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13x - 130 = 65
=> 13x = 65 + 130
=> 13x = 195
=> x = 195 : 13
=> x = 15
Vậy x = 15
![](https://rs.olm.vn/images/avt/0.png?1311)
\(104-4.\left(x-150\right)=48\)
\(=>4.\left(x-150\right)=104-48\)
\(=>4.\left(x-150\right)=56\)
\(=>x-150=56:4\)
\(=>x-150=14\)
\(=>x=14+150\)
\(=>x=164\)
_________________
\(\left(13x+39\right)-65=26.4^0\)
\(=>\left(13x+39\right)-65=26.1=26\)
\(=>13x+39=26+65\)
\(=>13x+39=91\)
\(=>13x=91-39\)
\(=>13x=52\)
\(=>x=52:13\)
\(=>x=4\)
a) 104.4.(x-150) = 48
4.(x-150) = 104 - 48
4. (x-150) = 56
x-150 = 56 : 4
x-150 = 14
x = 14 + 150
x = 164
![](https://rs.olm.vn/images/avt/0.png?1311)
750 : {130 – [(5 . 14 – 65)3 + 3]}
= 750 : {130 – [(70 – 65)3 + 3]}
= 750 : {130 – [(5)3 + 3]}
= 750 : (130 – 128)
= 750 : 2
= 375
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(72.19 - 36^2:18= 72.19- 36.36:18\)
\(=72.19 - 72=72.(19 - 1)\)
\(=72.18=1296\)
b)
\(\begin{array}{l}750{\rm{ }}:{\rm{ }}\left\{ {130{\rm{ }} - {\rm{ }}[{{\left( {5.{\rm{ }}14{\rm{ }} - {\rm{ }}65} \right)}^3} + {\rm{ }}3} \right\}\\ = 750{\rm{ }}:{\rm{ }}\left\{ {130{\rm{ }} - {\rm{ }}\left[ {{{\left( {{\rm{70 }} - {\rm{ }}65} \right)}^3} + {\rm{ }}3} \right]} \right\}\\ = 750{\rm{ }}:{\rm{ }}\left\{ {130{\rm{ }} - {\rm{ }}\left[ {{5^3} + {\rm{ }}3} \right]} \right\}\\ = 750{\rm{ }}:{\rm{ }}\left\{ {130{\rm{ }} - {\rm{ }}\left[ {125 + {\rm{ }}3} \right]} \right\}\\ = 750{\rm{ }}:{\rm{ }}\left\{ {130{\rm{ }} - {\rm{ 128}}} \right\}\\ = 750:2 = 375.\end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Cho số m. Nếu chia m cho 65 dư 33, chia 13 dư 7và 130<m<300
m là 163
chúc bạn học tốt
k đúng cho mình nha
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Có :\(\widehat{xOz}< \widehat{xOy}\left(65^o< 130^o\right)\)
=> Oz nằm giữa Ox và Oy
\(\Rightarrow\widehat{xOz}+\widehat{yOz}=\widehat{xOy}\)
\(\Rightarrow65^o+\widehat{yOz}=130^o\)
\(\Rightarrow\widehat{yOz}=65^o\)
\(\Rightarrow\widehat{xOz}=\widehat{yOz}=\widehat{\frac{xOy}{2}}=\frac{130^o}{2}=65^o\)
=> Oz là tia phân giác góc xOy
b) Có \(\widehat{xOy}+\widehat{mOy}=\widehat{xOm}\)
\(\Rightarrow130^o+\widehat{mOy}=180^o\)
\(\Rightarrow\widehat{mOy}=50^o\)
Có :\(\widehat{xOz}+\widehat{mOz}=\widehat{xOm}\)
\(\Rightarrow65^o+\widehat{mOz}=180^o\)
\(\Rightarrow\widehat{mOz}=115^o\)
d) Do Ot là tia pg góc mOy
\(\Rightarrow\widehat{tOy}=\frac{\widehat{mOy}}{2}=\frac{50^o}{2}=25^o\)
Có \(\widehat{tOy}+\widehat{yOz}=\widehat{tOz}\)
\(\Rightarrow25^o+65^o=\widehat{tOz}\)
\(\Rightarrow\widehat{tOz}=90^o\)
=> tOz là góc vuông
#H
13x - 130 = 65
13x = 65 + 130
13x = 195
x = 195 : 13
x = 15
k mình nha
13x - 130 = 65
13x = 65 + 130
13x = 195
x = 195 : 13
x = 15
nha bn