a) ĐS: 5.

b) = = = √9.√25 = 3.5 = 15.

c) ĐS: 45

d) ĐS: 25

a. \(\sqrt{13^2-12^2}\)

=\(\sqrt{\left(13+12\right).\left(13-12\right)}\)

=\(\sqrt{25.1}\)

=\(\sqrt{25}.\sqrt{1}\)

=5.1

=5

b. \(\sqrt{17^2-8^2}\)

=\(\sqrt{\left(17+8\right).\left(17-8\right)}\)

=\(\sqrt{25.9}\)

=\(\sqrt{25}.\sqrt{9}\)

=5.3

=15

c. \(\sqrt{117^2-108^2}\)

=\(\sqrt{\left(117+108\right).\left(117-108\right)}\)

=\(\sqrt{225.9}\)

=\(\sqrt{225}.\sqrt{9}\)

=15.3

=45

d. \(\sqrt{313^2-312^2}\)

=\(\sqrt{\left(313+312\right).\left(313-312\right)}\)

=\(\sqrt{625.1}\)

=\(\sqrt{625}.\sqrt{1}\)

=25.1

=25

c.\(\sqrt{117^2-108^2}\)

a) \(\sqrt{13^2-12^2}\)=\(\sqrt{\left(13-12\right)\left(13+12\right)}\)=\(\sqrt{1x25}\)=5

Câu a: Ta có:

√132−122=√(13+12)(13−12)132−122=(13+12)(13−12)

                      =√25.1=√25=25.1=25

                      =√52=|5|=5=52=|5|=5.

Câu b: Ta có:

√172−82=√(17+8)(17−8)172−82=(17+8)(17−8)

                    =√25.9=√25.√9=25.9=25.9

                    =√52.√32=|5|.|3|=52.32=|5|.|3|.

                    =5.3=15=5.3=15.

Câu c: Ta có:

√1172−1082=√(117−108)(117+108)1172−1082=(117−108)(117+108)

                          =√9.225=9.225 =√9.√225=9.225

                          =√32.√152=|3|.|15|=32.152=|3|.|15|

                          =3.15=45=3.15=45.

Câu d: Ta có:

√3132−3122=√(313−312)(313+312)3132−3122=(313−312)(313+312)

                          =√1.625=√625=1.625=625

                          =√252=|25|=25=252=|25|=25.

a, \(\sqrt{13^2-12^2}=\sqrt{\left(13-12\right)\left(13+12\right)}=\sqrt{1.25}=5\)

b, \(\sqrt{17^2-8^2}=\sqrt{\left(17-8\right)\left(17+8\right)}=\sqrt{9.25}=3.5=15\)

c, \(\sqrt{117^2-108^2}=\sqrt{\left(117-108\right)\left(117+108\right)}=\sqrt{9.225}=3.15=45\)

d, \(\sqrt{313^2-312^2}=\sqrt{\left(313-312\right)\left(313+312\right)}=\sqrt{1.625}=25\)

a, \(\sqrt{13^2-12^2}=\sqrt{\left(13-12\right)\left(13+12\right)}\)

\(=\sqrt{1.25}=\sqrt{25}=5\)

b, \(\sqrt{17^2-8^2}=\sqrt{\left(17-8\right)\left(17+8\right)}\)

\(=\sqrt{9.25}=\sqrt{9}.\sqrt{25}=3.5=15\)

c, \(\sqrt{117^2-108^2}=\sqrt{\left(117-108\right)\left(117+108\right)}\)

\(=\sqrt{9.225}=\sqrt{9}.\sqrt{225}=3.15=45\)

d, \(\sqrt{313^2-312^2}=\sqrt{\left(313-312\right)\left(313+312\right)}\)

\(=\sqrt{1.625}=\sqrt{625}=25\)

Chúc bạn học tốt!!!

a, \(\sqrt{13^2-12^2}=\sqrt{\left(13-12\right)\left(13+12\right)}=\sqrt{25}=5\)

b, \(\sqrt{17^2-8^2}=\sqrt{\left(17-8\right)\left(17+8\right)}=\sqrt{9.25}=15\)

c, \(\sqrt{117^2-108^2}=\sqrt{\left(117-108\right)\left(117+108\right)}\)

\(=\sqrt{9.225}=45\)

d, \(\sqrt{313^2-312^2}=\sqrt{\left(313-312\right)\left(313+312\right)}=\sqrt{625}=25\)

a) \(\sqrt{117^2-108^2}=\sqrt{\left(117-108\right)\left(117+108\right)}=\sqrt{9\cdot225}=\sqrt{3^2\cdot15^2}=\left|3\cdot15\right|=45\)

b) \(\sqrt{9-4\sqrt{5}}+2=\sqrt{5-4\sqrt{5}+4}+2=\sqrt{\left(\sqrt{5}-2\right)^2}+2=\left|\sqrt{5}-2\right|+2=\sqrt{5}\)

\(a,\sqrt{117^2-108^2}\\ =\sqrt{\left(117-108\right)\left(117+108\right)}\\ =\sqrt{9.225}\\ =\sqrt{3^2}.\sqrt{15^2}\\ =3.15\\ =45\)

\(b,\sqrt{9-4\sqrt{5}}+2=\sqrt{5}\)

\(VT=\sqrt{9-4\sqrt{5}}+2\\ =\sqrt{\sqrt{5^2}-2.2\sqrt{5}+2^2}+2\\ =\sqrt{\left(\sqrt{5}-2\right)^2}+2\\ =\left|\sqrt{5}-2\right|+2\\ =\sqrt{5}-2+2\\ =\sqrt{5}=VP\left(dpcm\right)\)

a) \(\sqrt{117^2-108^2}\)

\(=\sqrt{\left(117-108\right)\left(117+108\right)}\)

\(=\sqrt{9\cdot225}\)

\(=\sqrt{3^2\cdot15^2}\)

\(=\sqrt{\left(3\cdot15\right)^2}\)

\(=3\cdot15\)

\(=45\)

b) VT: \(\sqrt{9-4\sqrt{5}}+2\)

\(=\sqrt{4-4\sqrt{5}+5}+2\)

\(=\sqrt{2^2-2\cdot2\cdot\sqrt{5}\cdot\left(\sqrt{5}\right)^2}+2\)

\(=\sqrt{\left(2-\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|+2\)

\(=\sqrt{5}-2+2=\sqrt{5}=VP\) (đpcm)

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

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

b, \(\sqrt{3-2\sqrt{2}}=\sqrt{\sqrt{2}^2-2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)

c, \(\sqrt{8-2\sqrt{15}}=\sqrt{\sqrt{5}^2-2\sqrt{5.3}+\sqrt{3}^2}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\)

Bài 1: 

a: Ta có: \(\sqrt{3x^2}=\sqrt{12}\)

\(\Leftrightarrow3x^2=12\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

b: Ta có: \(\sqrt{\left(x-2\right)^2}=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

làm bừa thui,ai tích mình mình tích lại

Số số hạng là : 

Có số cặp là :

50 : 2 = 25 ( cặp )

Mỗi cặp có giá trị là :

99 - 97 = 2 

Tổng dãy trên là :

25 x 2 = 50

Đáp số : 50