Ai giúp mình bài này với ạ <3
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\(3x+2.\left(x-3\right)-x=2010\)
\(\Leftrightarrow3x+2x-6-x=2010\)
\(\Leftrightarrow4x-6=2010\)
\(\Leftrightarrow4x=2016\)
\(\Leftrightarrow x=504\)
Vậy...
\(3\times+2\left(\times-3\right)-\times=2010\)
\(\Rightarrow2\times+2\times-6=2010\)
\(\Rightarrow4\times-6=2010\)
\(\Rightarrow4\times=2016\)
\(\Rightarrow\times=504\)
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\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)
Ta lấy vễ trên chia vế dưới
\(=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)
Ta lấy vế trên chia vế dưới
\(=2^3.3=24\)
\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)
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1: \(\overrightarrow{AG}=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)
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Lời giải:
$a+b+c=0$
$\Rightarrow a+b=-c$
$\Rightarrow (a+b)^2=(-c)^2$
$\Rightarrow a^2+b^2-c^2=-2ab$
$\Rightarrow \frac{ab}{a^2+b^2-c^2}=\frac{ab}{-2ab}=\frac{-1}{2}$
Tương tự với các phân thức còn lại suy ra:
$A=\frac{-1}{2}+\frac{-1}{2}+\frac{-1}{2}=\frac{-3}{2}$