\(\sqrt{2\sqrt{6}+6+2\sqrt{3}+2\sqrt{2}}\)

=\(\sqrt{2+3+1+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}}\)

=\(\sqrt{\left(\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+1^2+2\sqrt{2}.\sqrt{3}+2\sqrt{3}.1+2\sqrt{2}.1}\)

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

=\(\sqrt{2}+\sqrt{3}+1\)(Vì \(\sqrt{2}+\sqrt{3}+1>0\))

Chắc đúng á   :=)

\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right)\)

\(\Rightarrow B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{779}{780}\)

\(\Rightarrow B=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{1558}{1560}\)

\(\Rightarrow B=\frac{1.4}{2.3}.\frac{2.5}{3.4}\frac{3.6}{4.5}...\frac{38.41}{39.40}\)

\(\Rightarrow B=\frac{\left(1.2.3...38\right)\left(4.5.6...41\right)}{\left(2.3.4...39\right)\left(3.4.5...40\right)}\)

\(\Rightarrow B=\frac{1.41}{39.3}=\frac{41}{117}\)

Vậy B=\(\frac{41}{117}\)

Ai thấy đúng thì k nha

B= \(\frac{41}{117}\)

4.(1.2.3)=1.2.3.4-0.1.2.3

4.(2.3.4)=2.3.4.5-1.2.3.4

4.(3.4.5)=3.4.5.6-2.3.4.5

        ...

4(n-1)n(n+1)=(n-1)n(n+1)(n+2)-(n-2)(n-1)n(n+1)

=> 4B=(n-1)n(n+1)(n+2)=> B=(n-1)n(n+1)(n+2) chia hết cho 4.

4(n-1)n(n+1

Câu hỏi của Nguyễn Thùy Linh - Toán lớp 7 - Học toán với OnlineMath

Em tham khảo tại đây nhé.