\(<=>15\sqrt{x+3} - 3\sqrt{x+3}=4\sqrt{x+3} <=> 15\sqrt{x+3} - 3\sqrt{x+3}-4\sqrt{x+3}=0 <=> 8\sqrt{x+3}=0 <=>\sqrt{x+3}=0 => x+3=0 =>x=-3\)

\(\Leftrightarrow6\sqrt{x+1}=12\)

hay x=3

Ta có: \(A=\dfrac{10\sqrt{6}-12}{\sqrt{6}-5}-3\sqrt{\dfrac{2}{3}}+\dfrac{15}{\sqrt{6}-1}\)

\(=\dfrac{-2\sqrt{6}\left(5-\sqrt{6}\right)}{5-\sqrt{6}}-\sqrt{\dfrac{2}{3}\cdot9}+\dfrac{15\left(\sqrt{6}+1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}\)

\(=-2\sqrt{6}-\sqrt{6}+3\left(\sqrt{6}+1\right)\)

\(=-3\sqrt{6}+3\sqrt{6}+3\)

=3

g: \(=\left|\sqrt{6}-1\right|=\sqrt{6}-1\)

h: \(=\left|2\sqrt{3}-1\right|=2\sqrt{3}-1\)

l: \(=\left|2-\sqrt{3}\right|-2=2-\sqrt{3}-2=-\sqrt{3}\)

j: \(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)

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

1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)

 

a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)

\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)

\(=33\sqrt{3}\cdot\sqrt{3}\)

=99

b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)

\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)

\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)

c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=36-36\sqrt{2}+18\sqrt{3}\)

d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)

\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)

a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)

   \(=28.3+9.3-4.3=99\)

b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)

  \(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)

d,Ta có:\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

        \(=3\sqrt{75\sqrt{2}}+5\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

        \(=15\sqrt{3\sqrt{2}}+20\sqrt{3\sqrt{2}}-16\sqrt{3\sqrt{2}}\)

        \(=19\sqrt{3\sqrt{2}}\)

\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}\)+\(\frac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)\(=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\)+\(\frac{2\sqrt{3}\left(\sqrt{3}+\sqrt{3}\right)}{\sqrt{3}+\sqrt{2}}\)\(=\sqrt{3}+2\sqrt{3}\)=\(3\sqrt{3}.\)

a: Sửa đề: \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

\(=4-\sqrt{15}+\sqrt{15}=4\)

b: \(A=2-\sqrt{3}+\sqrt{3}-1=1\)

c: \(C=3\sqrt{5}-2-3\sqrt{5}-2=-4\)

d: Sửa đề: \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

=6

a) \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

\(A=\left|4-\sqrt{15}\right|+\sqrt{15}\)

\(A=4-\sqrt{15}+\sqrt{15}\)

\(A=4\)

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

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

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

\(B=1\)

c) \(C=\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)

\(C=\sqrt{\left(3\sqrt{5}\right)^2-2\cdot3\sqrt{15}\cdot2+2^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot2+2^2}\)

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

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

\(C=3\sqrt{5}-2-3\sqrt{5}-2\)

\(C=-4\)

d) \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

\(D=\sqrt{\left(2\sqrt{5}\right)^2+2\cdot2\sqrt{5}\cdot3+3^2}-\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot3+3^3}\)

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

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

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

\(D=6\)

a) A = \(13-2\sqrt{42}=\left(\sqrt{7}-\sqrt{6}\right)^2\)

<=> \(\sqrt{A}=\sqrt{7}-\sqrt{6}\)

b) \(A=46+6\sqrt{5}=\left(\sqrt{45}+1\right)^2\)

<=> \(\sqrt{A}=\sqrt{45}+1\)

c) \(A=12-3\sqrt{15}=\dfrac{1}{2}\left(24-6\sqrt{15}\right)=\dfrac{1}{2}\left(\sqrt{15}-3\right)^2\)

<=> \(\sqrt{A}=\dfrac{1}{\sqrt{2}}\left(\sqrt{15}-3\right)\)

\(\frac{2\sqrt{12}-\sqrt{6}}{2\sqrt{6}-\sqrt{3}}+\frac{10+\sqrt{5}}{2\sqrt{15}+\sqrt{3}}\)

\(=\frac{\sqrt{2}\left(2\sqrt{6}-\sqrt{3}\right)}{2\sqrt{6}-\sqrt{3}}+\frac{\sqrt{5}\left(2\sqrt{5}+1\right)}{\sqrt{3}\left(2\sqrt{5}+1\right)}\)

\(=\sqrt{2}+\frac{\sqrt{5}}{\sqrt{3}}\)

\(=\frac{\sqrt{6}+\sqrt{5}}{\sqrt{3}}\)

p/s: chúc bạn học tốt