\(a,\left\{{}\begin{matrix}IB=IA\\IA=IC\end{matrix}\right.\Rightarrow IB=IA=IC=\dfrac{1}{2}BC\\ \Rightarrow\Delta ABC\text{ vuông tại }A\\ \Rightarrow\widehat{BAC}=90^0\\ b,\left\{{}\begin{matrix}IO\text{ là p/g }\widehat{AIB}\\IO'\text{ là p/g }\widehat{AIC}\end{matrix}\right.\\ \text{Mà }\widehat{AIB}+\widehat{AIC}=180^0\\ \Rightarrow IO\bot IO'\Rightarrow\widehat{OIO'}=90^0\\ c,\text{Áp dụng HTL: }IA=\sqrt{OA\cdot O'A}=6\left(cm\right)\\ \Rightarrow BC=2IA=12\left(cm\right)\)

Ta có \(2^{p-1}\equiv1\left(\text{mod }p\right)\)

Ta có \(n.2^n\equiv m\left(p-1\right).2^{m\left(p-1\right)}\left(\text{mod }p\right)\Rightarrow n.2^n\equiv-m\equiv1\left(\text{mod }p\right)\)

\(\Rightarrow m=kp-1\left(k\in N\text{*}\right)\)

Vậy với \(n=\left(kp-1\right)\left(p-1\right)\left(k\in N\text{*}\right)\) thì \(n.2^n-1⋮p\)

\(a,=\left(6x^3+3x^2-10x^2-5x+4x+2\right):\left(2x+1\right)\\ =\left[3x^2\left(2x+1\right)-5x\left(2x+1\right)+2\left(2x+1\right)\right]:\left(2x+1\right)\\ =3x^2-5x+2\\ b,Sửa:\left(2x^3-21x^2+67x-60\right):\left(x-5\right)\\ =\left(2x^3-10x^2-11x^2+55x+12x-60\right):\left(x-5\right)\\ =\left[2x^2\left(x-5\right)-11x\left(x-5\right)+12\left(x-5\right)\right]:\left(x-5\right)\\ =2x^2-11x+12\)

Sửa: CMR: \(\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3=\dfrac{a^2}{bc}\)

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\\ \Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3\left(1\right)\\ \dfrac{a}{b}=\dfrac{b}{c}=k\Rightarrow a=bk;b=ck\Rightarrow a=ck^2\\ \Rightarrow\dfrac{a^2}{bc}=\dfrac{c^2k^4}{ck\cdot c}=k^3=\left(\dfrac{a}{b}\right)^3\left(2\right)\\ \left(1\right)\left(2\right)\RightarrowĐpcm\)

thiếu ý đầu rùi e :v