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![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
\(-\frac{80}{49}< -\frac{49}{49}=1=-\frac{73}{73}< -\frac{29}{73}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x + 5) / 95 + (x +10)/90 + (x + 15)/85 + (x + 20)/80 = -4
<=> (x + 5)/95 + (x + 5)/90 + 5/90 + (x + 5)/85 + 10/85+ (x + 5)/80 + 15/80 = -4
<=> (x + 5)(1/95+1/90+1/85+1/80) =-4 -5/90-10/85-15/85
<=> (x + 5)(1/95+1/90+1/85+1/80)= -1-(1 + 5/90 )-(1 + 10/85) - (1 + 15/80)
<=>(x + 5)(1/95+1/90+1/85+1/80) = -1 - 95/90 - 95/85 - 95/80
<=>(x + 5)(1/95+1/90+1/85+1/80) = -95 (1/95+1/90+1/85+1/80)
<=> x + 5 = -95 => x = -100
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{-29}{73}=\frac{-29\cdot49}{73\cdot49}=\frac{-1421}{3577}\)
\(\frac{-80}{49}=\frac{-80\cdot73}{49\cdot73}=\frac{-5840}{3577}\)
Vì \(\frac{-1421}{3577}>\frac{-5840}{3577}\) nên \(-\frac{29}{73}>\frac{-80}{49}\)
Cách 2:
Ta có: \(-\frac{29}{73}>-1\)
\(-\frac{80}{49}< -1\)
Do đó: \(-\frac{80}{49}< -1< \frac{-29}{73}\)
hay \(-\frac{80}{49}< \frac{-29}{73}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
D={6;7;8;...;73}
Tập D có số phần tử là:
(73-6) :1+1=67(phần tử)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a. $=(-25)(-4)(-35)=100(-35)=-3500$
b. $=16-10=6$
c. $=180-(-16)-(-36)=180+16+36=232$
d. $=250-200:[1(-3)^2+(-8)]$
$=250-200:(9-8)=250-200=50$
2.
$60+2(12-x)=-48$
$2(12-x)=60-(-48)=60+48=108$
$12-x=108:2=54$
$x=12-54=-42$
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a. $=(-25)(-4)(-35)=100(-35)=-3500$
b. $=16-10=6$
c. $180-(-16)-(-36)=180+16+36=196+36=232$
d. $=250-200:[2000.(-3).2-6]$
$=250-200:[2000.(-6)+(-6)]$
$=250-200:[(-6)(2000+1)]=250-200[(-6).2001]$
$=250+200.6.2001=250+2401200=2401450$
Bài 2:
$60+2(12-x)=-48$
$2(12-x)=-48-60=-108$
$12-x=-108:2=-54$
$x=12-(-54)=66$
![](https://rs.olm.vn/images/avt/0.png?1311)
B= {73;74;75:76;77;78;79}
b= {73;74;75:76;77;78;79}