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1: \(\left(2\sqrt{5}-5\right)^2=45-20\sqrt{5}\)
\(\left(\sqrt{5}-3\right)^2=14-6\sqrt{5}\)
mà \(45-20\sqrt{5}< 14-6\sqrt{5}\)
nên \(2\sqrt{5}-5< \sqrt{5}-3\)
3: \(\left(2+\sqrt{3}\right)^2=7+4\sqrt{3}\)
\(\left(\sqrt{2}+\sqrt{5}\right)^2=7+2\sqrt{10}\)
mà 4 căn 3>2 căn 10
nên \(2+\sqrt{3}>\sqrt{2}+\sqrt{5}\)
a, Điều kiện x ∉ {\(\frac{5}{3};\frac{1}{7}\)}
\(\sqrt{3x-5}=\sqrt{7x-1}\)
\(\left(\sqrt{3x-5}\right)^2=\left(\sqrt{7x-1}\right)^2\)
\(\left|3x-5\right|=\left|7x-1\right|\)
\(3x-5=7x-1\)
\(-4x=4\) => x = -1
a) \(A=\sqrt{4-\sqrt{15}}-\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\)\(\sqrt{2}A=\sqrt{8-2\sqrt{15}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{5}-\sqrt{3}-\left(\sqrt{3}+1\right)=\sqrt{5}-1\)
\(\Rightarrow\)\(A=\frac{\sqrt{5}-1}{\sqrt{2}}\)
b) tương tự câu a
c) \(\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}-\sqrt{6-2\sqrt{5+\sqrt{13-4\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}-\sqrt{6-2\sqrt{5+\sqrt{\left(\sqrt{12}-1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{5-\left(\sqrt{12}+1\right)}}-\sqrt{6-2\sqrt{5+\left(\sqrt{12}-1\right)}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}-\sqrt{6-2\sqrt{4+2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}-\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}-\sqrt{6-2\left(\sqrt{3}+1\right)}\)
\(=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)=2\)
\(a.\sqrt{4-\sqrt{15}}-\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{5-2.\sqrt{5}.\sqrt{3}+3}-\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{3}-1}{\sqrt{2}}=\dfrac{\sqrt{5}-2\sqrt{3}-1}{\sqrt{2}}\)
\(b.\sqrt{4+\sqrt{15}}+\sqrt{7-\sqrt{45}}=\dfrac{\sqrt{5+2\sqrt{5}.\sqrt{3}+3}+\sqrt{9-2.3\sqrt{5}+5}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}+3-\sqrt{5}}{\sqrt{2}}=\dfrac{3+\sqrt{3}}{\sqrt{2}}\)
\(c.\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}-\sqrt{6-2\sqrt{5+\sqrt{13-4\sqrt{3}}}}=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}-\sqrt{6-2\sqrt{3+2\sqrt{3}+1}}=\sqrt{3+2\sqrt{3}+1}-\sqrt{3-2\sqrt{3}+1}=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b1. a)
Gỉa sử căn bậc 2 + căn bậc 3 lớn hơn hoặc bằng căn bậc 10
=> ( căn bậc 2 + căn bậc 3 )2 lớn hơn hoặc bằng căn bậc 102
2+ 2 * căn bậc 3 + 3 lớn hơn hoặc bằng 10
5 + 2 căn 6 lớn hơn hoặc bằng 10
2 căn 6 lớn hơn hoặc bằng 5
( 2 căn 6 )2 lớn hơn hoặc bằng 52
4 * 6 lớn hơn 25
24 lớn hơn hoặc bằng 25 (sai)
Vậy căn bậc 2 + căn bậc 3 nhỏ hơn căn bậc 10
\(a=\sqrt{\sqrt[3]{x^6}+\sqrt[3]{x^4y^2}}+\sqrt{\sqrt[3]{y^6}+\sqrt[3]{y^4x^2}}\)
\(=\sqrt{\sqrt[3]{x^4}\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)}+\sqrt{\sqrt[3]{y^4}\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)}\)
\(=\sqrt{\sqrt[3]{x^2}+\sqrt[3]{y^2}}\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}\right)\)\(\Rightarrow a=\left(\sqrt{\sqrt[3]{x^2}+\sqrt[3]{y^2}}\right)^3\)
\(\Rightarrow\sqrt[3]{a^2}=\sqrt[3]{x^2}+\sqrt[3]{y^2}\)
a) \(A=\sqrt{a-2-2\sqrt{a-3}}-\sqrt{a+1-4\sqrt{a-3}}=\sqrt{\left(a-3\right)-2\sqrt{a-3}+1}-\sqrt{\left(a-3\right)-4\sqrt{a-3}+4}=\sqrt{\left(\sqrt{a-3}-1\right)^2}-\sqrt{\left(\sqrt{a-3}-2\right)^2}\)Ta có 3≤a≤4\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\left(\sqrt{a-3}-1\right)^2}=1-\sqrt{a-3}\\\sqrt{\left(\sqrt{a-3}-2\right)^2}=2-\sqrt{a-3}\end{matrix}\right.\)
Vậy A=\(1-\sqrt{a-3}-\left(2-\sqrt{a-3}\right)=1-\sqrt{a-3}-2+\sqrt{a-3}=-1\)b) B=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\times\sqrt{2003-2\sqrt{2005-2\sqrt{2004}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.3+9}}}\times\sqrt{2003-2\sqrt{2004-2\sqrt{2004}+1}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\times\sqrt{2003-2\sqrt{\left(\sqrt{2004}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\times\sqrt{2003-2\sqrt{2004}+2}=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\times\sqrt{2004-2\sqrt{2004}+1}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\times\sqrt{\left(\sqrt{2004}-1\right)^2}=\sqrt{\sqrt{5}-\sqrt{5}+1}\times\left(\sqrt{2004}-1\right)=\sqrt{1}\times\left(\sqrt{2004}-1\right)=\sqrt{2004}-1\)
Ta có: \(B=21\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=21\cdot\left[2+\sqrt{3}+3-\sqrt{5}+2\sqrt{\left(2+\sqrt{3}\right)\left(3-\sqrt{5}\right)}\right]-6\cdot\left[2-\sqrt{3}+3+\sqrt{5}+2\cdot\sqrt{\left(2-\sqrt{3}\right)\left(3+\sqrt{5}\right)}\right]-15\sqrt{15}\)
\(=21\cdot\left(5+\sqrt{3}-\sqrt{5}+\sqrt{\left(4+2\sqrt{3}\right)\left(6-2\sqrt{5}\right)}\right)-6\cdot\left[5-\sqrt{3}+\sqrt{5}+\sqrt{\left(4-2\sqrt{3}\right)\left(6+2\sqrt{5}\right)}\right]-15\sqrt{15}\)
\(=21\cdot\left[5+\sqrt{3}-\sqrt{5}+\left(\sqrt{3}+1\right)\left(\sqrt{5}-1\right)\right]-6\cdot\left[5-\sqrt{3}+\sqrt{5}+\left(\sqrt{3}-1\right)\left(\sqrt{5}+1\right)\right]-15\sqrt{15}\)
\(=21\cdot\left(5+\sqrt{3}-\sqrt{5}+\sqrt{15}-\sqrt{3}+\sqrt{5}-1\right)-6\cdot\left(5-\sqrt{3}+\sqrt{5}+\sqrt{15}+\sqrt{3}-\sqrt{5}-1\right)-15\sqrt{15}\)
\(=21\cdot\left(4+\sqrt{15}\right)-6\left(4+\sqrt{15}\right)-15\sqrt{15}\)
\(=84+21\sqrt{15}-24-6\sqrt{15}-15\sqrt{15}\)
\(=60\)