{3993}{3993^2}< \dfrac{3993}{1997^2+1996^2}\)
\(\Rightarrow\dfrac{1}{3993}< \dfrac{1997^2-1996^2}{1997^2+1996^2}\)
rút gọn phân số:
\(a.\dfrac{-315}{540}\)
\(b.\dfrac{25.13}{26.35}\)
\(c.\dfrac{3.13-13.18}{15.40-80}\)
\(d.\dfrac{-1997.1996+1}{\text{(}-1995\text{)}.\left(-1997\right)+1996}\)
\(\dfrac{x+1}{1998}\)+ \(\dfrac{x+2}{1997}\)=\(\dfrac{x+3}{1996}\)+\(\dfrac{x+4}{1995}\)
=>(x+1/1998+1)+(x+2/1997+1)=(x+3/1996+1)+(x+4/1995+1)
=>x+1999=0
=>x=-1999
Tính nhanh
\(\dfrac{1996\cdot1997+1998\cdot3+1994}{1997\cdot1999-1997\cdot1997}\)
\(\dfrac{1996\cdot1997+\left(1997+1\right)\cdot3+1994}{1997\cdot\left(1999-1997\right)}\) = \(\dfrac{1996\cdot1997+3\cdot1997+1997}{1997\cdot2}\)
\(\dfrac{1997\cdot\left(1996+3+1\right)}{1997\cdot2}\) = \(\dfrac{2000}{2}\)= \(1000\)
So sánh các phân số sau:
a) \(\dfrac{7}{{10}}\) và \(\dfrac{{11}}{{15}}\)
b) \(\dfrac{{ - 1}}{8}\) và \(\dfrac{{ - 5}}{{24}}\)
a)
Ta có: \(BCNN\left( {10,15} \right) = 30\) nên
\(\begin{array}{l}\dfrac{7}{{10}} = \dfrac{{7.3}}{{10.3}} = \dfrac{{21}}{{30}}\\\dfrac{{11}}{{15}} = \dfrac{{11.2}}{{15.2}} = \dfrac{{22}}{{30}}\end{array}\)
Vì \(21 < 22\) nên \(\dfrac{{21}}{{30}} < \dfrac{{22}}{{30}}\) do đó \(\dfrac{7}{{10}} < \dfrac{{11}}{{15}}\).
b)
Ta có: \(BCNN\left( {8,24} \right) = 24\) nên
\(\dfrac{{ - 1}}{8} = \dfrac{{ - 1.3}}{{8.3}} = \dfrac{{ - 3}}{{24}}\)
Vì \( - 3 > - 5\) nên \(\dfrac{{ - 3}}{{24}} > \dfrac{{ - 5}}{{24}}\) do đó \(\dfrac{{ - 1}}{8} > \dfrac{{ - 5}}{{24}}\).
Đề bài: So sánh các số hữu tỉ sau:
a)\(\dfrac{-13}{40}và\dfrac{12}{-40}\)
b)\(\dfrac{-5}{6}và\dfrac{-91}{104}\)
c)\(\dfrac{-15}{21}và\dfrac{-36}{44}\)
d)\(\dfrac{-16}{30}và\dfrac{-35}{84}\)
e)\(\dfrac{-5}{91}và\dfrac{-501}{9191}\)
f)\(\dfrac{-11}{3^7.7^3}và\dfrac{-78}{3^7.7^4}\)
giúp mik nha!!!
a: \(\dfrac{-13}{40}< \dfrac{-12}{40}\)
\(\dfrac{-5}{6}>\dfrac{-91}{104}\)
Bài 1:
a) Không quy đồng hãy so sánh: b) Tính nhanh:
\(\dfrac{2003}{2001}\) và \(\dfrac{1999}{1997}\) \(\dfrac{5}{9}\) x \(\dfrac{1}{4}\) +\(\dfrac{4}{9}\) x\(\dfrac{3}{12}\)
2003 / 2001 = 1 + 2/2001
1999/1997 = 1 + 2/1997
vì 2/ 2001 < 2/1997
nên 1 + 2/2001 < 1 + 2/1997
hay 2003 < 1999/1997
b, = 5/9 x 1/4 + 4/9 x 1/4
= 1/4 x ( 5/9 + 4/9 )
= 1/4 x 1
= 1/4
* Ý a mk k nhớ cách làm ^^, xl *
\(b,\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{3}{12}\)
\(=\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{1}{4}\)
\(=\dfrac{1}{4}\times\left(\dfrac{5}{9}+\dfrac{5}{9}\right)\)
\(=\dfrac{1}{4}\times\dfrac{9}{9}=\dfrac{1}{4}\times1=\dfrac{1}{4}\)
so sánh hai phân số : 1997/1996 và 1998/1997
$1997/1996$ $>$ $1998/1997$
$=>$ Phân số nào có mẫu số bé hơn thì phân số đó lớn hơn.