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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm BCNN(36; 48; 168)
36 = 22.32; 48 = 24.3; 168 = 23.3.7
BCNN(36; 48; 168) = 24.32.7 = 1008
Chọn D 1008
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải;
$x+5\vdots x^2+1(1)$
$\Rightarrow x(x+5)\vdots x^2+1$
$\Rightarrow (x^2+1)+(5x-1)\vdots x^2+1$
$\Rightarrow 5x-1\vdots x^2+1(2)$
Từ $(1); (2)\Rightarrow 5(x+5)-(5x-1)\vdots x^2+1$
$\Rightarrow 26\vdots x^2+1$
$\Rightarrow x^2+1\in \left\{1; 2; 13; 26\right\}$
$\Rightarrow x^2\in \left\{0; 1; 12; 25\right\}$
$\Rightarrow x\in \left\{0; \pm 1; \pm 5\right\}$ (do $x$ nguyên)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(a+8b⋮11\)
\(\Leftrightarrow2.\left(a+8b\right)⋮11\)
\(\Leftrightarrow2a+16b⋮11\)
\(\Leftrightarrow2a+5b+11b⋮11\)
Mà \(11b⋮11\)
nên \(2a+5b⋮11\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. $(3x+9)^{40}=49(3x+9)^{38}$
$(3x+9)^{40}-49(3x+9)^{38}$
$(3x+9)^{38}[(3x+9)^2-49]=0$
$\Rightarrow (3x+9)^{38}=0$ hoặc $(3x+9)^2-49=0$
Nếu $(3x+9)^{38}=0$
$\Rightarrow 3x+9=0$
$\Rightarrow x=-3$
Nếu $(3x+9)^2-49=0$
$\Rightarrow (3x+9)^2=49=7^2=(-7)^2$
$\Rightarrow 3x+9=7$ hoặc $3x+9=-7$
$\Rightarrow x=\frac{-2}{3}$ hoặc $x=\frac{-16}{3}$
b/
Xét $A=2^x+2^{x+1}+2^{x+2}+....+2^{x+2015}$
$2A=2^{x+1}+2^{x+2}+2^{x+3}+....+2^{x+2016}$
$\Rightarrow 2A-A=(2^{x+1}+2^{x+2}+2^{x+3}+....+2^{x+2016})-(2^x+2^{x+1}+2^{x+2}+....+2^{x+2015})$
$\Rightarrow A=2^{x+2016}-2^x$
Vậy $2^{x+2016}-2^x=2^{2019}-8$
$\Rightarrow 2^x(2^{2016}-1)=2^3(2^{2016}-1)$
$\Rightarrow 2^x=2^3$
$\Rightarrow x=3$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{-22}{10}=-\dfrac{22:2}{10:2}=-\dfrac{11}{5}\)
\(\dfrac{11}{-5}=-\dfrac{11}{5}\)
\(\Rightarrow\dfrac{-22}{10}=\dfrac{11}{-5}\)
b) Ta có: \(\dfrac{20}{-52}=-\dfrac{20:4}{52:4}=-\dfrac{5}{13}\)
Mà: \(-\dfrac{5}{13}\ne\dfrac{5}{-14}\)
\(\Rightarrow\dfrac{20}{-52}\ne\dfrac{5}{-14}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
C=(1-5-9+13)+(17-21-25+29)+...+(2013-2017-2021+2025)-2029
C=0+0+0+...+0-2029
C=-2029
C = 1 - 5 - 9 + 13 + 17 - 21 - 25 + 29 + ... + 2013 - 2017 - 2021 + 2025 - 2029
= (1 - 5 - 9 + 13) + (17 - 21 - 25 + 29) + ... + (2001 - 2005 - 2009 + 2013) - 2017 - 2021 + 2025 - 2029
= 0 + 0 + ... + 0 - 4042
= -4042
![](https://rs.olm.vn/images/avt/0.png?1311)
(-5)3.\(x^2\) = - 1125
\(x^2\) = (-1125) : (-53)
\(x^2\) = 9
\(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-3; 3}
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