\(=-5\cdot\dfrac{1}{2}+0-5\cdot\dfrac{6}{5}=-\dfrac{5}{2}-6=-\dfrac{17}{2}\)

\(=\left(\dfrac{3}{2}\cdot\dfrac{2}{5}+2\cdot\dfrac{1}{5}\right):\dfrac{3}{8}=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)\cdot\dfrac{8}{3}=\dfrac{8}{3}\)

a, \(=7\sqrt{2}-6\sqrt{2}+\frac{1}{2}.2\sqrt{2}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

b, \(=4\sqrt{a}+4\sqrt{10a}-9\sqrt{10a}=4\sqrt{a}-5\sqrt{10a}\)

c, \(=6+\sqrt{15}-\sqrt{60}=6+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)

Rút gọn

a) Ta có: \(\sqrt{98}-\sqrt{72}+\frac{1}{2}\sqrt{8}\)

\(=\sqrt{2}\left(\sqrt{49}-\sqrt{36}+\frac{1}{2}\sqrt{4}\right)\)

\(=\sqrt{2}\left(7-6+\frac{1}{2}\cdot2\right)\)

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

b) Ta có: \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\)

\(=\sqrt{a}\left(\sqrt{16}+2\sqrt{40}-3\sqrt{90}\right)\)

\(=\sqrt{a}\left(4+4\sqrt{10}-9\sqrt{10}\right)\)

\(=\sqrt{a}\left(4-5\sqrt{10}\right)\)

\(=4\sqrt{a}-5\sqrt{10a}\)

c) Ta có: \(\left(2\sqrt{3}+\sqrt{5}\right)\cdot\sqrt{3}-\sqrt{60}\)

\(=6+\sqrt{15}-\sqrt{60}\)

\(=6-\sqrt{15}\)

Bài 1: Tính

a) Ta có: \(\left(\sqrt{3}+2\right)^2\)

\(=\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot2+2^2\)

\(=3+4\sqrt{3}+4\)

\(=7+4\sqrt{3}\)

b) Ta có: \(-\left(\sqrt{2}-1\right)^2\)

\(=-\left[\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2\right]\)

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

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

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

Bài 2: Tính

a) Ta có: \(0.5\cdot\sqrt{100}-\sqrt{\frac{25}{4}}\)

\(=\frac{1}{2}\cdot10-\frac{5}{2}\)

\(=5-\frac{5}{2}\)

\(=\frac{5}{2}\)

b) Ta có: \(\left(\sqrt{1\frac{9}{16}}-\sqrt{\frac{9}{16}}\right):5\)

\(=\left(\sqrt{\frac{25}{16}}-\frac{3}{4}\right)\cdot\frac{1}{5}\)

\(=\left(\frac{5}{4}-\frac{3}{4}\right)\cdot\frac{1}{5}\)

\(=\frac{2}{4}\cdot\frac{1}{5}\)

\(=\frac{1}{10}\)

Bài 3: So sánh

a) Ta có: \(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{18}\)

\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{12}\)

mà \(\sqrt{18}>\sqrt{12}\)(Vì 18>12)

nên \(3\sqrt{2}>2\sqrt{3}\)

\(\Leftrightarrow\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)

b) Ta có: \(\left(15-2\sqrt{10}\right)^2\)

\(=225-2\cdot15\cdot2\sqrt{10}+\left(2\sqrt{10}\right)^2\)

\(=225-60\sqrt{10}+40\)

\(=265-60\sqrt{10}\)

\(=135+130-60\sqrt{10}\)

Ta có: \(\left(3\sqrt{15}\right)^2=3^2\cdot\left(\sqrt{15}\right)^2=9\cdot15=135\)

Ta có: \(130-60\sqrt{10}\)

\(=\sqrt{16900}-\sqrt{36000}< 0\)(Vì 16900<36000)

\(\Leftrightarrow130-60\sqrt{10}+135< 135\)(cộng hai vế của BĐT cho 135)

\(\Leftrightarrow\left(15-2\sqrt{10}\right)^2< \left(3\sqrt{15}\right)^2\)

\(\Leftrightarrow15-2\sqrt{10}< 3\sqrt{15}\)

\(\Leftrightarrow\frac{15-2\sqrt{10}}{3}< \frac{3\sqrt{15}}{3}=\sqrt{15}\)

hay \(\frac{15-2\sqrt{10}}{3}< \sqrt{15}\)

thank !! dù mik ko hỉu câu a cho lắm

câu a dùng hằng đẳng thức. Phần a là phần easy nhất mà k hỉu

\(=4\cdot5-2\cdot\dfrac{2}{3}=20-\dfrac{4}{3}=\dfrac{56}{3}\)

Jdkdk

Jidkri

\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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