cần giúp gấp vs ạ, em cảm ơn
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\(2\left(\overrightarrow{IA}+\overrightarrow{AB}\right)+3\left(\overrightarrow{IA}+\overrightarrow{AC}\right)=\overrightarrow{0}\Leftrightarrow5\overrightarrow{IA}+2\overrightarrow{AB}+3\overrightarrow{AC}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{AI}=\dfrac{2}{5}\overrightarrow{AB}+\dfrac{3}{5}\overrightarrow{AC}\)
\(\overrightarrow{JB}+\overrightarrow{BA}+3\overrightarrow{JB}+3\overrightarrow{BC}=\overrightarrow{0}\Leftrightarrow\overrightarrow{BJ}=-\dfrac{1}{4}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{BC}=-\dfrac{1}{4}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{BA}+\dfrac{3}{4}\overrightarrow{AC}\)
\(=-\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{AI}.\overrightarrow{BJ}=\left(\dfrac{2}{5}\overrightarrow{AB}+\dfrac{3}{5}\overrightarrow{AC}\right)\left(-\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AC}\right)\)
\(=-\dfrac{2}{5}AB^2+\dfrac{9}{20}AC^2-\dfrac{3}{10}\overrightarrow{AB}.\overrightarrow{AC}\)
\(=-\dfrac{3}{5}a^2+\dfrac{9}{20}a^2-\dfrac{3}{10}a^2.cos60^0=-\dfrac{3}{10}a^2\)
b.
Từ câu a ta có
\(\overrightarrow{AI}=\dfrac{2}{5}\overrightarrow{AB}+\dfrac{3}{5}\overrightarrow{AC}\) (1)
\(\overrightarrow{JA}+3\overrightarrow{JC}=\overrightarrow{0}\Leftrightarrow\overrightarrow{JA}+3\overrightarrow{JA}+3\overrightarrow{AC}=\overrightarrow{0}\Leftrightarrow\overrightarrow{JA}=-\dfrac{3}{4}\overrightarrow{AC}\) (2)
Cộng vế (1) và (2):
\(\overrightarrow{JA}+\overrightarrow{AI}=-\dfrac{3}{4}\overrightarrow{AC}+\dfrac{2}{5}\overrightarrow{AB}+\dfrac{3}{5}\overrightarrow{AC}\)
\(\Leftrightarrow\overrightarrow{JI}=\dfrac{2}{5}\overrightarrow{AB}-\dfrac{3}{20}\overrightarrow{AC}\)
\(\Rightarrow IJ^2=\overrightarrow{JI}^2=\left(\dfrac{3}{5}\overrightarrow{AB}-\dfrac{3}{20}\overrightarrow{AC}\right)^2=\dfrac{9}{25}AB^2+\dfrac{9}{400}AC^2-\dfrac{9}{50}\overrightarrow{AB}.\overrightarrow{AC}\)
\(=\dfrac{9}{25}a^2+\dfrac{9}{400}a^2-\dfrac{9}{50}.a^2.cos60^0=...\)
m: \(\left(-7x+7\right)\left(2x+100\right)=0\)
=>\(\left[{}\begin{matrix}-7x+7=0\\2x+100=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}-7x=-7\\2x=-100\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-50\end{matrix}\right.\)
n: \(2x\left(x+2023\right)\left(-4x+8\right)=0\)
=>\(2\cdot x\left(x+2023\right)\cdot\left(-4\right)\left(x-2\right)=0\)
=>x(x-2)(x+2023)=0
=>\(\left[{}\begin{matrix}x=0\\x-2=0\\x+2023=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2023\end{matrix}\right.\)
o: \(-2024x\left(4x-4\right)\left(-2x+6\right)=0\)
=>\(x\left(4x-4\right)\left(-2x+6\right)=0\)
=>\(x\cdot4\left(x-1\right)\cdot\left(-2\right)\left(x-3\right)=0\)
=>x(x-1)(x-3)=0
=>\(\left[{}\begin{matrix}x=0\\x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=3\end{matrix}\right.\)
p: -17-2x=199
=>2x=-17-199=-216
=>x=-216/2=-108
q: -24+2x=100
=>2x=100+24=124
=>\(x=\dfrac{124}{2}=62\)
r: \(119-\left(x+5\right)=-21\)
=>\(x+5=119-\left(-21\right)=119+21=140\)
=>x=140-5=135
s: \(-24+\left(-7+x\right)=45\)
=>\(\left(x-7\right)-24=45\)
=>x-31=45
=>x=45+31=76
t: \(-146-\left(-5+x\right)=-6\)
=>\(x-5=-146-\left(-6\right)=-140\)
=>x=-140+5=-135
u: \(-29+4\left(1-5x\right)=-45\)
=>4(1-5x)=-45+29=-16
=>1-5x=-4
=>5x=1+4=5
=>\(x=\dfrac{5}{5}=1\)
v: \(24-5\left(-3+2x\right)=59\)
=>\(24-5\left(2x-3\right)=59\)
=>5(2x-3)=24-59=-35
=>2x-3=-7
=>2x=-4
=>x=-4/2=-2
-17-2x=199
<=>-2x=216
<=>x=-108
-24+2x=100
<=>2x=124
<=>x=62
-24+(-7+x)=45
<=>-7+x=69
<=>x=76
-146-(-5+x)=-6
<=>-146+5-x=-6
<=>x=-135
-29+4(1-5x)=-45
<=>-29+4+20x=-45
<=>20x=-20
<=>x=-1
24-5(-3+2x)=59
<=>24+15-10x=59
<=>-10x=20
<=>x=-2