Question

[Nguyễn Bảo Ngọc]

Bài 1: Tính

a) \((\sqrt{3}+2)^\text{2}\)

b) -\((\sqrt{2}-1)^\text{2}\)

Bài 2: Tính

a) \(0,5\sqrt{100} - \sqrt{\dfrac{25}{4}}\)

b) \((\sqrt{1\dfrac{9}{16}}- \sqrt{\dfrac{9}{16}}) : 5\)

Bài 3 : So sánh

a) \(\sqrt{3\sqrt{2}} và \sqrt{2\sqrt{3}}\)

b) \(\dfrac{15 - 2\sqrt{10}}{3} và \sqrt{15}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

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

Answer 2

[Tú Nguyễn]

phần a của 3 bài đều easy mà cả 3 bài đều easy