D = ( \(\sqrt{3}-1\)) . \(\sqrt{6+2\sqrt{2}.\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
Rút gọn
rút gọn
\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
Ta có \(\sqrt{18-\sqrt{128}}\)
= \(\sqrt{18-8\sqrt{2}}\)
= \(\sqrt{16-2×4×\sqrt{2}+2}\)
= \(4-\sqrt{2}\)
Từ đó cái ban đầu
= \(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
= \(\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
= \(\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
= \(\sqrt{6+2\sqrt{3}-2}\)
= \(\sqrt{4+2\sqrt{3}}\)
= \(\sqrt{3}+1\)
Rút gọn biểu thức:
\(a,\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
\(b,\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
a) \(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2.4\sqrt{2}+2}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}\)\(=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}=\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}=\sqrt{6-2\left(1+\sqrt{3}\right)}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}=1+\sqrt{3}\)
b) Tương tự a) đ/s =5
Rút gọn biểu thức sau :
\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
Ta có :
\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-\sqrt{128}}}}\)
Ta có :
\(18-\sqrt{128}=18-8\sqrt{2}=16-2.4.\sqrt{2}+2=\left(4-\sqrt{2}\right)^2\)
Vậy
\(\sqrt{18-\sqrt{128}}=4-\sqrt{2}\)
Thay vào ta có
\(\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-\sqrt{128}}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}\)
\(=\sqrt{6-2\sqrt{4+2\sqrt{3}}}\)
Lại có :
\(4+2\sqrt{3}=3+2.1.\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\)
Do đó :
\(\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
Vậy :
\(\sqrt{6-2\sqrt{4+2\sqrt{3}}}=\sqrt{6-2\left(\sqrt{3}+1\right)}\)
\(=\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3-2.1.\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}-1\)
Vậy : \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}=\sqrt{3}-1\)
Rút gọn : Sử dụng công thức \(\sqrt{A^2}=\left|A\right|\)
a) \(\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2}}\)
b) \(\sqrt{8}.\sqrt{3-\sqrt{5}}\)
c) \(\sqrt{15-6\sqrt{6}}-\sqrt{33-12\sqrt{6}}\)
d) \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
a) \(\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{4-2\sqrt{3}}}{2}=\frac{\sqrt{3-2\sqrt{3}+1}}{2}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}\)
\(=\frac{\left|\sqrt{3}-1\right|}{2}=\frac{\sqrt{3}-1}{2}\)
b) \(\sqrt{8}\cdot\sqrt{3-\sqrt{5}}=\sqrt{4}\cdot\sqrt{6-2\sqrt{5}}=2\sqrt{5-2\sqrt{5}+1}=2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\cdot\left|\sqrt{5}-1\right|=2\left(\sqrt{5}-1\right)=2\sqrt{5}-2\)
Rút gọn
a) \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
b) \(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{21}+\sqrt{18-\sqrt{128}}}}}\)
Các bn ơi giải giúp mik câu này với! Mik đang vội, cảm ơn nhìu!!
a)\(\sqrt{8-\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
b) \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\)
c) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
d) \(\left(\sqrt{2+1}\right)^3-\left(\sqrt{2-1}\right)^3\)
e) \(\sqrt{6-2\sqrt{\sqrt{2+\sqrt{12+\sqrt{18-\sqrt{128}}}}}}\)
Rút gọn biểu thức
Rút gọn:
a)\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{3}-1}}\)
\(=\sqrt{6+2\cdot\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
rút gọn biểu thức
A=\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{2+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
Ta có: A = (\(\sqrt{3}-1\))\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{2+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
= (\(\sqrt{3}-1)\)\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{2+2\sqrt{3}+\sqrt{16-8\sqrt{2}+2}}}}\)
= (\(\sqrt{3}-1\))\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{2+2\sqrt{3}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
= (\(\sqrt{3}-1\))\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{2+2\sqrt{3}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
= (\(\sqrt{3}-1\))\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{2+2\sqrt{3}+4-\sqrt{2}}}}\)
= (\(\sqrt{3}-1\))\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{2+2\sqrt{3}+4-\sqrt{2}}}}\)
= (\(\sqrt{3}-1\))\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{6+2\sqrt{3}-\sqrt{2}}}}\)
= (\(\sqrt{3}-1\))\(\sqrt{6+\sqrt{24-8\sqrt{6+2\sqrt{3}-\sqrt{2}}}}\)
Rút gọn bt
\(a,A=\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\)
\(b,C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(c,\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(d,\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)
\(\Leftrightarrow x^3=6-5x\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
c/
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=3-1=2\)
b/ \(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{25}}\)
\(=\sqrt{4+5}=3\)