\(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}=\sqrt{3+2+1+\sqrt{2^2.2.3}+\sqrt{2^2.3}+\sqrt{2^2.2}}\)

\(=\sqrt{\left(\sqrt{3}\right)^2+\left(\sqrt{2}\right)^2+1^2+2\sqrt{3}.\sqrt{2}+2\sqrt{3}.1+2\sqrt{2}.1}=\sqrt{\left(\sqrt{3}+\sqrt{2}+1\right)^2}\)

(áp dụng hằng đẳng thức (a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc)

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

giải ra chưa chỉ mình với

Bạn áp dụng hằng đẳng thức (a+b+c)^2= a^2+b^2+c^2+2(ab+ac+bc)

a: \(=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)

\(=\left(-3\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)

\(=\left(-3\sqrt{10}+10\right)\left(\dfrac{3}{10}\sqrt{10}+10\right)\)

\(=-9-30\sqrt{10}+3\sqrt{10}+100=91-27\sqrt{10}\)

b: \(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}\cdot\left(\dfrac{5}{2}\sqrt{2}+12\right)\)

\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\left(5\sqrt{3}+12\sqrt{6}\right)\)

\(=-60-144\sqrt{2}+30\sqrt{2}+144\)

\(=84-114\sqrt{2}\)

\(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(5\sqrt{\dfrac{1}{12}}+12\right)\)

\(=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\sqrt{6}-\left(\dfrac{5\sqrt{3}}{6}+12\right)\)

\(=6\sqrt{2}-18\sqrt{2}+12-\left(\dfrac{5\sqrt{3}+72}{6}\right)\)

\(=-12\sqrt{2}+12-\dfrac{5\sqrt{3}+72}{6}\)

\(=\dfrac{-72\sqrt{2}+72-5\sqrt{3}-72}{6}=\dfrac{5\sqrt{3}+72\sqrt{2}}{6}\simeq-18,4139\)

Ta có: \(-14,5\sqrt{2}\simeq-20,506\)

\(VT\ne VP\)

Đẳng thức không xảy ra

\(\text{c) }\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\)

Ta có : \(6< 6.25\Rightarrow\sqrt{6}< \sqrt{6.25}\Rightarrow\sqrt{6}< 2.5\)

\(12< 12.25\Rightarrow\sqrt{12}< \sqrt{12.25}\Rightarrow\sqrt{12}< 3.5\)

\(20< 20.25\Rightarrow\sqrt{20}< \sqrt{20.25}\Rightarrow\sqrt{20}< 4.5\)

\(30< 30.25\Rightarrow\sqrt{30}< \sqrt{30.25}\Rightarrow\sqrt{30}< 5.5\)

\(42< 42.25\Rightarrow\sqrt{42}< \sqrt{42.25}\Rightarrow\sqrt{42}< 6.5\)

\(50< 56.5\Rightarrow\sqrt{50}< \sqrt{56.25}\Rightarrow\sqrt{50}< 7.5\) \(\left(1\right)\)

Từ \(\left(1\right)\) suy ra :

\(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 2.5+3.5+4.5+5.5+6.5+7.5\)

\(\Rightarrow\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\) \(\left(ĐPCM\right)\)

Vậy \(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\)

\(\)\(\text{a) }\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 24\)

Ta có : \(1< 9\Rightarrow\sqrt{1}< \sqrt{9}\Rightarrow\sqrt{1}< 3\)

\(2< 9\Rightarrow\sqrt{2}< \sqrt{9}\Rightarrow\sqrt{2}< 3\)

\(3< 9\Rightarrow\sqrt{3}< \sqrt{9}\Rightarrow\sqrt{3}< 3\)

\(...\)

\(8< 9\Rightarrow\sqrt{8}< \sqrt{9}\Rightarrow\sqrt{8}< 3\) \(\left(1\right)\)

Từ \(\left(1\right)\) suy ra :

\(\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 3+3+...+3_{\left(\text{8 số hạng 3}\right)}\) \(\) \(\)

\(\) \(\Rightarrow\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 3\cdot8\)

\(\Rightarrow\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 24\) \(\left(ĐPCM\right)\)

Vậy \(\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 24\)

\(\text{b) }\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>10\)

Ta có : \(1< 100\Rightarrow\sqrt{1}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{1}}< \dfrac{1}{\sqrt{100}}\)

\(2< 100\Rightarrow\sqrt{2}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{2}}< \dfrac{1}{\sqrt{100}}\)

\(...\)

\(100=100\Rightarrow\sqrt{100}=\sqrt{100}\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}\) \(\left(1\right)\)

Từ \(\left(1\right)\) suy ra :

\(\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}_{\left(\text{100 số hạng}\dfrac{1}{\sqrt{100}}\right)}\)

\(\Rightarrow\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}\cdot100\)

\(\Rightarrow\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>\dfrac{10}{\sqrt{100}}\)

\(\Rightarrow\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>10\) \(\left(ĐPCM\right)\)

Vậy \(\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>10\)

\(\)

Cho phép mình chữa đề câu \(c\) thành như thế này nhé Fairy Tail

\(\text{c) }\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 33\)

Ta có : \(6< 9\Rightarrow\sqrt{6}< \sqrt{9}\Rightarrow\sqrt{6}< 3\)

\(12< 16\Rightarrow\sqrt{12}< \sqrt{16}\Rightarrow\sqrt{12}< 4\)

\(20< 25\Rightarrow\sqrt{20}< \sqrt{25}\Rightarrow\sqrt{20}< 5\)

\(30< 36\Rightarrow\sqrt{30}< \sqrt{36}\Rightarrow\sqrt{30}< 6\)

\(42< 49\Rightarrow\sqrt{42}< \sqrt{49}\Rightarrow\sqrt{42}< 7\)

\(50< 64\Rightarrow\sqrt{50}< \sqrt{64}\Rightarrow\sqrt{50}< 8\) \(\left(1\right)\)

Từ \(\left(1\right)\) suy ra :

\(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 3+4+5+6+7+8\)

\(\Rightarrow\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 33\) \(\left(ĐPCM\right)\)

Vậy \(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 33\)

a: \(\left(3+\sqrt{2}\right)^2=3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2\)

\(=9+6\sqrt{2}+2=11+6\sqrt{2}\)

b: \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)

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

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

c: \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\sqrt{7}-1-\sqrt{7}-1=-2\)

d: \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)

\(=\sqrt{45-2\cdot3\sqrt{5}\cdot2+4}-\sqrt{45+2\cdot3\sqrt{5}\cdot2+4}\)

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

\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)

a) \(\left(3+\sqrt{2}\right)^2=9+6\sqrt{2}+2=11+6\sqrt{2}\)

b) \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)

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

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

c) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\sqrt{7}-1-\sqrt{7}-1=-2\)

d) \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)

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

\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)