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

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

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

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

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

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

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 => \(\sin B=\cos C=\frac{2\sqrt{7}}{7}\)

\(D=\sqrt{x-2}+\sqrt{4-x}\ge\sqrt{x-2+4-x}\)

\(=\sqrt{2}\)

dấu "=" xảy ra khi: \(\orbr{\begin{cases}\sqrt{x-2}=0\\\sqrt{4-x}=0\end{cases}\orbr{\begin{cases}x=2\\x=4\end{cases}}}\)

vậy MIN \(D=\sqrt{2}\)

\(D=\sqrt{x-2}+\sqrt{4-x}\le\frac{x-2+1+4-x+1}{2}=4\)

dấu "=" xảy ra khi \(x=3\)

vậy \(MAX:D=4\)

\(D=\sqrt{x-2}+\sqrt{4-x}\)

\(\Rightarrow D^2=x-2+2\sqrt{\left(x-2\right)\left(4-x\right)}+4-x=2+2\sqrt{\left(x-2\right)\left(4-x\right)}\)

*GTNN

Với 2 ≤ x ≤ 4 => \(2\sqrt{\left(x-2\right)\left(4-x\right)}\ge0\Leftrightarrow2+2\sqrt{\left(x-2\right)\left(4-x\right)}\ge2\)

hay D² ≥ 2 => D ≥ √2 . Dấu "=" xảy ra <=> x = 2 hoặc x = 4 (tm)

*GTLN

Áp dụng bất đẳng thức AM-GM ta có :

\(2\sqrt{\left(x-2\right)\left(4-x\right)}\le x-2+4-x=2\Rightarrow2+2\sqrt{\left(x-2\right)\left(4-x\right)}\le4\)

hay D² ≤ 4 => D ≤ 2 . Dấu "=" xảy ra <=> x = 3 (tm)

Vậy \(\hept{\begin{cases}Min_D=\sqrt{2}\Leftrightarrow x=2orx=4\\Max_D=2\Leftrightarrow x=3\end{cases}}\)

\(A=\sqrt{x-2+2\sqrt{x-3}}\)

\(A=\sqrt{x-3+2\sqrt{x-3}+1}\)

\(A=\sqrt{\left(\sqrt{x-3}+1\right)^2}\)

\(A=\left|\sqrt{x-3}+1\right|\)

\(A=\sqrt{x-3}+1\)

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

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

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

\(B=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)

kết hợp với đkxđ của x ta lập bảng xét dấu và phân TH

\(TH1:1\le x\le2\)

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

\(TH2:x>2\)

\(B=\sqrt{x-1}+1+\sqrt{x-1}-1\)

\(B=2\sqrt{x-1}\)

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

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

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

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

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

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

\(=\frac{3-\sqrt{5}}{2}+\frac{3+\sqrt{5}}{2}=\frac{3-\sqrt{5}+3+\sqrt{5}}{2}=3\)

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

\(=\sqrt{3}-\sqrt{3}+1-\frac{\sqrt{3}+1}{2}=1+\frac{\sqrt{3}+1}{2}=\frac{3+\sqrt{3}}{2}\)