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

10+2 căn 5

= 10+2 căn 10 . căn 2 trên 2+2 trên 4

= (căn 10+ căn 2 trên 2) ²

mik ko biết viết căn nhé, bạn tự dịch, còn kqua sai thì thôi nhé

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

Bấm máy giải pt bậc 2 với hệ số: \(1\) ; \(-14\); \(\dfrac{6^2.5}{4}\) nghiệm trả về sẽ cho biết có phân tích được hay không

\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)

\(\sqrt{8-2\sqrt{12}}=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}=\left|\sqrt{6}-\sqrt{2}\right|=\sqrt{6}-\sqrt{2}\)

\(\sqrt{21+6\sqrt{6}}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=\left|3\sqrt{2}-\sqrt{3}\right|=3\sqrt{2}-\sqrt{3}\)

\(\sqrt{15-6\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}=\left|3-\sqrt{6}\right|=3-\sqrt{6}\)

\(\sqrt{29-12\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}=\left|2\sqrt{5}-3\right|=2\sqrt{5}-3\)

\(\sqrt{41+12\sqrt{5}}=\sqrt{\left(6+\sqrt{5}\right)^2}=6+\sqrt{5}\)

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

BBieesn đổi hằng đẳng thức 

x²+4x+4

=x²+2.2x+2²

=(x+2)²

Ta có:

\(x^2+4x+4\)

\(=x^2+2.2x+2^2\)

\(=\left(x+2\right)^2\)

Ta có : x² + 4x + 4 = (x + 2)²

Em kéo xuống trang 40, mục số 3:

Một số mẹo nhỏ với Casio.pdf - Google Drive

\(\left(\sqrt{3+\sqrt{15}-\sqrt{3-\sqrt{5}}}\right)^2=3+\sqrt{15}-\sqrt{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(3+\sqrt{15}-\sqrt{3-\sqrt{5}}\right)}{\sqrt{2}}=\dfrac{3\sqrt{2}+\sqrt{30}-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{3\sqrt{2}+\sqrt{30}-\sqrt{5-2\sqrt{5}+1}}{\sqrt{2}}=\dfrac{3\sqrt{2}+\sqrt{30}-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\dfrac{3\sqrt{2}+\sqrt{30}-\left|\sqrt{5}-1\right|}{\sqrt{2}}=\dfrac{3\sqrt{2}+\sqrt{30}-\sqrt{5}+1}{\sqrt{2}}=\dfrac{\sqrt{2}\left(3\sqrt{2}+\sqrt{30}-\sqrt{5}+1\right)}{2}=\dfrac{6+2\sqrt{15}-\sqrt{10}+\sqrt{2}}{2}\)