x + 12 = 59 - 47
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47.(59 - 71) - 71.( -59 - 47) + 72.59
= 47.59 - 47.71 + 71.59 + 71.47 + 72.59
= (47.59 + 71.59 + 72.59 ) - (47.71 - 47.71)
= 59.(47 + 71 + 72) - 0
= 59.190
= 11210
47. (-12)-71.(-106)+72.59
=-564-7526+4248
=-8087+4248
=-3839
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3:
a: (a-b+c)-(c-b-a)
=a-b+c-c+b+a
=2a
b: \(-\left(a-b\right)+\left(b-c+a\right)-\left(a+b-c\right)\)
\(=-a+b+b-c+a-a-b+c\)
\(=b-a\)
c: \(-\left(a-b-c\right)-\left(-a+b+c\right)-\left(a-b+c\right)\)
\(=-a+b+c+a-b-c-a+b-c\)
\(=-a+b-c\)
2:
a: \(12-\left(-12\right)=12+12=24\)
b: \(-\left(35-49\right)+\left(27-49\right)=-35+49+27-49\)
=-35+27
=-8
c: \(47-\left(59-63\right)+\left(63-47\right)\)
\(=47-59+63+63-47\)
=126-59
=67
d: \(-\left(-20\right)-\left(-30\right)-70\)
=20+30-70
=50-70
=-20
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a) \(\dfrac{15}{59}>\dfrac{15}{60}=\dfrac{1}{4}\)
\(\dfrac{24}{97}< \dfrac{24}{96}=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{15}{59}>\dfrac{24}{97}\)
b) \(\dfrac{12}{47}>\dfrac{12}{48}=\dfrac{1}{4}\)
\(\dfrac{19}{77}< \dfrac{19}{76}=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{12}{47}>\dfrac{19}{77}\)
a: \(\dfrac{15}{59}>\dfrac{15}{60}=\dfrac{1}{4}\)
\(\dfrac{1}{4}=\dfrac{24}{96}>\dfrac{24}{97}\)
Do đó: \(\dfrac{15}{59}>\dfrac{24}{97}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\Leftrightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1+\dfrac{51-x}{49}+1=0\)
=>100-x=0
hay x=100
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: (59-x)/41 +(57-x)/43 +(55-x)/45 +(53-x)/47 +(51-x)/49 =-5
<=>[(59-x)/41 +1 ] +[(57-x)/43 +1] +[(55-x)/45 +1] +[(53-x)/47 +1] +[(51-x)/49 +1] =0
<=>(59-x-41)/41 + (57-x-43)/43 +(55-x-45)/45 +(53-x-47)/47 +(51-x-49)/49 =0
<=>(100-x)/41 + (100-x)/43 + (100-x)/45 +(100-x)/47 + (100-x)/49 =0
<=>(100-x).( 1/41 + 1/43 + 1/45 + 1/47 + 1/49 ) =0
mà (1/41 + 1/43 + 1/45 + 1/47 + 1/49) khác 0 nên 100-x =0 <=>x=100
vậy nghiệm của pt là x=100
x = 0
K mình mình sẽ K lại
x + 12 = 59 - 47
x + 12 = 12
x = 12 - 12
x = 0