Bài 3: Liên hệ giữa phép nhân và phép khai phương

1, \(\sqrt{8}-3\sqrt{32}+\sqrt{72}=2\sqrt{2}-12\sqrt{2}+6\sqrt{2}=-4\sqrt{2}\)

2,\(6\sqrt{12}-2\sqrt{48}+5\sqrt{75}-7\sqrt{108}=12\sqrt{3}-8\sqrt{3}+25\sqrt{3}-42\sqrt{3}=-13\sqrt{3}\)

3, \(\sqrt{20}+3\sqrt{45}-6\sqrt{80}-\dfrac{1}{3}\sqrt{125}=2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\dfrac{5}{3}.\sqrt{5}=-\dfrac{44}{3}.\sqrt{5}\)

4, \(2\sqrt{5}-\sqrt{125}-\sqrt{80}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}=-7\sqrt{5}\)

5, \(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}\)

Câu a : \(\sqrt{40^2-24^2}=\sqrt{\left(40+24\right)\left(40-24\right)}=\sqrt{64.16}=8.4=32\)

Câu b : \(\sqrt{52^2-48^2}=\sqrt{\left(52+48\right)\left(52-48\right)}=\sqrt{100.4}=10.2=20\)

\(\left(2\sqrt{2}-\sqrt{3}\right)^2=8-2\cdot2\sqrt{2}\cdot\sqrt{3}+3=11-4\sqrt{6}\)

\(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}=6+2\sqrt{9-5}=6+2.2=10\)

đề bài khó hỉu quá

1: \(=\left(a-3\right)\cdot\dfrac{\left|b\right|}{a-3}=\left|b\right|\)

2: \(\dfrac{1}{3+a}\cdot\sqrt{\dfrac{a^2+6a+9}{b^2}}\)

\(=\dfrac{1}{a+3}\cdot\dfrac{\left|a+3\right|}{b}=\pm\dfrac{1}{b}\)

3: \(=\left|a+1\right|-\dfrac{3a}{a-2}\cdot\dfrac{\left|a-2\right|}{3}\)

\(=\left|a+1\right|-a\)

4: \(=-6\sqrt{3}+6+28+6\sqrt{3}=34\)

a.\(\Leftrightarrow\left(\sqrt{4+\sqrt{15}}\right)^2\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)

\(\Leftrightarrow\sqrt{2}\sqrt{4+\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}\)

\(\Leftrightarrow\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4^2-15}\)

\(\Leftrightarrow\sqrt{5+2\sqrt{5.3}+3}\left(\sqrt{5}-\sqrt{3}\right)\)

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

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

b.T\(^2\)câu a ta tách (3+\(\sqrt{5}\))=\(\left(\sqrt{3+\sqrt{5}}\right)^2\);(\(\sqrt{10}-\sqrt{6}\))\(=\sqrt{2}\left(\sqrt{5}-1\right)\)

Rồi tính tiếp với đáp số =8

c.\(\Leftrightarrow\dfrac{\sqrt{2}\sqrt{\sqrt{5}+1}\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}{\sqrt{2}\sqrt{\sqrt{5}+1}\sqrt{\sqrt{5}+1}}-\sqrt{2-2\sqrt{2}+1}\)

\(\Leftrightarrow\dfrac{\sqrt{2\left(\sqrt{5}+1\right)\left(\sqrt{5}+2\right)}+\sqrt{\left(2\sqrt{5}+1\right)\left(\sqrt{5}-2\right)}}{\sqrt{2}(\sqrt{5}+1)}-\sqrt{\left(\sqrt{2}-1\right)^2}\)

\(\Leftrightarrow\dfrac{\sqrt{14+6\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{2}\left(\sqrt{5}+1\right)}-\left(\sqrt{2}-1\right)\)

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

\(\Leftrightarrow\dfrac{3+\sqrt{5}+\sqrt{5}-1}{\sqrt{2}\left(\sqrt{5}+1\right)}-\sqrt{2}+1\)

\(\Leftrightarrow\dfrac{2\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}-\sqrt{2}+1\Leftrightarrow\sqrt{2}-\sqrt{2}+1\)=1

27M² = 27(a²b²c²) \(\le\)(a² + b² + c²)³ = 225³

\(\Rightarrow\)M² \(\le\)75³

\(\Leftrightarrow-375\sqrt{3}\le M\le375\sqrt{3}\)

\(\sqrt{2x+5}=5\left(x\ge-\dfrac{5}{2}\right)\)

\(\Leftrightarrow2x+5=25\)

\(\Leftrightarrow2x=20\)

\(\Leftrightarrow x=10\left(n\right)\)

\(\sqrt{2x+5}=5\left(x\ge-\dfrac{5}{2}\right)\)

\(\Leftrightarrow2x+5=25\Leftrightarrow2x=20\Leftrightarrow x=10\left(TM\right)\)

KL.......

\(\sqrt{2x+5}=5\)

Như này à bn?

\(Q=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2+1}}-\left(\sqrt{2}+\sqrt{3}\right)\)

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

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

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

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

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