a/ Tam giác AMN cân tại A (gt). \(\Rightarrow\) \(\widehat{AMN}=\widehat{ANM};AM=AN.\)

Xét tam giác AMB và tam giác ANC có:

+ AM = AN (cmt).

+ \(\widehat{AMB}=\widehat{ANC}\left(\widehat{AMN}=\widehat{ANM}\right).\)

+ MB = NC (gt).

\(\Rightarrow\) Tam giác AMB = Tam giác ANC (c - g - c).

\(\Rightarrow\) AB = AC (cặp cạnh tương ứng).

Xét tam giác ABC có: AB = AC (cmt).

\(\Rightarrow\) Tam giác ABC cân tại A.

b/ Tam giác ABC cân tại A (cmt) \(\Rightarrow\) \(\widehat{ABC}=\widehat{ACB}.\)

Mà \(\widehat{ABC}=\widehat{MBH;}\widehat{ACB}=\widehat{NCK}\text{​​}\) (đối đỉnh).

\(\Rightarrow\) \(\widehat{MBH}=\widehat{NCK}.\)

Xét tam giác MBH và tam giác NCK \(\left(\widehat{BHM}=\widehat{CKN}=90^o\right)\)có:

+ MB = NC (gt).

+ \(\widehat{MBH}=\widehat{NCK}\left(cmt\right).\)

\(\Rightarrow\) Tam giác MBH = Tam giác NCK (cạnh huyền - góc nhọn).

c/ Tam giác MBH = Tam giác NCK (cmt).

\(\Rightarrow\) \(\widehat{BMH}=\widehat{CNK}\) (cặp góc tương ứng).

Xét tam giác OMN có: \(\widehat{NMO}=\widehat{MNO}\) (do \(\widehat{BMH}=\widehat{CNK}\)).

\(\Rightarrow\) Tam giác OMN tại O.

a) \(A=x^4+4x+7=\left(x^2+4x+4\right)+3=\left(x+2\right)^2+3\ge3\)

\(minA=3\Leftrightarrow x=-2\)

b) \(B=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(C=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

\(maxC=7\Leftrightarrow x=2\)

d) \(D=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)

\(maxD=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)

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\(1,\\ a,=6x^4y^4-x^3y^3+\dfrac{1}{2}x^4y^2\\ b,=4x^3+5x^2-8x^2-10x+12x+15\\ =4x^3-3x^2+2x+15\\ 2,\\ a,=7\left(x^2-6x+9\right)=7\left(x-3\right)^2\\ b,=\left(x-y\right)^2-36=\left(x-y-6\right)\left(x-y+6\right)\\ 3,\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow x\left(x-0,6\right)\left(x+0,6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,6\\x=-0,6\end{matrix}\right.\)

Kẻ AH⊥BC

ta có: \(VP=AB^2+BC^2-2.AB.BC.cosB=AB^2+BC^2-2.AB.BC.\dfrac{BH}{AB}=AB^2+BC^2-2.BH.BC=AB^2-BH^2+BC^2-2.BH.BC+BH^2=AH^2+\left(BC-BH\right)^2=AH^2+CH^2=AC^2=VT\)

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