mình xin rút khỏi câu lạc bộ

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\(16=\sqrt{16^2}\)

\(16^2=\left(15+1\right).\left(17-1\right)=15.17-15+17=15.17+2\)

Mà \(15.17+2>15.17\)

\(\Leftrightarrow\sqrt{15.17+2}>\sqrt{15.17}\)

\(\Leftrightarrow16>\sqrt{15}.\sqrt{17}\)

a) \(\text{Xét }\Delta\text{ vuông }HBA\text{ và }\Delta\text{ vuông }ABC,\text{có: }\)

\(\widehat{BHA}=\widehat{BAC}=90^o\)

\(\widehat{ABC}\text{ chung }\)

\(\Rightarrow\Delta\text{ vuông }HBA\text{ đồng dạng }\Delta\text{ vuông }ABC\)

\(\Rightarrow\frac{HB}{AB}=\frac{HA}{AC}=\frac{AB}{BC}\)

\(\frac{BH}{AB}=\frac{AB}{BC}\Leftrightarrow BC.HB=AB^2\)

b) cm tương tự câu a:

t/g vuông BAC đồng dạng t/g vuông AHC

\(\Rightarrow\frac{AB}{AH}=\frac{BC}{AC}=\frac{AC}{HC}\)

\(\frac{BC}{AC}=\frac{AC}{HC}\Leftrightarrow AC^2=BC.HC\)

c) \(\frac{HA}{AC}=\frac{AB}{BC}\left(\text{câu a}\right)\)

\(\Leftrightarrow HA.BC=AC.AB\)

d) \(\left\{{}\begin{matrix}\frac{AH}{AC}=\frac{HB}{AB}\\\frac{AB}{AH}=\frac{AC}{HC}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AH=\frac{HB.AC}{AB}\\AH=\frac{AB.HC}{AC}\end{matrix}\right.\)

\(\Leftrightarrow AH^2=\frac{HB.AC}{AB}\cdot\frac{AB.HC}{AC}=HB.HC\)

e) ĐKXĐ: \(x^2-9\ge0\Leftrightarrow\left(x-3\right).\left(x+3\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

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

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

\(TH1:x-3=0\Leftrightarrow x=3\)

\(TH2:\sqrt{x-3}=-\sqrt{x+3}\Leftrightarrow x=3\text{ và }x=-3\left(loai\right)\)

Vậy giá trị x cần tìm là 3

d) ĐKXĐ: \(x+1\ge0\Leftrightarrow x\ge-1\)

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

\(\Leftrightarrow\sqrt{\left(x+1\right)^2}-\sqrt{x+1}=0\)

\(\Leftrightarrow\sqrt{x+1}.\left(\sqrt{x+1}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\\\sqrt{x+1}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(TMĐKXĐ\right)\\x=0\left(TMĐKXĐ\right)\end{matrix}\right.\)

Vậy giá trị x thỏa mãn là \(\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

c) ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ge0\\x-1\ge0\end{matrix}\right.\Rightarrow x\ge1\)

\(\sqrt{x+2}.\sqrt{x-1}=2\)

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

\(\Leftrightarrow x^2+x-2=4\)

\(\Leftrightarrow\left(x^2+x+\frac{1}{4}\right)=4+\frac{9}{4}=\frac{25}{4}\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{5}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{5}{2}\\x+\frac{1}{2}=-\frac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TMDKXD\right)\\x=-3\left(KTMDKXD\right)\end{matrix}\right.\)

Vậy giá trị x thỏa mãn là: x=2

b) \(\sqrt{36^2-12x+1}=5\)

\(\Leftrightarrow\sqrt{\left(6x-1\right)^2}=5\)

\(\Leftrightarrow\left|6x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\frac{2}{3}\end{matrix}\right.\)

Vậy giá trị x thỏa mãn là \(\left[{}\begin{matrix}x=1\\x=-\frac{2}{3}\end{matrix}\right.\)