a: \(\sqrt{6+\sqrt{20}}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
b: \(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
=>A^2=2+căn 3+2-căn 3+2=6
=>A=căn 6>căn 2+1
1) có bao nhiêu giá trị nguyên của x để biểu thức
\(M=\sqrt{x+4}+\sqrt{2-x}\) có nghĩa
2) so sánh
a) \(\sqrt{33}-\sqrt{17}\) và \(6-\sqrt{15}\)
b) \(4\sqrt{5}\) và \(5\sqrt{3}\)
c) \(\sqrt{3\sqrt{2}}\) và \(\sqrt{2\sqrt{3}}\)
d) \(\sqrt{10}+\sqrt{17}+1\) và \(\sqrt{61}\)
giúp mk nhé mk cần gấp
Bài 1 :tính giá trị của biểu thức
a) \(\left(\sqrt{5}+\sqrt{2}\right)\left(3\sqrt{2}-1\right)\)
b) \(3\sqrt{50}-2\sqrt{75}-4\dfrac{\sqrt{54}}{\sqrt{3}}-3\sqrt{\dfrac{1}{3}}\)
c) \(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4+2\sqrt{3}}\)
d) \(\sqrt{48-2\sqrt{135}}-\sqrt{45}+\sqrt{18}\)
e)\(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\dfrac{6}{2-\sqrt{10}}-\dfrac{20}{\sqrt{10}}\)
Bài 2 :Tính:
a) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}\)
b) \(\left(2\sqrt{3}+4\right)\left(\sqrt{3}-2\right)\)
c) \(\sqrt{3+2\sqrt{2}}+\sqrt{\left(\sqrt{2}-2\right)^2}\)
d)\(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}+\sqrt{6}\)
e)\(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)
f) \(\dfrac{1}{5}\sqrt{50}-2\sqrt{96}-\dfrac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\dfrac{1}{6}}\)
Rút gọn các biểu thức sau:
a) \(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}-\sqrt{20}\)
b) \(\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\dfrac{10}{1+\sqrt{6}}\)
1,Rút gọn B=\(\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{40}}\)
2,Tính
a, A=\(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
1) Rút gọn:
a) A = \(\sqrt{5-2\sqrt{3-\sqrt{3}}}-\sqrt{3+\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
b) B = \(\sqrt{13+\sqrt{2}+5\sqrt{1+2\sqrt{2}}}+\sqrt{13+\sqrt{2}+5\sqrt{1+2\sqrt{2}}}\)
c) C = \(\dfrac{\sqrt{21+3\sqrt{5}}+\sqrt{21-3\sqrt{5}}}{\sqrt{21}+6\sqrt{11}}+\sqrt{11-6\sqrt{2}}\)
d) D = \(\left(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\right).\sqrt{\dfrac{2+2\sqrt{5}}{2+\sqrt{5}}}\)
e) E = \(\dfrac{\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}}{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right):\sqrt{\sqrt{13}+2}}\)
Rút gọn biểu thức :
a,\(\frac{2+\sqrt{3}}{2-\sqrt{3}};\frac{5+2\sqrt{6}}{5-2\sqrt{6}}\)
b,\(\frac{\sqrt{3}-1}{\sqrt{3}+1}\)
c,\(\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
d,\(\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}\)
Giair phương trình:
1) \(\sqrt[5]{32-x^2}-\sqrt[5]{1-x^2}=4\)
2) \(\sqrt{x}+\sqrt[4]{20-x}=4\)
3) \(x^3+1=2\sqrt{3x-1}\)
4) \(\sqrt[3]{x-1}+3=\sqrt[4]{82-x}\)
5)
\(a.\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)
\(b.\sqrt{5x^2+14x+9}-\sqrt{x^2-x-20}=5\sqrt{x+1}\)
\(B=\dfrac{5\left(\sqrt{6}-1\right)}{\sqrt{6}+1}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}+\sqrt{3-2\sqrt{2}}\)
Giải phương trình:
1. \(\sqrt{\dfrac{42}{5-x}}+\sqrt{\dfrac{60}{7-x}}=6\)
2. \(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
3. \(x^2+x+12\sqrt{x+1}=36\)
4. \(\sqrt{x+2}-\sqrt{x-6}=2\)
5. \(\sqrt[3]{x-1}-\sqrt[3]{x-3}=\sqrt[3]{2}\)
6. \(5\sqrt{1+x^3}=2\left(x^2+2\right)\)
6. \(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\)
Tính \(\frac{2\sqrt{3}-4}{\sqrt{3}-1}+\frac{2\sqrt{2}-1}{\sqrt{2}-1}-\frac{1+\sqrt{6}}{\sqrt{2}+3}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
