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Chứng minh: \(\left(\sqrt[3]{3+2.\sqrt{2}}+\sqrt[3]{3-2.\sqrt{2}}\right)^8>3^6\)
Cần gấp.
Chứng minh đẳng thức
\(\left(3+\sqrt{5\:}\right)\left(\sqrt{10\:\:\:\:\:\:}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
Bài 1. Rút gọn
a. \(2\sqrt{8}-3\sqrt{18}+\sqrt{32}\)
b. \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(1+\sqrt{2}\right)^2}\)
c. \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
d. \(\sqrt{2-\sqrt{3}+\sqrt{2+\sqrt{3}}}\)
Bài 2. Giải phương trình
a. \(x\sqrt{8}-6\sqrt{2}=0\)
b. \(\sqrt{2x+1}-3=0\)
c. \(\sqrt{x^2-4x+4}-3=0\)
d. \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25+2}=0\)
Chứng minh : \(\frac{1}{\left(\sqrt{2}+\sqrt{5}\right)^3}+\frac{1}{\left(\sqrt{5}+\sqrt{8}\right)^3}\)\(+...+\frac{1}{\left(\sqrt{2006}+\sqrt{2009}\right)^3}\)\(< \frac{11}{135}\)
chứng minh rằng
\(\left(3-2\sqrt{6}-\sqrt{33}\right)\times\left(\sqrt{6}+\sqrt{4}+4\right)=24\)
Rút gọn :\(A=\frac{\sqrt{6+2.\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)}-\sqrt{6-2.\left(\sqrt{6}-\sqrt{3}-\sqrt{2}\right)}}{\sqrt{2}}\)
Tính GTBT chứa căn:
a,\(\left(\sqrt{14}-3\sqrt{2}\right)^2\)+\(6\sqrt{28}\)
b,\(\left(\sqrt{6}-\sqrt{5}\right)^2\)-\(2\sqrt{120}\)
c,\(\left(2\sqrt{3}-3\sqrt{2}\right)^2+2\sqrt{6}+3\sqrt{24}\)
Rút gọn:
\(A=\left(\frac{4x\sqrt{x}+3x+9}{x+5\sqrt{x}+6}-\frac{3-\sqrt{x}}{2+\sqrt{x}}\right)\div\left(\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{3+4\sqrt{x}}{x+5\sqrt{x}+6}\right)\)
\(B=\left(x-\sqrt{x}-2\right)\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{4-\sqrt{x}}{x-2\sqrt{x}}\right)\)
Giải phương trình:
\(\sqrt{6y-y^2-6}=\left(3-y\right)\sqrt{y^2-2\left(3+\sqrt{3}\right)y+12+6\sqrt{3}}\)
