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1)
a. \(\sqrt{\dfrac{25}{7}}.\sqrt{\dfrac{7}{9}}=\sqrt{\dfrac{25.7}{7.9}}=\sqrt{\dfrac{25}{9}}=\dfrac{5}{3}\)
b. \(\left(\sqrt{\dfrac{9}{2}}+\sqrt{\dfrac{1}{2}}-\sqrt{2}\right).\sqrt{2}=3+1-2=2\)
c. \(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=4-12+10=2\)
d. \(\left(\sqrt{\dfrac{2}{3}}-\sqrt{\dfrac{3}{2}}\right)^2=\dfrac{2}{3}+\dfrac{3}{2}-2\sqrt{\dfrac{2}{3}.\dfrac{3}{2}}=\dfrac{1}{6}\)
2)
a. \(\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b. \(\sqrt{8-2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)
c. \(1+\sqrt{6-2\sqrt{5}}=1+\sqrt{5-2\sqrt{5}+1}=1-\sqrt{\left(\sqrt{5}-1\right)^2}=1-\sqrt{5}+1=2-\sqrt{5}\)
d. \(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}+\sqrt{2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{2}=\sqrt{5}-\sqrt{2}+\sqrt{2}=\sqrt{5}\)
3. \(a.A=x^2+2x+16=\left(\sqrt{2}-1\right)^2+2.\left(\sqrt{2}-1\right)+16=2-2\sqrt{2}+1+2\sqrt{2}-2+16=17\)
\(b.B=x^2+12x-14=\left(5\sqrt{2}-6\right)^2+12.\left(5\sqrt{2}-6\right)-14=50+36-60\sqrt{2}+60\sqrt{2}-72-14=0\)
Help me nha 
@Phùng Khánh Linh@Nhã Doanh@Liana@Yukru Cảm ơn trước nhé 
a)
\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\\ =\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\\ =\sqrt{5}+1-\sqrt{5}+1\\ =2\)
b) Sửa đề:
\(\sqrt{7+2\sqrt{6}}+\sqrt{7-2\sqrt{6}}-2\sqrt{6}\\ =\sqrt{\left(\sqrt{6}+1\right)^2}+\sqrt{\left(\sqrt{6}-1\right)^2}-2\sqrt{6}\\ =\sqrt{6}+1+\sqrt{6}-1-2\sqrt{6}\\ =0\)
c)
\(\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}-2\sqrt{5}\\ =\sqrt{9+2\sqrt{20}}+\sqrt{9-2\sqrt{20}}-2\sqrt{5}\\ =\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}-2\sqrt{5}\\ =\sqrt{5}+2+\sqrt{5}-2-2\sqrt{5}\\ =0\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
\(=3+2\sqrt{2}+3-2\sqrt{2}\)
\(=6\)
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)
\(=2+\sqrt{5}-\sqrt{5}+2\)
\(=4\)
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)
\(=1+\sqrt{5}-\sqrt{5}+1\)
\(=2\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(A=\sqrt{3}+2+2-\sqrt{3}\)
A = 2 + 2
A = 4
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(B=\sqrt{2}+3+3-\sqrt{2}\)
B = 3 + 3
B = 6
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(C=3+2\sqrt{2}+3-2\sqrt{2}\)
C = 3 + 3
C = 6
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(D=\sqrt{5}+2-\sqrt{5}+2\)
D = 2 + 2
D = 4
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(E=\sqrt{5}+1-\sqrt{5}+1\)
E = 1 + 1
E = 2
a) Ta có: \(\sqrt{11-2\sqrt{10}}\)
\(=\sqrt{10-2\cdot\sqrt{10}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{10}-1\right)^2}\)
\(=\left|\sqrt{10}-1\right|=\sqrt{10}-1\)
b) Ta có: \(\sqrt{9-2\sqrt{14}}\)
\(=\sqrt{7-2\cdot\sqrt{7}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{2}\right|\)
\(=\sqrt{7}-\sqrt{2}\)
c) Ta có: \(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\)
\(=\sqrt{3}+1+\sqrt{3}-1\)
\(=2\sqrt{3}\)
d) Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5+2\cdot\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)
\(=\sqrt{5}-2-\left(\sqrt{5}+2\right)\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
e) Ta có: \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}\right)-\sqrt{2}\cdot\left(\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\frac{\sqrt{7-2\cdot\sqrt{7}\cdot1+1}-\sqrt{7+2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\left(\sqrt{7}+1\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)
\(=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
g) Ta có: \(\sqrt{3}+\sqrt{11+6\sqrt{2}}+\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{3}+\sqrt{9+2\cdot3\cdot\sqrt{2}+2}+\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}\)
\(=\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\left|3+\sqrt{2}\right|+\left|\sqrt{2}+\sqrt{3}\right|\)
\(=\sqrt{3}+3+\sqrt{2}+\sqrt{2}+\sqrt{3}\)
\(=3+2\sqrt{3}+2\sqrt{2}\)
h) Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{\left(\sqrt{3}+2\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\cdot\left(\sqrt{3}+2\right)}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\sqrt{3}-20}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\left(5-\sqrt{3}\right)}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
k) Ta có: \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
\(=\sqrt{49-2\cdot7\cdot\sqrt{45}+45}-\sqrt{49+2\cdot7\cdot\sqrt{45}+45}\)
\(=\sqrt{\left(7-\sqrt{45}\right)^2}-\sqrt{\left(7+\sqrt{45}\right)^2}\)
\(=\left|7-\sqrt{45}\right|-\left|7+\sqrt{45}\right|\)
\(=7-\sqrt{45}-\left(7+\sqrt{45}\right)\)
\(=7-\sqrt{45}-7-\sqrt{45}\)
\(=-2\sqrt{45}=-6\sqrt{5}\)
i) Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow A^2=\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)
\(=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\cdot\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\cdot\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\cdot\sqrt{6-2\sqrt{5}}\)
\(=8+2\cdot\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=8+2\cdot\left(\sqrt{5}-1\right)\)
\(=8+2\sqrt{5}-2\)
\(=6+2\sqrt{5}\)
\(=\left(\sqrt{5}+1\right)^2\)
\(\Leftrightarrow A=\sqrt{5}+1\)
\(A=\sqrt{19-3\sqrt{40}}-\sqrt{19+3\sqrt{40}}=\sqrt{19-2\sqrt{90}}-\sqrt{19+2\sqrt{90}}=\sqrt{10-2.\sqrt{10}.3+9}-\sqrt{10+2.\sqrt{10}.3+9}=\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{\left(\sqrt{10}+3\right)^2}=\sqrt{10}-3-\sqrt{10}-3=-6\)\(B=\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}=\sqrt{18-2.\sqrt{18}.\sqrt{3}+3}+\sqrt{6+2.\sqrt{3}.\sqrt{6}+3}-\sqrt{24+12\sqrt{3}}=\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{\sqrt{3}}\right)^2}-\sqrt{\left(\sqrt{18}+\sqrt{6}\right)^2}=\sqrt{18}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{18}-\sqrt{6}=0\)
\(C=\sqrt{6+2\sqrt{2\sqrt{3-\sqrt{4+2\sqrt{3}}}}}\)
\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\) \(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\) \(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(D=\sqrt{\frac{8+2\sqrt{15}}{2}}-\sqrt{\frac{14-6\sqrt{5}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{2}}\)
\(=\frac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\frac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
\(E=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
\(F=\sqrt{\frac{24-6\sqrt{7}}{2}}-\sqrt{\frac{24+6\sqrt{7}}{2}}\) \(=\sqrt{\frac{21-2\sqrt{21\cdot3}+3}{2}}-\sqrt{\frac{21+2\sqrt{21\cdot3}+3}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{21}-\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{21}+\sqrt{3}\right)^2}{2}}\)
\(=\frac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=\frac{-2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
\(G=\left(3+\sqrt{3}\right)\cdot\sqrt{12-6\sqrt{3}}\) \(=\left(3+\sqrt{3}\right)\cdot\sqrt{\left(3-\sqrt{3}\right)^2}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
\(H=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(3-\sqrt{5}\right)^2}\) \(=\sqrt{5}-2-3-\sqrt{5}=-5\)
\(I=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{\left(2\sqrt{3}-1\right)^2}\)
\(=2\sqrt{2}-1-2\sqrt{3}+1=2\sqrt{2}-2\sqrt{3}\)
1.a)
\(2\sqrt{3}=\sqrt{12}>\sqrt{9}=3.\)
\(3\sqrt{2}=\sqrt{18}>\sqrt{16}=4.\)
Suy ra VT > 7
1.b)
\(\sqrt{16}+\sqrt{25}=4+5=9\)
2.a)
\(\sqrt{21-6\sqrt{6}}=\sqrt{\left(3\sqrt{2}\right)^2-6\sqrt{6}+3}=3\sqrt{2}-\sqrt{3}\)
b)\(\sqrt{9-2\sqrt{14}}=\sqrt{\frac{18-4\sqrt{14}}{2}}=\frac{\sqrt{14}-2}{\sqrt{2}}=\sqrt{7}-1\)
Các câu còn lại bạn làm tương tự nhé!
c) \(\sqrt{4-\sqrt{7}}=\frac{1}{\sqrt{2}}.\sqrt{8-2\sqrt{7}}=\frac{1}{\sqrt{2}}\sqrt{7-2\sqrt{7}+1}\)
\(=\frac{1}{\sqrt{2}}\sqrt{\left(\sqrt{7}-1\right)^2}=\frac{\sqrt{2}\left(\sqrt{7}-1\right)}{2}\)
d) \(\sqrt{4+2\sqrt{3}-\sqrt{4-2\sqrt{3}}}=\sqrt{4+2\sqrt{3}-\sqrt{3-2\sqrt{3}+1}}\)
\(=\sqrt{4+2\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{4+2\sqrt{3}-\sqrt{3}+1}=\sqrt{5+\sqrt{3}}\)