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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

\(\left(2+\sqrt{6}\right)\sqrt{7-4\sqrt{3}}\)

\(=\left(2+\sqrt{6}\right)\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\left(2+\sqrt{6}\right)\left(2-\sqrt{3}\right)\)

\(=4-2\sqrt{3}+2\sqrt{6}-3\sqrt{2}\)

\(B=x^2-2xy+2y^2-4y=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)-4\)

\(=\left(x-y\right)^2+\left(y-2\right)^2-4\ge-4\)

\(pt\Leftrightarrow\sqrt{\left(x+3\right)\left(x-3\right)}+\sqrt{\left(x+3\right)^2}=0\)

\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x-3}+\sqrt{x+3}\right)=0\)

Suy ra x=-3 là nghiệm pt trong ngoặc >0 

cảm ơn bạn...bài này mình làm r ạ!!