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\(\dfrac{-11}{3^7\cdot7^3}=\dfrac{1}{3^7\cdot7^3}\cdot\left(-11\right)\)
\(\dfrac{-78}{3^7\cdot7^4}=\dfrac{-78}{3^7\cdot7^3\cdot7}=\dfrac{1}{3^7\cdot7^3}\cdot\dfrac{-78}{7}\)
mà \(-11>-\dfrac{78}{7}\)
nên \(\dfrac{-11}{3^7\cdot7^3}>\dfrac{-78}{3^7\cdot7^4}\)
Lời giải:
a. $\frac{3}{-7}=\frac{-27}{63}$
$\frac{-5}{9}=\frac{-35}{63}$
Do $\frac{27}{63}< \frac{35}{63}$ nên $\frac{-27}{63}> \frac{-35}{63}$
$\Rightarrow \frac{3}{-7}> \frac{-5}{9}$
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b.
$-0,625=\frac{-625}{1000}=\frac{-5}{8}=\frac{-125}{200}$
$\frac{-19}{50}=\frac{-76}{200}> \frac{-125}{200}$
$\Rightarrow -0,625> \frac{-19}{50}$
c.
$-2\frac{5}{9}=-(2+\frac{5}{9})=\frac{-23}{9}=-(\frac{-23}{-9})$
Câu 1 :
\(\dfrac{-25}{37}\&\dfrac{-20}{31}\)
Ta thấy \(\dfrac{-25}{37}< \dfrac{-20}{37}\)
mà \(\dfrac{-20}{37}< \dfrac{-20}{31}\)
\(\Rightarrow\dfrac{-25}{37}< \dfrac{-20}{31}\)
Câu 2 :
\(\dfrac{2}{3}\&\dfrac{5}{7}\)
\(\dfrac{2}{3}:\dfrac{5}{7}=\dfrac{2}{3}.\dfrac{7}{5}=\dfrac{14}{15}< 1\)
\(\Rightarrow\dfrac{5}{7}>\dfrac{2}{3}\) Câu 3 : \(\dfrac{8}{13}\&\dfrac{5}{7}\)Ta thấy \(\dfrac{8}{13}:\dfrac{5}{7}=\dfrac{8}{13}.\dfrac{7}{5}=\dfrac{56}{65}< 1\)
\(\Rightarrow\dfrac{8}{13}< \dfrac{5}{7}\)
\(\dfrac{12}{1.4.7}+\dfrac{12}{4.7.10}+\dfrac{12}{7.10.13}+...+\dfrac{12}{54.57.60}\)
\(=2\left(\dfrac{1}{1.4}-\dfrac{1}{4.7}+\dfrac{1}{4.7}-\dfrac{1}{7.10}+\dfrac{1}{7.10}-\dfrac{1}{10.13}+...+\dfrac{1}{54.57}-\dfrac{1}{57.60}\right)\)\(=2\left(\dfrac{1}{1.4}-\dfrac{1}{57.60}\right)\)
\(=2\left(\dfrac{1}{4}-\dfrac{1}{57.60}\right)=\dfrac{1}{2}-\dfrac{1}{2.57.60}< \dfrac{1}{2}\left(đpcm\right)\)
1)mik ko biết trục số ở đâu nên tham khảo:
2
-0,75 <5/3
Ta có: \(\frac{-11}{3^7.7^3}=\frac{-11}{\frac{3^7.7^4.1}{7}}=-\frac{77}{3^7.7^4}=\frac{-78+1}{3^7.7^4}=-\frac{78}{3^7.7^4}+\frac{1}{3^7.7^4}\)
Do \(3^7.7^4>3^4.7^4\) => \(\frac{78}{3^7.7^4}< \frac{78}{3^4.7^4}\) => \(-\frac{78}{3^7.7^4}>-\frac{78}{3^4.7^4}\)=> \(-\frac{78}{3^7.7^4}+\frac{1}{3^7.7^4}>-\frac{78}{3^4.7^4}\)
=> \(-\frac{11}{3^7.7^4}>-\frac{78}{3^4.7^4}\)
Ta có: \(\frac{-1987}{-1986}=\frac{1986+1}{1986}=1+\frac{1}{1986}>1\)
\(\frac{-1984}{-1985}=\frac{1985-1}{1985}=1-\frac{1}{1985}< 1\)
=> \(\frac{-1987}{-1986}>\frac{-1984}{-1985}\)
Ta có: \(\frac{x}{5}< \frac{5}{4}< \frac{x+2}{5}\)
<=> \(x< \frac{25}{4}< x+2\)
Xét: \(x+2>\frac{25}{4}\) => \(x>\frac{17}{4}\)
=> \(\frac{17}{4}< x< \frac{25}{4}\)
Do x thuộc Z => x \(\in\){5; 6}
a: -3/100=-9/300; -2/3=-200/300
=>-3/100>-2/3
b: -3/5=-9/15
-2/3=-10/15
=>-3/5>-2/3
c: -5/4<-1<-3/8
d: -2/3=-8/12; -3/4=-9/12
=>-2/3>-3/4
e: -267/268>-1
-1>-1347/1343
=>-267/268>-1347/1343
a,
Ta có:
\(\dfrac{3}{7}=1-\dfrac{4}{7}\)
\(\dfrac{11}{15}=1-\dfrac{4}{15}\)
So sánh phân số \(\dfrac{4}{7}\) và \(\dfrac{4}{15}\)
Vì \(7< 15\) nên \(\dfrac{1}{7}>\dfrac{1}{15}\)
\(\Rightarrow1-\dfrac{4}{7}< 1-\dfrac{4}{15}\)
Vậy \(\dfrac{3}{7}< \dfrac{11}{15}\)
b)
\(\dfrac{-11}{6}< -1< \dfrac{-8}{9}\) nên \(\dfrac{-11}{6}< \dfrac{-8}{9}\)
c)
\(\dfrac{305}{25}=\dfrac{305:5}{25:5}=\dfrac{61}{5}\)
Ta có:
Mẫu số chung 2 phân số: 80
\(\dfrac{297}{16}=\dfrac{297*5}{16*5}=\dfrac{1485}{80}\)
\(\dfrac{61}{5}=\dfrac{61*16}{5*16}=\dfrac{976}{80}\)
Vì \(1485>976\) nên\(\dfrac{1485}{80}>\dfrac{976}{80}\)
Vậy \(\dfrac{297}{16}>\dfrac{305}{25}\)
d,
$\frac{-205}{317}=\frac{-205:-1}{317:-1}=\frac{205}{-317}$
Ta có:
Mẫu số chung 2 phân số: -35187
\(\dfrac{205}{-317}=\dfrac{205*111}{-317*111}=\dfrac{22755}{-35187}\)
\(\dfrac{-83}{111}=\dfrac{-83*-317}{111*-317}=\dfrac{26311}{-35187}\)
Vì \(22755< 26311\) nên\(\dfrac{22755}{-35187}< \dfrac{26311}{-35187}\)
Vậy \(\dfrac{-205}{317}< \dfrac{-83}{111}\)
Câu d, mình làm sai, cho mình sửa lại:
\(\dfrac{-205}{317}=\dfrac{-22755}{35187}\)
\(\dfrac{-83}{111}=\dfrac{-26311}{35187}\)
Vậy là \(-22755>-26311\) hay \(\dfrac{-205}{317}>\dfrac{-83}{111}\)
-11>-78
nên \(-\dfrac{11}{3^7\cdot7^4}>-\dfrac{78}{3^7\cdot7^4}\)