Bài 1:
a.
\(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(2-\sqrt{2})^2}\)
\(=|\sqrt{3}+1|-|2-\sqrt{2}|=\sqrt{3}+1-(2-\sqrt{2})=\sqrt{3}+\sqrt{2}-1\)
b.
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\sqrt{20}-\sqrt{9})^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-(\sqrt{20}-\sqrt{9})}}=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}}=\sqrt{\sqrt{5}-(\sqrt{5}-1)}=1\)
c.
\(=\sqrt{6+2\sqrt{5}-\sqrt{(\sqrt{20}-\sqrt{9})^2}}\)
\(=\sqrt{6+2\sqrt{5}-(\sqrt{20}-\sqrt{9})}=\sqrt{6+2\sqrt{5}-2\sqrt{5}+3}=3\)
d.
\(\sqrt{2+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{2+\sqrt{5-\sqrt{(\sqrt{12}+1)^2}}}\)
\(=\sqrt{2+\sqrt{5-(\sqrt{12}+1)}}=\sqrt{2+\sqrt{4-2\sqrt{3}}}=\sqrt{2+\sqrt{(\sqrt{3}-1)^2}}\)
\(=\sqrt{2+(\sqrt{3}-1)}=\sqrt{1+\sqrt{3}}\)
Bài 2:
a. Để $A$ có nghĩa thì \(\left\{\begin{matrix} a\geq 0\\ b\geq 0\\ a\neq b\\ ab>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a>0\\ b>0\\ a\neq b\end{matrix}\right.\)
b.
\(A=\frac{a+b+2\sqrt{ab}-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{ab}(\sqrt{a}+\sqrt{b})}{\sqrt{ab}}\)
\(=\frac{(\sqrt{a}-\sqrt{b})^2}{\sqrt{a}-\sqrt{b}}-(\sqrt{a}+\sqrt{b})=(\sqrt{a}-\sqrt{b})-(\sqrt{a}+\sqrt{b})=-2\sqrt{b}\) không phụ thuộc vào giá trị của $a$
Ta có đpcm.
Bài 3:
a. ĐKXĐ:
\(\left\{\begin{matrix} x\geq 0\\ x\sqrt{x}-1\neq 0\\ \sqrt{x}-1\neq 0\\ x-1\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ x\neq 1\end{matrix}\right.\)
b.
\(B=\left[\frac{2\sqrt{x}+x}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\right].\frac{x+\sqrt{x}+1}{x-1}\)
\(=\frac{2\sqrt{x}+x-x-\sqrt{x}-1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\frac{x+\sqrt{x}+1}{x-1}=\frac{\sqrt{x}-1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\frac{x+\sqrt{x}+1}{x-1}=\frac{1}{x-1}\)
Bài 4.
a. Để $C$ có nghĩa thì:
\(\left\{\begin{matrix}
x\geq 0\\
x-9\neq 0\\
2-\sqrt{x}\neq 0\\
3+\sqrt{x}\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x\geq 0\\
x\neq 9\\
x\neq 4\end{matrix}\right.\)
b.
\(C=\frac{x-9-(x-3\sqrt{x})}{x-9}:\frac{(\sqrt{x}-3)(3+\sqrt{x})+(\sqrt{x}-2)(2-\sqrt{x})+(9-x)}{(2-\sqrt{x})(3+\sqrt{x})}\)
\(=\frac{-9+3\sqrt{x}}{x-9}:\frac{(\sqrt{x}-2)(2-\sqrt{x})}{(2-\sqrt{x})(3+\sqrt{x})}\)
\(=\frac{3(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}:\frac{\sqrt{x}-2}{\sqrt{x}+3}=\frac{3}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{\sqrt{x}-2}=\frac{3}{\sqrt{x}-2}\)
c.
$Để $C=4\Leftrightarrow \sqrt{x}-2=\frac{3}{4}$
$\Leftrightarrow x=\frac{121}{16}$ (thỏa mãn)
Vậy.........
Bài 5.
a.
ĐKXĐ: \(\left\{\begin{matrix} x\geq 0\\ 3+\sqrt{x}\neq 0\\ 9-x\neq 0\\ x-3\sqrt{x}\neq 0\\ \sqrt{x}\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x>0\\ x\neq 9\\ \end{matrix}\right.\)
b.
\(D=\frac{\sqrt{x}(\sqrt{x}-3)-(x+9)}{(\sqrt{x}+3)(\sqrt{x}-3)}:\frac{3\sqrt{x}+1-(\sqrt{x}-3)}{\sqrt{x}(\sqrt{x}-3)}\)
\(=\frac{-3(\sqrt{x}+3)}{(\sqrt{x}+3)(\sqrt{x}-3)}:\frac{2(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}-3)}=\frac{-3}{\sqrt{x}-3}.\frac{\sqrt{x}(\sqrt{x}-3)}{2(\sqrt{x}+2)}\)
\(=\frac{-3\sqrt{x}}{2(\sqrt{x}+2)}\)
c.
$D< -1$
$\Leftrightarrow D+1< 0\Leftrightarrow \frac{4-\sqrt{x}}{2(\sqrt{x}+2)}< 0$
$\Leftrightarrow 4-\sqrt{x}< 0$
$\Leftrightarrow x> 16$
Kết hợp đkxđ suy ra $x>16$
Bài 6.
$\frac{HB}{HC}=\frac{1}{4}\Leftrightarrow HC=4HB$
Theo hệ thức lượng trong tam giác vuông:
$AH^2=BH.CH$
$\Leftrightarrow 14^2=BH.4BH$
$\Leftrightarrow 49=BH^2\Rightarrow BH=7$ (cm)
$CH=4BH=28$ (cm)
$BC=BH+CH=28+7=35$ (cm)
Theo hệ thức lượng trong tam giác vuông:
$AB^2=BH.BC=7.35\Rightarrow AB=7\sqrt{5}$ (cm)
$AC^2=CH.BC=28.35\Rightarrow AC=14\sqrt{5}$ (cm)
Chu vi tam giác $ABC$ là:
$AB+BC+AC=21\sqrt{5}+35$ (cm)
Bài 7:
Áp dụng định lý Pitago:
$AB=\sqrt{BD^2-AD^2}=\sqrt{10-1}=3$ (cm)
Theo tính chất tia phân giác:
$\frac{AD}{DC}=\frac{AB}{BC}$
$\Leftrightarrow \frac{1}{DC}=\frac{3}{BC}$
$\Leftrightarrow BC=3DC$
Tiếp tục áp dụng định lý Pitago:
$AB^2+AC^2=BC^2$
$3^2+(1+DC)^2=(3DC)^2$
$\Leftrightarrow 8DC^2-2DC-10=0$
$\Rightarrow DC=1,25$ (cm)
$BC=3DC=3.1,25=3,75$ (cm)
Bài 8:
Áp dụng hệ thức lượng trong tam giác vuông với tam giác $BCD$ vuông tại $C$, đường cao $CA$:
$CA^2=AB.AD$
$20^2=15.AD$
$AD=\frac{80}{3}$ (cm)
$CD^2=DA.DB=DA(DA+AB)=\frac{80}{3}(\frac{80}{3}+15)=\frac{10000}{9}$
$\Rightarrow CD=\frac{100}{3}$ (cm)
Bài 9:
Theo hệ thức lượng trong tam giác:
$\frac{1}{DE^2}=\frac{1}{AD^2}+\frac{1}{DC^2}=\frac{1}{32^2}+\frac{1}{60^2}$
$\Rightarrow DE=\frac{480}{17}$ (cm)
Áp dụng định lý Pitago:
$AE=\sqrt{AD^2-DE^2}=\sqrt{32^2-(\frac{480}{17})^2}=\frac{256}{17}$ (cm)
$EC=\sqrt{DC^2-DE^2}=\sqrt{60^2-(\frac{480}{17})^2}=\frac{900}{17}$ (cm)
Tiếp tục áp dụng hệ thức lượng trong tam giác vuông:
$FE=\frac{AE^2}{DE}=\frac{256^2}{17^2}: \frac{480}{17}=\frac{2048}{255}$ (cm)
$FD=DE+EF=\frac{480}{17}+\frac{2048}{255}=\frac{544}{15}$ (cm)
$AF=\sqrt{FD^2-AD^2}=\sqrt{(\frac{544}{15})^2-32^2}=\frac{256}{15}$ (cm)
$FB=AB-AF=60-\frac{256}{15}=\frac{644}{15}$ (cm)
Bài 10:
a.
Áp dụng định lý Pitago: $AC=\sqrt{BC^2-AB^2}=\sqrt{9^2-5^2}=2\sqrt{14}$ (cm)
$\cos B=\frac{AB}{BC}=\frac{5}{9}$
$\Rightarrow \widehat{B}=56,25^0$
$\widehat{C}=90^0-\widehat{B}=90^0-56,25^0=33,75^0$
b.
$\widehat{C}=90^0-\widehat{B}=60^0$
$\frac{\sqrt{3}}{2}=\cos 30^0=\cos B=\frac{AB}{BC}$
$\Rightarrow AB=8.\frac{\sqrt{3}}{2}=4\sqrt{3}$ (cm)
$\frac{1}{2}=\sin 30^0=\sin B=\frac{AC}{BC}\Rightarrow AC=\frac{BC}{2}=4$ (cm)
Bài 10:
a.
Áp dụng định lý Pitago: $AC=\sqrt{BC^2-AB^2}=\sqrt{9^2-5^2}=2\sqrt{14}$ (cm)
$\cos B=\frac{AB}{BC}=\frac{5}{9}$
$\Rightarrow \widehat{B}=56,25^0$
$\widehat{C}=90^0-\widehat{B}=90^0-56,25^0=33,75^0$
b.
$\widehat{C}=90^0-\widehat{B}=60^0$
$\frac{\sqrt{3}}{2}=\cos 30^0=\cos B=\frac{AB}{BC}$
$\Rightarrow AB=8.\frac{\sqrt{3}}{2}=4\sqrt{3}$ (cm)
$\frac{1}{2}=\sin 30^0=\sin B=\frac{AC}{BC}\Rightarrow AC=\frac{BC}{2}=4$ (cm)
Bài 11:
Kẻ $CH\perp AB$ với $H\in AB$
$\frac{CH}{CB}=\sin B=\sin 45^0$
$\Rightarrow CH=CB\sin 45^0=4\sqrt{2}$ (cm)
b.
$\frac{CH}{AC}=\sin A=\sin 30^0$
$\Rightarrow AC=\frac{CH}{\sin 30^0}=\frac{4\sqrt{2}}{\sin 30^0}=8\sqrt{2}$ (cm)
a.
$BH=\sqrt{CB^2-CH^2}=\sqrt{8^2-(4\sqrt{2})^2}=4\sqrt{2}$ (cm)
$AH=\sqrt{AC^2-CH^2}=\sqrt{(8\sqrt{2})^2-(4\sqrt{2})^2}=4\sqrt{6}$ (cm)
$AB=AH+HB=4\sqrt{2}+4\sqrt{6}$ (cm)
c.
$S_{ABC}=\frac{CH.AB}{2}=\frac{4\sqrt{2}(4\sqrt{2}+4\sqrt{6})}{2}=16+16\sqrt{3}$ (cm vuông)
Bài 12:
a.
Áp dụng định lý Pitago:
$AH=\sqrt{AB^2-BH^2}=\sqrt{13^2-5^2}=12$
$\sin B=\frac{AH}{AB}=\frac{12}{13}$
$\sin C=\sin \widehat{BAH}=\frac{BH}{AB}=\frac{5}{13}$
b.
$AH^2=BH.CH=12\Rightarrow AH=2\sqrt{3}$
$AB=\sqrt{BH^2+AH^2}=\sqrt{3^2+(2\sqrt{3})^2}=\sqrt{21}$
$\sin B=\frac{AH}{AB}=\frac{2\sqrt{3}}{\sqrt{21}}=\frac{2\sqrt{7}}{7}$
$\sin C=\sin \widehat{BAH}=\frac{BH}{AB}=\frac{3}{\sqrt{21}}$
Bài 13:
a.
$\widehat{B}=90^0-\widehat{C}=60^0$
$\cos 30=\cos C=\frac{AC}{BC}=\frac{5,4}{BC}$
$\Rightarrow BC=\frac{5,4}{\cos 30}=\frac{18\sqrt{3}}{5}$
$\tan 30=\tan C=\frac{AB}{AC}=\frac{AB}{5,4}$
$\Rightarrow AB=\frac{9\sqrt{3}}{5}$
b.
$1=\tan 45^0=\tan C =\frac{AB}{AC}$
$\Rightarrow AC=AB=10$ (cm)
$\frac{\sqrt{2}}{2}=\sin 45^0=\sin C=\frac{AB}{BC}$
$\Rightarrow BC=10\sqrt{2}$ (cm)
Bài 14:
a.
$c=\sqrt{a^2-b^2}=\sqrt{15^2-10^2}=5\sqrt{5}$ (cm)
$\cos C=\frac{b}{a}=\frac{10}{15}=\frac{2}{3}$
$\Rightarrow \widehat{C}=48,19^0$
$\widehat{B}=90^0-\widehat{C}=41,81^0$
b.
$a=\sqrt{b^2+c^2}=\sqrt{12^2+7^2}=\sqrt{193}$ (cm)
$\tan C=\frac{c}{b}=\frac{7}{12}$
$\Rightarrow \widehat{C}=30,26^0$
$\widehat{B}=90^0-\widehat{C}=59,74^0$
Bài 15:
Kẻ $AH\perp BC$ ($H\in BC$)
$AH=\sin C.AC=\sin 50.35=26,81$ (cm)
$CH=\cos C.AC=\cos 50.35=22,5$ (cm)
$BH=\frac{AH}{\tan 60^0}=\frac{AH}{26,81}{\tan 60^0}=15,48$ (cm)
$BC=BH+CH=37,98$ (cm)
$S_{ABC}=\frac{AH.BC}{2}=\frac{26,81.37,98}{2}=699$ (cm vuông)