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chết mình gửi nhầm đề bài là thế này
tính : 11 + 13 + 15 +17 +........+97 + 99 = .......
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
Ta có: \(\dfrac{y-5}{7-y}=\dfrac{2}{-3}\)
\(\Rightarrow\left(y-5\right).\left(-3\right)=2\left(7-y\right)\)
\(\Rightarrow-3y+15=14-2y\)
\(\Rightarrow-3y+2y=-15+14\)
\(\Rightarrow-1y=-1\)
Vậy y=1
Ta có:y-5/7-y=2/-3
=>(y-5).(-3)=(7-y).2
=>-3y+15=14-2y
=>-3y+2y=14-15
=>-y=-1
=>y=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2^{19}.27^3+15.4^9.9^4\right):\left(6^9.2^{10}+12^{10}\right)\)
\(=\left[2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4\right]:\left[2^9.3^9.2^{10}+2^{10}.6^{10}\right]\)
\(=\left(2^{19}.3^9+3.5.2^{18}.3^8\right):\left(2^{19}.3^9+2^{10}.2^{10}.3^{10}\right)\)
\(=\left(2^{19}.3^9+5.3^9.2^{18}\right):\left(2^{19}.3^9+2^{20}.3^{10}\right)\)
\(=2^{18}.3^9.\left(1.2+5\right):2^{19}.3^9.\left(1+2.3\right)\)
\(=\left(2^{18}.3^9.7\right):\left(2^{18}.2.3^9.7\right)\)
\(=1:2\)
\(=0.5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{-24}{-6}=\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}=4\)
=>x=12; y2=1; z3=-8
=>x=12; \(y\in\left\{1;-1\right\}\); z=-2
b: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{y}{-3}=\dfrac{z}{-17}=\dfrac{t}{9}\)
=>x/5=y/-3=z/-17=t/9=-2
=>x=-10; y=6; z=34; t=-18
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì A = \(\overline{155a710b4c16}\) \(⋮\) 11 nên (1+5+7+0+4+1) - (5+a+1+b+c+6) \(⋮\) 11
18 - 12 - (a+b+c) \(⋮\) 11
6 - (a+b+c) \(⋮\) 11
suy ra: (a+b+c)\(\in\){6; 17; 28;...}
Vì a; b; c < 5 hay a+b+c < 15 nên a+b+c = 6.
Vậy a+b+c = 6.
![](https://rs.olm.vn/images/avt/0.png?1311)
x O y z t
Vì Oz là tia đối của Ox
Ot là tia đối của Oy
Nên \(\widehat{xOy}=\widehat{zOt}\) ( đối đỉnh)
Mà \(\widehat{xOy}=80^o\)
Suy ra : \(\widehat{zOt}=80^o\)
Lời giải:
$A=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{97}+3^{98}+3^{99})$
$=(3+3^2+3^3)+3^3(3+3^2+3^3)+....+3^{96}(3+3^2+3^3)$
$=(3+3^2+3^3)(1+3^3+...+3^{96})$
$=39(1+3^3+...+3^{96})\vdots 39$
Tức là số dư là 0