Mình giải theo kiểu lớp 6 nhá !

=\(33.\left(\frac{34}{15}+\frac{34}{35}+\frac{34}{63}+\frac{34}{99}\right)\)

=\(33.\left[34.\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\right]\)

=\(33.\left[34.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\right]\)

=\(33.\left[34.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).\frac{1}{2}\right]\)

=\(33.\left[34\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).\frac{1}{2}\right]\)

=\(33.\left[34.\left(\frac{1}{3}-\frac{1}{11}\right).\frac{1}{2}\right]\)

=\(33.\left(34.\frac{8}{33}.\frac{1}{2}\right)\)

=\(33.\frac{136}{33}\)

=\(\frac{33.136}{33}\)(*)

=\(136\)

(Bạn có thể bỏ bước có dấu *)

 (  3434/3535 + 3434/6363 + 3434/9999 ) × 55

= (  34/35 + 34/63 + 34/99 ) × 55

= 34 × ( 1/35 + 1/63 + 1/99 ) × 55

=  34 × 3/55 × 55

= 102/55 × 55

= 102

Mình ko chắc lắm đâu nhá

Ta có:
\(\dfrac{3434}{7878}=\dfrac{3434:202}{7878:202}=\dfrac{17}{39}\)
\(\Rightarrow\dfrac{3434}{7878}< \dfrac{20}{39}\)
\(-------\)
Ta có:
\(\dfrac{123123}{405405}=\dfrac{123123:1001}{405405:1001}=\dfrac{123}{405}\)
\(\Rightarrow\dfrac{123123}{405405}>\dfrac{123}{1109}\)
\(---------\)
Ta có:
\(\dfrac{666666}{323232}=\dfrac{666666:80808}{323232:80808}=\dfrac{8,25}{4}\)
\(\Rightarrow\dfrac{666666}{323232}< \dfrac{25}{4}\)

tại sao ko sử dụng máy tính

Vì \(\sqrt{2,5+6,5}\ge0\Rightarrow\sqrt{2,5+6,5}< \sqrt{2,5+6,5}+1\)

VÌ A = 1/2010 > 1/2011 > 1/2012  (1)

     B = 1/2009 <1/1007 (2) 

TỪ (1) VÀ (2)  => 1/2010 < 1/1007

VẬY A < B

Sao bạn biết

Giả sử A > B

<=> 19 >\(5\sqrt{3}+6\sqrt{2}\)

<=> (6 + 3 - \(2\sqrt{3}\sqrt{6}\)

) + (10 - 5\(\sqrt{3}\))>0

<=> (\(\sqrt{6}-\sqrt{3}\))² + (10 - \(5\sqrt{3}\))>0

Mà 10 - 5\(\sqrt{3}\)> 10 - 5\(\sqrt{4}\) = 0

Vậy A > B