\(\sqrt{20}\cdot\sqrt{72}\cdot\sqrt{4,9}=\sqrt{20\cdot72\cdot4,9}=\sqrt{2\cdot10\cdot72\cdot4,9}\\ =\sqrt{144\cdot49}=\sqrt{144}\cdot\sqrt{49}=12\cdot7=84\)

Bài 2:

a) \(\sqrt{3a^3}\cdot\sqrt{12a}=\sqrt{3a^3\cdot12a}=\sqrt{36a^4}=6a^2\)

b) \(\sqrt{2a\cdot32ab^2}=\sqrt{64a^2b^2}=8ab\)

a. =\(\sqrt{20.72.4,9}=\sqrt{2.72.49}=\sqrt{144.49}=12.7=84\)

b. \(\sqrt{\frac{999}{111}}=\sqrt{9}=3\)

c. = \(\sqrt{9472+27256}=\sqrt{36728}\approx191,645\)

d. = \(\sqrt{\frac{\left(149+76\right)\left(149-76\right)}{\left(457+348\right)\left(457-348\right)}}=\sqrt{\frac{225.73}{805.109}}=\sqrt{\frac{3285}{17549}}\approx136,817\)

a) \(\sqrt{200}-\sqrt{32}+\sqrt{72}\)

\(=\sqrt{10^2\cdot2}-\sqrt{4^2\cdot2}+\sqrt{6^2\cdot2}\)

\(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}\)

\(=\left(10-4+6\right)\sqrt{2}\)

\(=12\sqrt{2}\)

b) \(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}\)

\(=4\cdot2\sqrt{5}-3\cdot5\sqrt{5}+5\cdot3\sqrt{5}-3\sqrt{5}\)

\(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}\)

\(=\left(8-15+15-3\right)\sqrt{5}\)

\(=5\sqrt{5}\)

c) \(\left(2\sqrt{8}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\sqrt{20}-2\sqrt{2}\right)\)

\(=\left(2\cdot2\sqrt{2}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\cdot2\sqrt{5}-2\sqrt{2}\right)\)

\(=\left(3\sqrt{5}-3\sqrt{2}\right)\left(72-10\sqrt{5}-2\sqrt{2}\right)\)

c) \(\sqrt{20}-\sqrt{45}+3\sqrt{8}+\sqrt{72}\)

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

\(=-\sqrt{5}+12\sqrt{2}\)

d) \(\dfrac{3}{\sqrt{3}+1}\)

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

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

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

e) \(\dfrac{2}{\sqrt{10}-\sqrt{7}}\)

\(=\dfrac{2\left(\sqrt{10}+\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}\)

\(=\dfrac{2\left(\sqrt{10}+\sqrt{7}\right)}{3}\)

\(=\dfrac{2\sqrt{10}+2\sqrt{7}}{3}\)

c)

\(=\sqrt{4.5}-\sqrt{9.5}+\sqrt{8.9}+\sqrt{72}\\ =2\sqrt{5}-3\sqrt{5}+\sqrt{72}+\sqrt{72}\\ =-\sqrt{5}+2\sqrt{72}\\ =-\sqrt{5}+2\sqrt{36.2}\\ =-\sqrt{5}+12\sqrt{2}\)

a) \(\Rightarrow\dfrac{x^2}{\sqrt{5}}=\sqrt{20}\Rightarrow x^2=\sqrt{20.5}=\sqrt{100}=10\)

\(\Rightarrow x=\pm\sqrt{10}\)

b)ĐKXĐ: \(x\ge0\)

 \(\Rightarrow3\sqrt{2x}+\sqrt{2}-6\sqrt{2}+4=0\)

\(\Rightarrow3\sqrt{2x}=5\sqrt{2}-4\)

\(\Rightarrow18x=50+16-40\sqrt{2}\)

\(\Rightarrow x=\dfrac{66-40\sqrt{2}}{18}\)

\(a,\Leftrightarrow\dfrac{x^2}{\sqrt{5}}=\sqrt{20}=2\sqrt{5}\Leftrightarrow x^2=2\sqrt{5}\cdot\sqrt{5}=10\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}\\x=-\sqrt{10}\end{matrix}\right.\)

\(b,ĐK:x\ge0\\ PT\Leftrightarrow3\sqrt{2x}+\dfrac{1}{7}\cdot7\sqrt{2}-6\sqrt{2}+4=0\\ \Leftrightarrow3\sqrt{2x}=5\sqrt{2}-4\\ \Leftrightarrow\sqrt{2x}=\dfrac{5\sqrt{2}-4}{3}\\ \Leftrightarrow2x=\dfrac{66-40\sqrt{2}}{9}\\ \Leftrightarrow x=\dfrac{66-40\sqrt{2}}{18}=\dfrac{33-20\sqrt{2}}{9}\left(tm\right)\)

\(\sqrt{2}+\sqrt{6}+\sqrt{12}+...+\sqrt{110}\)\(=\sqrt{1.2}+\sqrt{2.3}+\sqrt{3.4}+...+\sqrt{10.11}\)

\(< \frac{1+2}{2}+\frac{2+3}{2}+\frac{3+4}{2}+...+\frac{10+11}{2}\)\(=\frac{1}{2}\left[\left(1+2+3+...+10\right)+\left(2+3+4+...+11\right)\right]\)\(=\frac{1}{2}\left(\frac{11.10}{2}+\frac{13.10}{2}\right)=\frac{1}{2}\left(55+65\right)=60\)

Vậy \(\sqrt{2}+\sqrt{6}+\sqrt{12}+...+\sqrt{110}< 60.\)

chị ơi toán lớp 7 hả

\(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{56}+\sqrt{72}+\sqrt{90}+\sqrt{110}\) < 60 nha.

a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)

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

\(=3\sqrt{5}+12\sqrt{2}\)

b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)

\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)

\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)

\(=9+3\sqrt{5}-4\sqrt{5}+4\)

\(=13-\sqrt{5}\)

c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)

\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)

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

\(=-\sqrt{5}\)

e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)

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

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

f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{5}+1-\sqrt{5}+2\)

=3

e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)

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

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

f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{5}+1-\sqrt{5}+2\)

=3

a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)

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

\(=3\sqrt{5}+12\sqrt{2}\)

b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)

\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)

\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)

\(=9+3\sqrt{5}-4\sqrt{5}+4\)

\(=13-\sqrt{5}\)

\(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)

\(=\sqrt{4.5}-\sqrt{9.5}+3\sqrt{18}+\sqrt{4.18}\)

\(=2\sqrt{5}-3\sqrt{5}+3\sqrt{18}+2\sqrt{18}\)

\(=-\sqrt{5}+5\sqrt{18}\)

\(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)

\(=2\sqrt{5}-3\sqrt{5}+3\sqrt{18}+2\sqrt{18}\)

\(=-\sqrt{5}+5\sqrt{18}\)