Xét dạng tổng quát:

\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}.\frac{1}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)

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

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

Áp dụng vào bài toán:

\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}}\)

\(< 2\left(1-\frac{1}{\sqrt{2}}\right)+2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+...+\left(\frac{1}{\sqrt{2003}}-\frac{1}{\sqrt{2004}}\right)\)

\(< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2003}}-\frac{1}{\sqrt{2004}}\right)\)

\(< 2\left(1-\frac{1}{\sqrt{2004}}\right)\)

\(< 2-\frac{2}{\sqrt{2004}}< 2\)

=>đpcm

\(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\right)\left(3\sqrt{\frac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)

\(=\left(\frac{3}{2}\sqrt{6}+2\frac{\sqrt{2.3}}{3}-4\frac{\sqrt{2.3}}{2}\right)\left(3\frac{\sqrt{2.3}}{3}-2\sqrt{3}-\sqrt{6}\right)\)

\(=\sqrt{6}\left(\frac{3}{2}+\frac{2}{3}-\frac{4}{2}\right)\left(\sqrt{6}-\sqrt{6}-2\sqrt{3}\right)\)

\(=\sqrt{6}.\frac{1}{6}.\left(-2\sqrt{3}\right)\)

\(=-\sqrt{2}\)

Q ở đâu z bạn?

diện tích hình thang ABCD là:

(4+9)*5:2=32,5(cm2)

              đáp số:32,5cm2

cick cho mk nhé!

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

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

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

b,=\(\sqrt{4-4\sqrt{3}+3}=\sqrt{\left(2-\sqrt{3}\right)^2}\)

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

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

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

=\(\sqrt{2}-1\)(vì căn 2 > 1)

\(\sqrt{2x^2+5x-2}-\sqrt{2x^2+5x-9}=1\)

<=> \(\sqrt{2x^2+5x-2}=1+\sqrt{2x^2+5x-9}\)(1)

ĐK : \(\orbr{\begin{cases}x\ge\frac{\sqrt{97}-5}{4}\\x\le\frac{-\sqrt{97}-5}{4}\end{cases}}\)

Đặt t = 2x² + 5x - 2

(1) <=> \(\sqrt{t}=1+\sqrt{t-7}\)( t ≥ 7 )

Bình phương hai vế

<=> \(t=t+2\sqrt{t-7}-6\)

<=> \(t+2\sqrt{t-7}-t=6\)

<=> \(2\sqrt{t-7}=6\)

<=> \(\sqrt{t-7}=3\)

<=> t - 7 = 9

<=> t = 16 ( tm )

=> 2x² + 5x - 2 = 16

<=> 2x² + 5x - 2 - 16 = 0

<=> 2x² + 5x - 18 = 0

<=> 2x² - 4x + 9x - 18 = 0

<=> 2x( x - 2 ) + 9( x - 2 ) = 0

<=> ( x - 2 )( 2x + 9 ) = 0

<=> \(\orbr{\begin{cases}x-2=0\\2x+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{9}{2}\end{cases}}\)( tm )

Vậy phương trình có hai nghiệm x₁ = 2 ; x₂ = -9/2

\(\sqrt{2x^2+5x-2}-\sqrt{2x^2+5x-9}=1\)

\(\Leftrightarrow\sqrt{2x^2+5x-2}-\sqrt{2x^2+5x-2-7}=1\)

Đặt : \(\sqrt{2x^2+5x-2}=t\)

\(\Leftrightarrow t-\sqrt{t^2-7}=1\)

Gải được t thế vào tìm được x =2 nha bạn

b)\(\sqrt{4x-8}+2\sqrt{9x-18}-\sqrt{x-2}=14\)

Đk:\(x\ge2\)

\(pt\Leftrightarrow\sqrt{4x-8}-4+2\sqrt{9x-18}-12-\left(\sqrt{x-2}-2\right)=0\)

\(\Leftrightarrow\frac{4x-8-16}{\sqrt{4x-8}+4}+\frac{4\left(9x-18\right)-144}{2\sqrt{9x-18}+12}-\frac{x-2-4}{\sqrt{x-2}+2}=0\)

\(\Leftrightarrow\frac{4\left(x-6\right)}{\sqrt{4x-8}+4}+\frac{36\left(x-6\right)}{2\sqrt{9x-18}+12}-\frac{x-6}{\sqrt{x-2}+2}=0\)

\(\Leftrightarrow\left(x-6\right)\left(\frac{4}{\sqrt{4x-8}+4}+\frac{36}{2\sqrt{9x-18}+12}-\frac{1}{\sqrt{x-2}+2}\right)=0\)

Thấy: \(\frac{4}{\sqrt{4x-8}+4}+\frac{36}{2\sqrt{9x-18}+12}-\frac{1}{\sqrt{x-2}+2}=0\) vô nghiệm

Nên x-6=0 suy ra x=6

a. => | x-2 | = 8 

         x=10

=> |x-2|=-8

      x=-6