A O B C E D K H I

a) Do H là trung điểm ED nên \(OH⊥DE\) .

Theo tính chất hai tiếp tuyến cắt nhau ta cũng có \(OK⊥DC\)

Vậy thì \(\Delta HOA\sim\Delta IKA\left(g-g\right)\Rightarrow\frac{OA}{OK}=\frac{AH}{AI}\Rightarrow AI.AO=AK.AH\)

b) Ta thấy \(AD.AE=AB^2=AI.AO=AK.AH\)

Vậy nên \(\frac{1}{AD}+\frac{1}{AE}=\frac{AD+AE}{AD.AE}=\frac{AD+AE}{AH.AK}=\frac{2AH}{AH.AK}=\frac{2}{AK}.\)

\(\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)

\(\left(\sqrt{x}-\sqrt{y}\right)^2\)

\(x-1-2\sqrt{x-1}+1=\left(\sqrt{x-1}-1\right)^2\)

\(\left(\sqrt{15}x-4\right)^2\)

\(a\sqrt{a}-b\sqrt{b}\)

\(=\sqrt{a^3}-\sqrt{b^3}\)

\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\)

\(x+y-2\sqrt{xy}\)

\(=\left(\sqrt{x}-\sqrt{y}\right)^2\)

Ta có \(P=\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{8}-2}-\frac{\sqrt{15}-\sqrt{3}}{2-2\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{3}}\)

\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{2\left(1-\sqrt{5}\right)}\right).\left(\sqrt{7}-\sqrt{3}\right)\)

\(=\left(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}\right).\left(\sqrt{7}-\sqrt{3}\right)=\frac{\sqrt{7}+\sqrt{3}}{2}.\left(\sqrt{7}-\sqrt{3}\right)\)

\(=\frac{7-3}{2}=2\)

Vậy \(P=2\)

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Cho tam giác ABC vuông tại A, AH là đường cao. Biết AB=15cm,HC=16cm.Tính BC,AH,HB,AC.

\(\frac{1}{x^2-x}:\frac{\sqrt{x}-1}{x\sqrt{x}+x+\sqrt{x}}\)

\(x^2-3x+\frac{7}{2}=\sqrt{\left(x^2-2x+2\right)\left(x^2-4x+5\right)}\)

\(\Leftrightarrow2x^2-6x+7=2\sqrt{\left(x^2-2x+2\right)\left(x^2-4x+5\right)}\)

Đặt \(\hept{\begin{cases}\sqrt{x^2-2x+2}=a>0\\\sqrt{x^2-4x+5}=b>0\end{cases}}\)

\(\Rightarrow a^2+b^2=2ab\)

\(\Leftrightarrow\left(a-b\right)^2=0\)

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt{x^2-2x+2}=\sqrt{x^2-4x+5}\)

\(\Leftrightarrow2x=3\)

\(\Leftrightarrow x=\frac{3}{2}\)

\(x^2-3x+\frac{7}{2}=\sqrt{\left(x^2-2x+2\right)\left(x^2-4x+5\right)}\)

\(\Leftrightarrow x^2-3x+\frac{7}{2}-\frac{5}{4}=\sqrt{\left(x^2-2x+2\right)\left(x^2-4x+5\right)}-\frac{5}{4}\)

\(\Leftrightarrow\frac{4x^2-12x+9}{4}=\frac{\left(x^2-2x+2\right)\left(x^2-4x+5\right)-\frac{25}{16}}{\sqrt{\left(x^2-2x+2\right)\left(x^2-4x+5\right)}+\frac{5}{4}}\)

\(\Leftrightarrow\frac{\left(2x-3\right)^2}{4}-\frac{x^4-6x^3+15x^2-18x+10-\frac{25}{16}}{\sqrt{\left(x^2-2x+2\right)\left(x^2-4x+5\right)}+\frac{5}{4}}=0\)

\(\Leftrightarrow\frac{\left(2x-3\right)^2}{4}-\frac{\frac{16x^4-96x^3+240x^2-288x+135}{16}}{\sqrt{\left(x^2-2x+2\right)\left(x^2-4x+5\right)}+\frac{5}{4}}=0\)

\(\Leftrightarrow\frac{\left(2x-3\right)^2}{4}-\frac{\frac{\left(2x-3\right)^2\left(4x^2-12x+15\right)}{16}}{\sqrt{\left(x^2-2x+2\right)\left(x^2-4x+5\right)}+\frac{5}{4}}=0\)

\(\Leftrightarrow\left(2x-3\right)^2\left(\frac{1}{4}-\frac{\frac{4x^2-12x+15}{16}}{\sqrt{\left(x^2-2x+2\right)\left(x^2-4x+5\right)}+\frac{5}{4}}\right)=0\)

\(\Rightarrow x=\frac{3}{2}\)

Bài làm của mk cho ai khùng thôi, bn tham khảo cx dc :v