a: \(=\sqrt{25}=5\)

b: \(=3\cdot5=15\)

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

b) \(\sqrt{17^2-8^2}=\sqrt{\left(17-8\right)\left(17+8\right)}=\sqrt{9.25}=\sqrt{9}.\sqrt{25}=3.5=15\)

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

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)\(\sqrt{\left(13+12\right)\left(13-12\right)}=\sqrt{25}+\sqrt{1}=5+1=6\)=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

\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)

k: \(\sqrt[3]{\left(4-2\sqrt{3}\right)\left(\sqrt{3}-1\right)}\)

\(=\sqrt[3]{\left(\sqrt{3}-1\right)^3}\)

\(=\sqrt{3}-1\)

câu đầu bạn xem lại đề đi nha 

các phần còn lại

b)B=\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)=\(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)=\(\sqrt{7}-1-\left(\sqrt{7}+1\right)=-2\)

c)tính từng căn nha

\(\sqrt{13-4\sqrt{3}}=\sqrt{12-2\sqrt{12}+1}=\sqrt{\left(\sqrt{12}-1\right)^2}=\sqrt{12}-1=2\sqrt{3}-1\)

\(\sqrt{22-12\sqrt{2}}=\sqrt{18-4\sqrt{18}+4}=\sqrt{\left(\sqrt{18}-2\right)^2}=\sqrt{18}-2=3\sqrt{2}-3\)

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

thay vào tính C đc C=2

d)có \(\sqrt{9+4\sqrt{2}}=\sqrt{8+2\sqrt{8}+1}=\sqrt{\left(\sqrt{8}+1\right)^2}=\sqrt{8}+1\)\(\Rightarrow6\sqrt{2+\sqrt{9+4\sqrt{2}}}=6\sqrt{2+\sqrt{8}+1}=6\sqrt{2+2\sqrt{2}+1}\)

=\(6\sqrt{\left(\sqrt{2}+1\right)^2}=6\left(\sqrt{2}+1\right)=6\sqrt{2}+6\)\(\Rightarrow D=\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{17-6\sqrt{2}-6}=\sqrt{11-6\sqrt{2}}=\sqrt{9-6\sqrt{2}+2}\)

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

a: \(5\sqrt{2}-8\sqrt{3}+30\sqrt{3}-6\sqrt{3}=5\sqrt{2}+16\sqrt{3}\)

b: \(=14\sqrt{3}-\dfrac{3}{32}\cdot8\sqrt{3}+\dfrac{4}{18}\cdot9\sqrt{3}-\dfrac{1}{10}\cdot10\sqrt{3}\)

\(=14\sqrt{3}-\dfrac{3}{4}\sqrt{3}+2\sqrt{3}-1\sqrt{3}=\dfrac{57}{4}\sqrt{3}\)

c: \(=\dfrac{-1}{2}\cdot6\sqrt{3}+\dfrac{1}{15}\cdot5\sqrt{3}-\dfrac{1}{22}\cdot11\sqrt{3}+2\sqrt{3}\)

\(=-3\sqrt{3}+\dfrac{1}{3}\sqrt{3}-\dfrac{1}{2}\sqrt{3}+2\sqrt{3}=-\dfrac{7}{6}\sqrt{3}\)

d: \(=\dfrac{5}{8}\cdot4\sqrt{3}-\dfrac{1}{33}\cdot11\sqrt{3}+\dfrac{3}{14}\cdot7\sqrt{3}-\dfrac{1}{4}\cdot8\sqrt{3}\)

\(=\dfrac{5}{2}\sqrt{3}-\dfrac{1}{3}\sqrt{3}+\dfrac{3}{2}\sqrt{3}-2\sqrt{3}=\dfrac{5}{3}\sqrt{3}\)

Bài 1.

1. \(\sqrt{-3x+6}\) có nghĩa khi \(-3x+6\ge0\Leftrightarrow-3x\ge-6\Rightarrow x\le2\)

2.

\( a){\left( {\sqrt 7 - \sqrt 5 } \right)^2} + 2\sqrt {35} = 7 - 2\sqrt {35} + 5 + 2\sqrt {35} = 12\\ b)3\sqrt 8 - \sqrt {50} - \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} = 6\sqrt 2 - 5\sqrt 2 - \sqrt 2 + 1 = 1 \)

Bài 2.

\( M = \dfrac{{\sqrt a + 3}}{{\sqrt a - 2}} - \dfrac{{\sqrt a - 1}}{{\sqrt a + 2}} + \dfrac{{4\sqrt a - 4}}{{4 - a}}\\ M = \dfrac{{\left( {\sqrt a + 2} \right)\left( {\sqrt a + 3} \right) - \left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right) - \left( {4\sqrt a - 4} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}\\ M = \dfrac{{4\sqrt a + 8}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}\\ M = \dfrac{{4\left( {\sqrt a + 2} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}\\ M = \dfrac{4}{{\sqrt a - 2}} \)

Bài 3.

1.

\( a)\sqrt {{{313}^2} - {{312}^2}} + \sqrt {{{17}^2} - {8^2}} = \sqrt {\left( {313 - 312} \right)\left( {313 + 312} \right)} + \sqrt {\left( {17 - 8} \right)\left( {17 + 8} \right)} \\ = \sqrt {625} + \sqrt {9.25} = 25 + 3.5 = 25 + 15 = 40\\ b)\dfrac{{2 + \sqrt 2 }}{{1 + \sqrt 2 }} = \dfrac{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}}{{1 + \sqrt 2 }} = \sqrt 2 \)

2. \(\left\{{}\begin{matrix}2x+y=3\\3x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+2y=6\\3x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x=7\\2x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm duy nhất \(\left(1;1\right)\)

3.

\( \sqrt {9\left( {x - 1} \right)} = 21\\ \Leftrightarrow 3\sqrt {x - 1} = 21\\ \Leftrightarrow \sqrt {x - 1} = 7\\ \Leftrightarrow x - 1 = 49\\ \Leftrightarrow x = 50 \)
Thử lại $x=50$ là nghiệm