\(\dfrac{10^2+11^2+12^2}{13^2+14^2}=\dfrac{100+121+144}{169+196}\)

\(=\dfrac{365}{365}=1\)

D=(102+112+122):(132+142)

D =(100+121+144) / (169+196)

D =365 / 365

D = 1

Đề là \(D=\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\) cậu thêm mũ vào nhé.

=365 : 365

=1

(10² +11²+12²):(13² +14²)

=(100+121+144):(169+196)

=365:365

=1

( 100 + 121 + 144 ) : ( 169 + 196 )

= 365 : 365 = 1

\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)

\(=\left(100+121+144\right):\left(169+196\right)\)

\(=\left(221+144\right):365\)

\(=365:365\)

\(=1\)Vậy kết quả = 1

\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)

\(=\left(100+121+144\right):\left(169+196\right)\)

\(=365:365\)

\(=1\)

\(a,\dfrac{121.75.130.169}{39.60.11.198}=\dfrac{11.11.25.3.13.10.169}{13.3.6.10.11.11.18}=\dfrac{25.169}{6.18}\)

còn bài tính hợp lí và tính nhanh thì sao bạn

\(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)

\(=\left(\sqrt{12+2\sqrt{\left(\sqrt{13+1}\right)^2}}-\sqrt{\left(\sqrt{11+1}\right)^2}\right)\left(\sqrt{11}+\sqrt{13}\right)\)

\(=\left(\sqrt{12+2\sqrt{13+2}}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)

\(=\left(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)

\(=\left(\sqrt{13}+1-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)\(=\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{11}+\sqrt{13}\right)=13-11=2\)

sao dấu= thứ 2 lại ra như vậy