\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)

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

\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)

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

\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)

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

\(=\sqrt{4-3}=1\)

1. How many

2. How many

3. How much

4. How many

5. How many

(Bài này điền từ hả bạn?)

B. 2-3-4-1

Ta có \(x-y+2z=-20\)

\(\Leftrightarrow x-y+\dfrac{1}{2}\cdot4z=-20\Leftrightarrow x-y+\dfrac{1}{2}\cdot3x=-20\)

\(\Leftrightarrow\dfrac{5}{2}x-y=-20\Leftrightarrow5x-2y=-40\)

Theo đề bài: \(\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{5x}{15}=\dfrac{2y}{10}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\dfrac{5x}{15}=\dfrac{2y}{10}=\dfrac{5x-2y}{15-10}=-\dfrac{40}{5}=-8\)

\(\Rightarrow\dfrac{x}{3}=-8\Rightarrow x=-24\)

\(\Rightarrow\dfrac{y}{5}=-8\Rightarrow y=-40\)

\(\Rightarrow3\cdot\left(-24\right)=4z\Rightarrow z=-18\)

\(\dfrac{x-1}{15}=\dfrac{60}{x-1}\) (ĐK: \(x\ne1\))

\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=15\cdot60\)

\(\Leftrightarrow\left(x-1\right)^2=900\)

\(\Leftrightarrow\left(x-1\right)^2=30^2\)

\(\Leftrightarrow\left|x-1\right|=30\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=31\\x=-29\end{matrix}\right.\)(T/m)

\(\dfrac{15}{2x-1}=\dfrac{5}{3}\) (ĐK: \(x\ne\dfrac{1}{2}\))

\(\Leftrightarrow3\cdot15=5\left(2x-1\right)\)

\(\Leftrightarrow45=10x-5\)

\(\Leftrightarrow10x=50\)

\(\Leftrightarrow x=5\) (T/m)

a) O là tâm hbh ABCD

\(\rightarrow\) O là trung điểm AC \(\rightarrow\overrightarrow{OA}+\overrightarrow{OC}=\overrightarrow{0}\)

    O là trung điểm BD \(\rightarrow\overrightarrow{OB}+\overrightarrow{OD}=\overrightarrow{0}\)

\(\Rightarrow\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\left(\overrightarrow{OA}+\overrightarrow{OC}\right)+\left(\overrightarrow{OB}+\overrightarrow{OD}\right)=\overrightarrow{0}+\overrightarrow{0}=\overrightarrow{0}\) (đpcm)

b) Vì ABCD là hbh \(\rightarrow\overrightarrow{AB}=\overrightarrow{DC};\overrightarrow{AD}=\overrightarrow{BC}\)

Vì E là trung điểm AD \(\rightarrow\overrightarrow{EA}=\overrightarrow{DE}\)

 \(2\overrightarrow{CE}=\overrightarrow{CA}+\overrightarrow{CD}\rightarrow2\overrightarrow{EC}=-\left(\overrightarrow{CA}+\overrightarrow{CD}\right)=\overrightarrow{AC}+\overrightarrow{DC}\)

Ta có: \(\overrightarrow{EA}+\overrightarrow{EB}+2\overrightarrow{EC}=\overrightarrow{DE}+\overrightarrow{EB}+\overrightarrow{AC}+\overrightarrow{DC}\)

\(=\overrightarrow{DB}+\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{AB}=2\overrightarrow{AB}+\overrightarrow{DC}\)

\(=2\overrightarrow{AB}+\overrightarrow{AB}=3\overrightarrow{AB}\) (đpcm)

c) Sửa đề : CM: vectơ AB + 2 lần vectơ EA + 4 lần vectơ ED = vectơ AC

Vì E là trung điểm AD \(\rightarrow\overrightarrow{ED}=\overrightarrow{AE};\overrightarrow{AD}=2\overrightarrow{AE}\)

\(\overrightarrow{AB}+2\overrightarrow{EA}+4\overrightarrow{ED}=\overrightarrow{AB}-2\overrightarrow{AE}+4\overrightarrow{AE}=\overrightarrow{AB}+2\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{AC}\)

Ta có: MB=3MC \(\rightarrow MB=\dfrac{3}{4}BC\)

mà M nằm giữa B và C \(\rightarrow\overrightarrow{BM}=\dfrac{3}{4}\overrightarrow{BC}\)

\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{BC}\)

\(=\overrightarrow{AB}+\dfrac{3}{4}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\overrightarrow{AB}-\dfrac{3}{4}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AC}=\dfrac{1}{4}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AC}\)

Ta có: \(\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}=\dfrac{\sqrt{5+2\cdot\sqrt{5}\cdot1+1}}{\sqrt{5}+1}=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{5}+1}\)

\(=\dfrac{\sqrt{5}+1}{\sqrt{5}+1}=1\) (1)

\(\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot\sqrt{2}+2}}{\sqrt{3}-\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{\sqrt{3}-\sqrt{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=1\) (2)

Từ (1)(2) \(\Rightarrow\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}=\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}\) (đpcm)

\(A=\dfrac{2021^2\left(2020-2019\right)}{\left(2020-1\right)\left(2020^3+1\right)}\cdot\dfrac{2019^2\left(2020^2+2021\right)}{2020^3-1}\)

\(A=\dfrac{2021^2}{2019\left(2020+1\right)\left(2020^2-2020\cdot1+1^2\right)}\cdot\dfrac{2019^2\left(2020^2+2021\right)}{\left(2020-1\right)\left(2020^2+2020\cdot1+1^2\right)}\)

\(A=\dfrac{2021^2\cdot2019^2\cdot\left(2020^2+2021\right)}{2019\cdot2021\left(2020^2-2019\right)\cdot2019\cdot\left(2020^2+2021\right)}\)

\(A=\dfrac{2021^2}{2021\left(2020^2-2019\right)}\)

\(A=\dfrac{2021}{2020^2-2019}\)