\(\sqrt{8-a}+\sqrt{5+a}=5\left(Đk:-5\le a\le8\right)\)

\(8-a+5+a+2\sqrt{\left(8-a\right)\left(5+a\right)}=25\)

\(\sqrt{\left(8-a\right)\left(5+a\right)}=6\)

a: \(A=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\cdot\sqrt{6-2\sqrt{5}}\)

\(=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)\)

\(=18-6\sqrt{5}+6\sqrt{5}-10=8\)

b: \(B=\left(\sqrt{5}+\sqrt{3}\right)\cdot\sqrt{2}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)

\(=2\left(5-3\right)=2\cdot2=4\)

a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

\(=1+\sqrt{2}\)

b: \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{6a^2}{24}}=\sqrt{\dfrac{a^2}{4}}=\dfrac{a}{2}\)

c: \(\sqrt{5a\cdot45a}-3a=-15a-3a=-18a\)

Tham khảo: https://hoc24.vn/cau-hoi/cho-adfrac12sqrtsqrt2dfrac18-dfrac18sqrt2-tinh-xa2sqrta4a21.1843666947439

\(=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)

\(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}\right)-2}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)

\(=\frac{\sqrt{a-4}+2+\sqrt{a-4}-2}{1-\frac{4}{a}}\)

\(=\frac{2a}{\sqrt{a-4}}\) 

bài này không phải của lớp 7

=\(\dfrac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\dfrac{8}{a}-\dfrac{16}{a^2}}}\)

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

=\(\dfrac{\sqrt{a-4}+2+\sqrt{a-4}-2}{1-\dfrac{4}{a}}\)

=\(\dfrac{2a}{\sqrt{a-4}}\)

Chúc Bạn Học Tốt

Tham khảo:

Tính giá trị biểu thức A = \(x^2+\sqrt{x^{^4}+x+1}\) với x =\(\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}-\dfrac{\sqrt{2}}{... - Hoc24

\(a=\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}-\dfrac{1}{8}\sqrt{2}\\ \Leftrightarrow a+\dfrac{\sqrt{2}}{8}=\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}\\ \Leftrightarrow\left(a+\dfrac{\sqrt{2}}{8}\right)^2=\dfrac{1}{4}\left(\sqrt{2}+\dfrac{1}{8}\right)\\ \Leftrightarrow a^2+\dfrac{a\sqrt{2}}{4}+\dfrac{1}{32}=\dfrac{\sqrt{2}}{4}+\dfrac{1}{32}\\ \Leftrightarrow a^2=\dfrac{\sqrt{2}-a\sqrt{2}}{4}=\dfrac{\sqrt{2}\left(1-a\right)}{4}\\ \Leftrightarrow a^4=\dfrac{a^2-2a+1}{8}\\ \Leftrightarrow a^4+a^2+1=\dfrac{a^2-2a+1}{8}+a^2+1=\dfrac{9a^2-2a+9}{8}\)

\(\Leftrightarrow a^2+\sqrt{a^4+a^2+1}=a^2+\dfrac{\sqrt{9a^2-2a+9}}{2\sqrt{2}}=\dfrac{2a^2\sqrt{2}+\sqrt{9a^2-2a+9}}{2\sqrt{2}}\)

a: \(x=4+\sqrt{3}+4-\sqrt{3}=8\)

Khi x=8 thì \(A=\dfrac{2-5\cdot2\sqrt{2}}{2\sqrt{2}+1}=\dfrac{2-10\sqrt{2}}{2\sqrt{2}+1}=-6+2\sqrt{2}\)

Tử:

\(M=\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}\)

\(M^2=a+4\sqrt{a-4}+2\sqrt{\left(a+4\sqrt{a-4}\right)\left(a-4\sqrt{a-4}\right)}+a-4\sqrt{a-4}\)

\(=2a+2\sqrt{a^2-16a+64}\)

\(=2a+2\sqrt{\left(a-8\right)^2}=2a+2a-16=4a-16\)

Mẫu:

\(\sqrt{1-\dfrac{8}{a}+\dfrac{16}{a^2}}=\sqrt{\left(1-\dfrac{4}{a}\right)^2}=1-\dfrac{4}{a}\)

Ta có:

\(\dfrac{4a-16}{1-\dfrac{4}{a}}=\dfrac{4\left(a-4\right)}{\dfrac{a-4}{a}}=4a\)

\(=\dfrac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\dfrac{8}{a}+\dfrac{16}{a^2}}}\)

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

\(=\dfrac{\sqrt{a-4}+2+\sqrt{a-4}-2}{1-\dfrac{4}{a}}\)

\(=\dfrac{2a}{\sqrt{a-4}}\)

Hok tốt!