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) 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: \(=\sqrt{25}=5\)

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, \(\sqrt{54}=\sqrt{9.6}=3\sqrt{6}\)

b, \(\sqrt{108}=\sqrt{36.3}=6\sqrt{3}\)

c, \(0,1\sqrt{20000}=0,1\sqrt{2.10000}=10\sqrt{2}\)

d, \(-0,05\sqrt{28800}=-0,05\sqrt{288.100}=-0,05.10.\sqrt{144.2}\)

\(=-0,5.12\sqrt{2}=-6\sqrt{2}\)

e, \(\sqrt{7.63.a^2}=\sqrt{7.7.9.a^2}=21\left|a\right|\)

a) $\sqrt{54}=\sqrt{9.6}=\sqrt{9}.\sqrt{6}=3\sqrt{6}$.

b) $\sqrt{108}=\sqrt{36.3}=\sqrt{36}.\sqrt{3}=6\sqrt{3}$.

c) $0,1\sqrt{20000}=0,1\sqrt{2.10000}=0,1\sqrt{2}.\sqrt{10000}$

$=0,1.100\sqrt{2}=10\sqrt{2}$.

d) $-0,05\sqrt{28800}=-0,05\sqrt{288.100}$

$=-0,05\sqrt{\mathrm{2.144.100}}=-0,05.\sqrt{2}.\sqrt{144}.\sqrt{100}$

$=-0,\mathrm{05.12.10}\sqrt{2}$

$=-6\sqrt{}$

a) \(\sqrt{54}=\sqrt{9}.\sqrt{6}=3\sqrt{6}\)

b) \(\sqrt{108}=\sqrt{9}.\sqrt{4}.\sqrt{3}=6\sqrt{3}\)

c) \(0,1\sqrt{20000}=0,1.\sqrt{2}.\sqrt{10000}=0,1.100.\sqrt{2}=10\sqrt{2}\)

d) \(-0,05\sqrt{28800}=-0,05\sqrt{400}.\sqrt{9}.\sqrt[]{4}.\sqrt{2}=-0,05.20.3.2.\sqrt{2}=-6\sqrt{2}\)

e) \(\sqrt{7.63.a^2}=21\sqrt{a^2}\left\{{}\begin{matrix}=-21a\\=21a\end{matrix}\right.\)

a) \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right).\sqrt{\dfrac{8-2\sqrt{15}}{2}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{25.6}-\sqrt{9.10}\right).\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right).\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}\)

\(=\left(\sqrt{10}+\sqrt{6}\right).\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\sqrt{2}.\left(\sqrt{5}+\sqrt{3}\right).\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)

\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)

a) Ta có: \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\sqrt{8-2\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(\sqrt{5}-\sqrt{3}\right)^2\cdot\left(4+\sqrt{15}\right)\)

\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)

\(=32+8\sqrt{15}-8\sqrt{15}-30\)

=2

a) \(=\sqrt{a}\left(\sqrt{a}-1\right)\)

b) \(=\left(\sqrt{a}\right)^2-2\sqrt{ab}+\left(\sqrt{b}\right)^2=\left(\sqrt{a}-\sqrt{b}\right)^2\)

c) \(=\left(\sqrt{x}\right)^2-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\)

d) \(=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

e) \(=\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)

f) \(=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\)

a: \(a-\sqrt{a}=\sqrt{a}\left(\sqrt{a}-1\right)\)

b: \(a-2\sqrt{ab}+b=\left(\sqrt{a}-\sqrt{b}\right)^2\)

c: \(x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\)

Bài 1 :

a) \(Cos30^o=Cos\left(2.15^o\right)=2cos^215^o-1\)

\(\Rightarrow cos^215^o=\dfrac{cos30^o+1}{2}\)

\(\Rightarrow cos^215^o=\dfrac{\dfrac{\sqrt[]{3}}{2}+1}{2}\)

\(\Rightarrow cos^215^o=\dfrac{\sqrt[]{3}+2}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{\sqrt[]{3}+2}}{2}\)

\(\Rightarrow cos15^o=\dfrac{2\sqrt[]{\sqrt[]{3}+2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{4\sqrt[]{3}+8}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{6+2.2\sqrt[]{2}\sqrt[]{6}+2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{\left(\sqrt[]{6}+\sqrt[]{2}\right)^2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{6}+\sqrt[]{2}^{ }}{4}\left(dpcm\right)\)

a)

 Dựng tam giác ABC vuông tại A với \(\widehat{C}=15^o\). Trên đoạn thẳng AC lấy điểm D sao cho \(\widehat{CBD}=15^o\). Không mất tính tổng quát, ta chuẩn hóa \(AB=1\). \(\Rightarrow\left\{{}\begin{matrix}BD=\dfrac{AB}{cos60^o}=2\\AD=AB.tan60^o=\sqrt{3}\end{matrix}\right.\)

 Dễ thấy tam giác DBC cân tại D \(\Rightarrow BD=CD=2\) \(\Rightarrow AC=AD+DC=2+\sqrt{3}\)

  \(\Rightarrow tanC=\dfrac{AB}{AC}=\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3}\) 

\(\Rightarrow\dfrac{sinC}{cosC}=2-\sqrt{3}\)

\(\Rightarrow sinC=\left(2-\sqrt{3}\right)cosC\)

Mà \(sin^2C+cos^2C=1\)

\(\Rightarrow\left(7-4\sqrt{3}\right)cos^2C+cos^2C=1\)

\(\Leftrightarrow\left(8-4\sqrt{3}\right)cos^2C=1\)

\(\Leftrightarrow cos^2C=\dfrac{1}{8-4\sqrt{3}}=\dfrac{2+\sqrt{3}}{4}\)

\(\Leftrightarrow cosC=\sqrt{\dfrac{2+\sqrt{3}}{4}}\) \(=\dfrac{\sqrt{2+\sqrt{3}}}{2}=\dfrac{\sqrt{8+4\sqrt{3}}}{4}\) \(=\dfrac{\sqrt{6}+\sqrt{2}}{4}\) 

\(\Rightarrow cos15^o=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)

b) \(A=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{32cos^415^o-10-8\sqrt[]{3}}}\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{32.\dfrac{1}{4^4}\left(\sqrt[]{6}+\sqrt[]{2}\right)^4-10-8\sqrt[]{3}}}\) \(\left(Cos15^o=\dfrac{\sqrt[]{6}-\sqrt[]{2}}{4}\right)\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{\dfrac{1}{8}\left(8+2\sqrt[]{12}\right)^2-10-8\sqrt[]{3}}}\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{\dfrac{1}{8}\left[64+32\sqrt[]{12}+48-80-64\sqrt[]{3}\right]}}\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\dfrac{1}{2}\sqrt[3]{32}}\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\dfrac{1}{2}.2\sqrt[3]{4}}=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{4}}\)

a) √54 = √9.6 = 3√6
b) √108 = √36.3 = 6√3
c) 0,1√20000 = 0,1√10000.2= 0,1.100√2 = 10√2
d) -0,05.√28800 = -0,05.√14400.2 = -0,05.120√2 = -6√2
e)√7.63.a2 = √7.7.9.a2 = 7.3|a| = 21|a|