Bài tập theo tuần Toán 9 - Tuần 4

$\begin{array}{l}\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\\ =\frac{2\sqrt{3}+2-2\sqrt{3}+2}{3-1}=\frac{4}{2}=2\end{array}$

$\begin{array}{l}\frac{2x}{y}\sqrt{\frac{y}{2x}}=\sqrt{\frac{4x^2.y}{y^2.2x}}=\sqrt{\frac{2x}{y}}=\frac{\sqrt{2x}}{\sqrt{y}}=\frac{\sqrt{2xy}}{y}\end{array}$

$\frac{3\sqrt{2}+2\sqrt{3}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{6}.\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=\sqrt{6}$

$\begin{array}{l}A=4.\left(\frac{1}{3-\sqrt{8}}+\frac{1}{\sqrt{7}-\sqrt{8}}-\frac{1}{\sqrt{6}-\sqrt{7}}-\frac{1}{\sqrt{6}-\sqrt{5}}-\frac{1}{2-\sqrt{5}}\right)\\ =4.\left[\frac{3+\sqrt{8}}{9-8}+\frac{\sqrt{7}+\sqrt{8}}{7-8}-\frac{\sqrt{6}+\sqrt{7}}{6-7}-\frac{\sqrt{6}+\sqrt{5}}{6-5}-\frac{2+\sqrt{5}}{4-5}\right]\\ =4.\left(3+\sqrt{8}-\sqrt{7}-\sqrt{8}+\sqrt{6}+\sqrt{7}-\sqrt{6}-\sqrt{5}+2+\sqrt{5}\right)\\ =4.5=20\end{array}$

$\begin{array}{l}\sqrt{2x+3}=1+\sqrt{2}\left(x\ge \frac{-3}{2}\right)\\ \Rightarrow 2x+3=3+2\sqrt{2}\\ \iff x=\sqrt{2}\left(tm\right)\end{array}$

$\begin{array}{l}\sin \alpha =\frac{4}{5}\Rightarrow {\cos }^2\alpha =1-{\sin }^2\alpha =1-{\left(\frac{4}{5}\right)}^2=\frac{9}{25}\Rightarrow \cos \alpha =\sqrt{\frac{9}{25}}=\frac{3}{5}\\ \tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{4}{5}:\frac{3}{5}=\frac{4}{3}\cot \alpha =\frac{1}{\tan \alpha }=\frac{3}{4}\end{array}$

$\begin{array}{l}{\sin }^{2}10°+{\sin }^{2}20°+\mathrm{.....}+{\sin }^{2}70°+{\sin }^2{80}^0\\ ={\sin }^{2}10°+{\sin }^{2}20°+\mathrm{.....}+{\cos }^{2}20°+{\cos }^2{10}^0\\ =\left({\sin }^{2}10°+{\cos }^2{10}^0\right)+\left({\sin }^{2}20°+{\cos }^{2}20°\right)+\mathrm{....}\\ =1+1+1+1=4\end{array}$

$\cos 78°=\sin 12°,\cos 68°=\sin 22°,\cos 50°=\sin 40°$

Vì $\sin 54°>\sin 40°>\sin 27°>\sin 22°>\sin 19°>\sin 12°$

$\Rightarrow \sin 54°>\cos 50°>\sin 27°>\cos 68°>\sin 19°>\cos 78°$

$\sqrt{\frac{1}{a^4}+\frac{1}{a}}=\sqrt{\frac{a^3+1}{a^4}}=\frac{\sqrt{a^3+1}}{a^2}$

$\sqrt{\frac{16x^3}{81y}}=\sqrt{\frac{16}{81}}.\frac{\sqrt{x^3}}{\sqrt{y}}=\frac{4x\sqrt{x}}{9\sqrt{y}}=\frac{4x\sqrt{xy}}{9y}$

$\sqrt{\frac{16x^3}{81y}}=\sqrt{\frac{16}{81}}.\frac{\sqrt{x^3}}{\sqrt{y}}=\frac{4x\sqrt{x}}{9\sqrt{y}}=\frac{4x\sqrt{xy}}{9y}$

$\begin{array}{l}C=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\mathrm{.......}-\frac{1}{\sqrt{48}-\sqrt{49}}\\ =\frac{\sqrt{1}+\sqrt{2}}{1-2}-\frac{\sqrt{2}+\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}-\mathrm{.....}-\frac{\sqrt{48}+\sqrt{49}}{48-49}\\ =-1-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}-\mathrm{.....}+\sqrt{48}+\sqrt{49}\\ =-1+\sqrt{49}=-1+7=6\end{array}$

$\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{{\sqrt{x}}^3-{\sqrt{y}}^3}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y$

${\sin }^2\alpha +{\cos }^2\alpha ={\left(\frac{AB}{BC}\right)}^2+{\left(\frac{AC}{BC}\right)}^2=\frac{A{B}^2+A{C}^2}{B{C}^2}=\frac{B{C}^2}{B{C}^2}=1$