\(x>25\Leftrightarrow\sqrt{x}>5\Leftrightarrow\sqrt{x}-5>0\) mới sử dụng được AM-GM nha

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\(P=\dfrac{x}{\sqrt{x}-5}=\dfrac{x-25}{\sqrt{x}-5}+\dfrac{25}{\sqrt{x}-5}=\sqrt{x}+5+\dfrac{25}{\sqrt{x}-5}=\left(\sqrt{x}-5+\dfrac{25}{\sqrt{x}-5}\right)+10\)

Áp dụng BĐT AM-GM có;
\(\sqrt{x}-5+\dfrac{25}{\sqrt{x}-5}\ge10\)

\(\Rightarrow P\ge10+10=20\)

Dấu "=" xảy ra khi\(\sqrt{x}-5=\dfrac{25}{\sqrt{x}-5}\)\(\Leftrightarrow x=100\)(tm)

Vậy...

Đề như này pk?

\(\dfrac{1+sinx}{1+cosx}=\dfrac{1}{2}\) (đk:\(cosx\ne-1\Leftrightarrow x\ne\pi+k2\pi\left(k\in Z\right)\))

\(\Leftrightarrow2\left(1+sinx\right)=1+cosx\)

\(\Leftrightarrow2sinx-cosx=-1\)

\(\Leftrightarrow\dfrac{2}{\sqrt{5}}sinx-\dfrac{1}{\sqrt{5}}cosx=-\dfrac{1}{\sqrt{5}}\)

Đặt \(cos\alpha=\dfrac{2}{\sqrt{5}}\Rightarrow sin\alpha=\dfrac{1}{\sqrt{5}}\) (vì \(cos^2\alpha+sin^2\alpha=1\))

\(\Rightarrow sin\left(-\alpha\right)=-\dfrac{1}{\sqrt{5}}\)

\(\Rightarrow sinx.cos\alpha-cosx.sin\alpha=sin\left(-\alpha\right)\)

\(\Leftrightarrow sin\left(x-\alpha\right)=sin\left(-\alpha\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\alpha=-\alpha+k2\pi\\x-\alpha=\pi+\alpha+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi+2\alpha+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)(thỏa mãn)

Vậy...

a)PQ \(\left\{{}\begin{matrix}quaP\left(1;-4\right)\\vtcp\overrightarrow{PQ}\left(1;7\right)\Rightarrow vtpt\overrightarrow{n}\left(7;-1\right)\end{matrix}\right.\)

\(\Rightarrow PQ:7x-y-11=0\)

b) Gọi pt đt tâm (O) có dạng (C):\(x^2+y^2=R^2\)

Do (C) tiếp xúc với đt \(2x+y-3=0\)

\(\Rightarrow R=d_{\left(O;\Delta\right)}=\dfrac{\left|2.0+0-3\right|}{\sqrt{2^2+1}}=\dfrac{3\sqrt{5}}{5}\)

\(\Rightarrow\left(C\right):x^2+y^2=\dfrac{9}{5}\)

c)\(I\in\left(\Delta\right)\Rightarrow I\left(t;3-2t\right)\)

\(IQ=R\Leftrightarrow\sqrt{\left(2-t\right)^2+4t^2}=3\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{2+\sqrt{29}}{5}\\t=\dfrac{2-\sqrt{29}}{5}\end{matrix}\right.\)\(\Rightarrow I\left(\dfrac{2+\sqrt{29}}{5};\dfrac{11-2\sqrt{29}}{5}\right);I\left(\dfrac{2-\sqrt{29}}{5};\dfrac{11+2\sqrt{29}}{5}\right)\)

Vậy pt đường tròn tâm I cần tìm là: \(\left(C\right)':\left(x-\dfrac{2+\sqrt{29}}{5}\right)^2+\left(y-\dfrac{11-2\sqrt{29}}{5}\right)^2=9\) hoặc \(\left(C\right)':\left(x-\dfrac{2-\sqrt{29}}{5}\right)^2+\left(y-\dfrac{11+2\sqrt{29}}{5}\right)^2=9\)

Có \(\widehat{ABD}+\widehat{A}=\widehat{A}+\widehat{ACE}=90^0\)

\(\Rightarrow\widehat{ABD}=\widehat{ACE}\)

\(\Rightarrow180^0-\widehat{ABD}=180^0-\widehat{ACE}\)

\(\Leftrightarrow\widehat{ABH}=\widehat{ACK}\)

Xét tam giác ABH và tam giác ACK có:

\(AB=CK\)

\(\widehat{ABH}=\widehat{ACK}\)

\(HB=AC\)

nên tam giác ABH= tam giác KCA (c.g.c)

\(\Rightarrow AH=AK\)