câu e và f cần giúp đỡ vs ạ
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\(21,\\ e,PT\Leftrightarrow\left|2x-5\right|=5-2x\Leftrightarrow\left[{}\begin{matrix}2x-5=5-2x\left(x\ge\dfrac{5}{2}\right)\\5-2x=5-2x\left(x< \dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\0x=0\left(tm\right)\end{matrix}\right.\\ \Leftrightarrow x\in R\\ f,\Leftrightarrow\left|x-\dfrac{1}{4}\right|=\dfrac{1}{4}-x\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{1}{4}-x\left(x\ge\dfrac{1}{4}\right)\\\dfrac{1}{4}-x=\dfrac{1}{4}-x\left(x< \dfrac{1}{4}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\left(tm\right)\\0x=0\left(tm\right)\end{matrix}\right.\\ \Leftrightarrow x\in R\)
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\(f,f\left(x\right)⋮g\left(x\right)\\ \Leftrightarrow4x^4-13x^3+23x^2+18x-k=\left(x+4\right)\cdot c\left(x\right)\)
Thay \(x=-4\left(\text{Bổ đề Bézout}\right)\)
\(\Leftrightarrow4\cdot\left(-4\right)^4-13\cdot\left(-4\right)^3+23\cdot\left(-4\right)^2+18\left(-4\right)-k=0\\ \Leftrightarrow1024+832+368-72-k=0\\ \Leftrightarrow k=2152\)
\(d,f\left(x\right)⋮g\left(x\right)\\ \Leftrightarrow x^4-8x^3+24x^2+7x+k=\left(x+4\right)\cdot a\left(x\right)\)
Thay \(x=-4\left(\text{Bổ đề Bézout}\right)\)
\(\Leftrightarrow\left(-4\right)^4-8\left(-4\right)^3+24\left(-4\right)^2+7\left(-4\right)+k=0\\ \Leftrightarrow256+512+384-28+k=0\\ \Leftrightarrow k=-1124\)
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\(e,\left(x-2\right)^2-16=0\\ \Leftrightarrow\left(x-6\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\\ f,x^2-5x-14=0\\ \Leftrightarrow\left(x-7\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ g,8x\left(x-3\right)+x-3=0\\ \Leftrightarrow\left(8x+1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{8}\\x=3\end{matrix}\right.\)
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xet 2 tg vuong aem va afm = nhau vi e = f =90o ; am chung; a1 =a2 vi t/c tg can
nen bạn co AE= AF bạn suy ra tg AEF cân vay AM la dg trg truc ( t/c tg cân)