2111-x/100+2231-x/110+2371-x/120+1231-x/130
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kẻ Oz//Ax thì \(\widehat{AOz}=180-\widehat{xAO}=50\\ BOz=AOB-AOz=120-50=70\)
suy ra BOz và OBy bù nhau nên Oz//By
mà Oz//Ax nên ta có đpcm
a)50 + x = 180 - 120
50 + x = 60
x = 60-50
x = 10
b)20-x = 150-140
20-x = 10
x = 20-10
x = 10
c) 25+x = 100-40
25+x=60
x= 60-25
x=35
d) 100+x =130-25
100+x = 105
x = 105-100
x = 5
e) 143+x = 174-22
143+x = 152
x = 152-143
x = 9
ỦNG HỘ MK NHÁ
120 + ( 50 + x ) = 180
50 + x = 180 - 120
50 + x = 60
x = 60 - 50
x = 10
140 + ( 20 - x ) = 150
20 - x = 150 - 140
20 - x = 10
x = 20 - 10
x = 10
100 - ( 25 + x ) = 40
25 + x = 100 - 40
25 + x = 60
x = 60 - 25
x = 35
130 - ( 100 + x ) = 25
100 + x = 130 - 25
100 + x = 105
x = 105 - 100
x = 5
174 - ( 143 + x ) = 22
143 + x = 174 - 22
143 + x = 152
x = 152 - 143
x = 9
k mk nha
Thanks nhìu
a) 120 + ( 50 + x ) = 180
50 + x = 180 - 120
50 + x = 60
x = 60 - 50
x = 10
b) 130 - ( 100 + x )= 25
100 + x = 130 - 25
100 + x = 105
x = 105 - 100
x = 5
a) 50 + x = 180 - 120
50 + x = 60
x = 60 - 50
x = 10
b)100 + x = 130 - 25
100 + x = 105
x = 105 - 100
x = 5.
Ta có : \(\frac{10-x}{100}+\frac{20-x}{110}+\frac{30-x}{120}=3\)
<=> \(\frac{10-x}{100}+\frac{20-x}{110}+\frac{30-x}{120}-3=0\)
<=> \(\left(\frac{10-x}{100}-1\right)+\left(\frac{20-x}{110}-1\right)+\left(\frac{30-x}{120}-1\right)\)= 0
<=> \(\left(\frac{-90-x}{100}\right)+\left(\frac{-90-x}{110}\right)+\left(\frac{-90-x}{120}\right)=0\)
<=> (-90-x) \(\left(\frac{1}{100}+\frac{1}{110}+\frac{1}{120}\right)=0\)
<=> -90- x = 0 vì \(\left(\frac{1}{100}+\frac{1}{110}+\frac{1}{120}\right)\ne0\) ( > 0)
<=> -x = 90
<=> x = -90
Vậy x = -90
(10-x)/100+(20-x)/110+(30-x)/120=3
=>(10-x)/100+(20-x)/110+(30-x)/120-3=0
=>(10-x)/100-1+(20-x)/110-1+(30-x)/120-1=0
=>(-90-x)/100+(-90-x)/110+(-90-x)/120=0
=.>(-90-x)(1/100+1/110+1/120)=0
=.>(-90-x)=0(vì(1/100+1/110+1/120)luôn>0)
=>x=-90