a) Có \(BC^2=15^2=225\)

\(AB^2+AC^2=9^2+12^2=81+144=225\)

do đó \(BC^2=AB^2+AC^2\)

Theo định lí Pythaogre đảo suy ra tam giác \(ABC\)vuông tại \(A\).

b) \(AH=\frac{AB.AC}{BC}=\frac{9.12}{15}=7,2\left(cm\right)\)

\(HB=\frac{AB^2}{BC}=\frac{9^2}{15}=5,4\left(cm\right)\)

\(HC=BC-HB=15-5,4=9,6\left(cm\right)\)

Ta có: E = \(\frac{x}{\sqrt{x}-1}=\frac{x-1+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1}{\sqrt{x}-1}=\sqrt{x}+1+\frac{1}{\sqrt{x}-1}\)

E = \(\sqrt{x}-1+\frac{1}{\sqrt{x}-1}+2\ge2\sqrt{\left(\sqrt{x}-1\right)\cdot\frac{1}{\sqrt{x}-1}}+2=2+2=4\)(bđt cosi)

Dấu "=" xảy ra <=> \(\sqrt{x}-1=\frac{1}{\sqrt{x}-1}\) <=> \(\left(\sqrt{x}-1\right)^2=1\)

<=> \(\orbr{\begin{cases}\sqrt{x}-1=1\\\sqrt{x}-1=-1\end{cases}}\) <=> \(\orbr{\begin{cases}x=4\left(tm\right)\\x=0\left(ktm\right)\end{cases}}\)

Vậy MinE = 4 <=>. x = 4

\(\sqrt{x-3}-2\ne0\)

\(\sqrt{x-3}\ne2\)

\(x\ne7\left(1\right)\)

\(\sqrt{x-3}\ge0\)

\(x\ge3\left(2\right)\)

\(\left(1\right);\left(2\right)< =>x\ge3;x\ne7\)

x>=3; x khác 7 

\(15.a,\sqrt{3}\left(\sqrt{4}+2\sqrt{9}\right)\frac{\sqrt{3}}{2}-\sqrt{150}\)

\(\left(2+2.3\right)\frac{3}{2}-\sqrt{150}\)

\(12-\sqrt{150}\)

\(6\left(2-\sqrt{25}\right)=6\left(-3\right)=-18\)

\(b,\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right).\sqrt{7}+2\sqrt{21}\)

\(\sqrt{196}-\sqrt{84}-7+2\sqrt{21}\)

\(14-\sqrt{21}\left(\sqrt{4}+2\right)-7\)

\(7-\sqrt{21}.4\)

\(7-\sqrt{84}\)

\(c,\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)\)

\(1+\sqrt{2}-\sqrt{3}+\sqrt{2}+2-\sqrt{6}+\sqrt{3}+\sqrt{6}-3\)

\(=2\sqrt{2}\)

\(d,\sqrt{3}\left(\sqrt{2}-\sqrt{3}\right)^2-\sqrt{3}+\sqrt{2}\)

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

\(\sqrt{3}\left(5-\sqrt{6}-1\right)+\sqrt{2}\)

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

\(4\sqrt{3}-\sqrt{18}+\sqrt{2}\)

\(\sqrt{2}\left(2\sqrt{3}-3+1\right)\)

\(\sqrt{2}\left(2\sqrt{3}-2\right)\)

\(2\sqrt{6}-\sqrt{6}=\sqrt{6}\)

\(\frac{\sqrt{10}+5\sqrt{2}}{\sqrt{5}+1}-6\sqrt{\frac{5}{2}}+\frac{12}{4-\sqrt{10}}\)

\(=\frac{\sqrt{10}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\frac{6\sqrt{10}}{2}+\frac{12\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}\)

\(=\sqrt{10}-3\sqrt{10}+2\left(4+\sqrt{10}\right)=-2\sqrt{10}+8+2\sqrt{10}=8\)

2. \(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{2}-1}+4\sqrt{\frac{3}{2}}+\frac{9}{3+\sqrt{3}}\)

\(=\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\frac{4\sqrt{6}}{2}+\frac{9\left(3-\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\)

\(=\sqrt{6}+2\sqrt{6}+\frac{9\left(3-\sqrt{6}\right)}{9-6}=3\sqrt{6}+3\left(3-\sqrt{6}\right)=9\)

chúc bn hok tốt

a) Xét ΔAHB có ^AHB = 90⁰ ( AH ⊥ BC ) => ΔAHB vuông tại H

Khi đó : \(\sin B=\sin\widehat{ABH}=\frac{AH}{AB}=\frac{5}{13};\cos B=\cos\widehat{ABH}=\frac{BH}{AB}=\frac{\sqrt{AB^2-AH^2}\left(pythagoras\right)}{AB}=\frac{12}{13}\)

ΔABC vuông tại A => ^B + ^C = 90⁰ => \(\sin C=\cos B=\frac{12}{13}\)

b) Áp dụng hệ thức lượng trong tam giác vuông cho ΔABC vuông tại A ta có :

\(AH^2=BH\cdot HC\Rightarrow AH=\sqrt{BH\cdot HC}=2\sqrt{3}\)

cmtt như a) ta có được ΔAHC vuông tại H

Khi đó : \(\sin C=\sin\widehat{ACH}=\frac{AH}{AC}=\frac{AH}{\sqrt{AH^2+HC^2}}=\frac{\sqrt{21}}{7};\cos C=\cos\widehat{ACH}=\frac{CH}{AC}=\frac{CH}{\sqrt{AH^2+HC^2}}=\frac{2\sqrt{7}}{7}\)ΔABC vuông tại A => ^B + ^C = 90⁰ => \(\sin B=\cos C=\frac{2\sqrt{7}}{7}\)