Question

[Nguyễn Minh Hùng]

Giúp em với ạ từ bài 1-5 ạ

Answer 1

[Nguyễn Ngọc Lộc]

Bài 3 :

\(\Leftrightarrow\sqrt{9x^2-6x+1}=\sqrt{\left(3x-1\right)^2}=\left|3x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy ...

Bài 5 :

Ta có :\(x-5\sqrt{x}+7=x-2.\sqrt{x}.\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\)

Thấy : \(\left(\sqrt{x}-\dfrac{5}{2}\right)^2\ge0\)

\(\Rightarrow\left(\sqrt{x}-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Rightarrow P=\dfrac{1}{x-5\sqrt{x}+7}=\dfrac{1}{\left(\sqrt{x}-\dfrac{5}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)

Vậy \(Max_P=\dfrac{4}{3}\Leftrightarrow\sqrt{x}-\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{25}{4}\)
 

Answer 2

[Nguyễn Lê Phước Thịnh]

Bài 1: 

a) Ta có: \(\sqrt{25}\cdot\sqrt{144}+\sqrt[3]{-27}-\sqrt[3]{216}\)

\(=5\cdot12-3-6\)

\(=60-9=51\)

b) Ta có: \(\sqrt{8.1\cdot360}\)

\(=\sqrt{8.1\cdot10\cdot36}\)

\(=\sqrt{81\cdot36}\)

\(=9\cdot6=54\)

Bài 2: 

a) Ta có: \(\sqrt{80}-\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{3\dfrac{1}{5}}\)

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

\(=3\sqrt{5}+2+\dfrac{4\sqrt{5}}{5}\)

\(=\dfrac{10+19\sqrt{5}}{5}\)

b) Ta có: \(\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}+\dfrac{3+6\sqrt{3}}{\sqrt{3}}-\dfrac{13}{\sqrt{3}+4}\)

\(=\dfrac{-\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\dfrac{\sqrt{3}\left(\sqrt{3}+6\right)}{\sqrt{3}}-\dfrac{13\left(4-\sqrt{3}\right)}{\left(4+\sqrt{3}\right)\left(4-\sqrt{3}\right)}\)

\(=-\sqrt{3}+\sqrt{3}+6-4+\sqrt{3}\)

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