a)\(\sqrt{\left(13+12\right)\left(13-12\right)}=\sqrt{25}+\sqrt{1}=5+1=6\)=6 ( hằng đẳng thức số 3) \(a^2-b^2=\left(a+b\right)\left(a-b\right)\)
b) tương tự
a) \(\sqrt{13^2-12^2}=\sqrt{\left(13-12\right)\left(13+12\right)}=\sqrt{25}=5\)
b) \(\sqrt{17^2-8^2}=\sqrt{\left(17-8\right)\left(17+8\right)}=\sqrt{25.9}=\sqrt{225}=15\)
c) \(\sqrt{117^2-108^2}=\sqrt{\left(117-108\right)\left(117+108\right)}=\sqrt{225.9}=\sqrt{2025}=45\)
d) \(\sqrt{313^2-312^2}=\sqrt{\left(313-312\right)\left(313+312\right)}=\sqrt{625}=25\)
mk nghi nhu vay ko biet co dung ko
dung thi bao mk nha
2
a) \(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\)
bien doi ve trai ta duoc
\(=2^2-\left(\sqrt{3}\right)^2\)
\(=4-3=1\)= ve phai
vay \(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\)
b) 2 so nghich dao cua nhau co h = 1
nen ta co \(\left(\sqrt{2006}-\sqrt{2005}\right).\left(\sqrt{2006}+\sqrt{2005}\right)=1\)
phan h ve trai ta co
\(\left(\sqrt{2006}-\sqrt{2005}\right).\left(\sqrt{2006}+\sqrt{2005}\right)\)
\(=\left(\sqrt{2006}\right)^2-\left(\sqrt{2005}\right)^2\)
\(=2006-2005\)
\(=1\)= ve phai
vay \(\left(\sqrt{2006}-\sqrt{2005}\right)\)va \(\left(\sqrt{2006}+\sqrt{2005}\right)\)la 2 so nghich dao cua nhau
mk vua lam nen ko biet co dung ko nua