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a)
\(A=\sqrt{26+15\sqrt{3}}=\sqrt{\frac{52+30\sqrt{3}}{2}}=\sqrt{\frac{27+25+2\sqrt{27.25}}{2}}\)
\(=\sqrt{\frac{(\sqrt{27}+\sqrt{25})^2}{2}}=\frac{\sqrt{27}+\sqrt{25}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}=\frac{3\sqrt{6}+5\sqrt{2}}{2}\)
b)
\(B\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{7+1+2\sqrt{7}}-\sqrt{7+1-2\sqrt{7}}-2\)
\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}-2=\sqrt{7}+1-(\sqrt{7}-1)-2=0\)
\(\Rightarrow B=0\)
c)
\(C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{3+5+2\sqrt{3.5}}\)
\(=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{5}+\sqrt{3})^2}=(\sqrt{5}-\sqrt{3})-(\sqrt{5}+\sqrt{3})=-2\sqrt{3}\)
d)
\(D=(\sqrt{6}-2)(5+2\sqrt{6})\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(2+3+2\sqrt{2.3})\sqrt{2+3-2\sqrt{2.3}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})^2\sqrt{(\sqrt{3}-\sqrt{2})^2}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})^2(\sqrt{3}+\sqrt{2})^2=\sqrt{2}[(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})]^2\)
\(=\sqrt{2}.1^2=\sqrt{2}\)
e)
\(E=(\sqrt{10}-\sqrt{2})\sqrt{3+\sqrt{5}}=(\sqrt{5}-1).\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=(\sqrt{5}-1)\sqrt{6+2\sqrt{5}}=(\sqrt{5}-1)\sqrt{5+1+2\sqrt{5.1}}\)
\(=(\sqrt{5}-1)\sqrt{(\sqrt{5}+1)^2}=(\sqrt{5}-1)(\sqrt{5}+1)=4\)
f)
\(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-3)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-(\sqrt{20}-3)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}}=\sqrt{\sqrt{5}-(\sqrt{5}-1)}=\sqrt{1}=1\)
Bài 4 :
\(a,\sqrt{x-1}=2\)
=> \(x-1=2^2=4\)
=>\(x=4+1=5\)
Vậy \(x\in\left\{5\right\}\)
\(b,\sqrt{x^2-3x+2}=2\)
=> \(x^2-3x+2=2\)
=> \(x^2-3x=2-2=0\)
=>\(x.\left(x-3\right)=0\)( phân tích đa thức thanh nhân tử )
=> \(\left[{}\begin{matrix}x=0\\x-3=0=>x=0+3=3\end{matrix}\right.\)
Vậy \(x\in\left\{0;3\right\}\)
MÌNH Biết vậy thôi ,
Bài 4 :
c) \(\sqrt{4x+1}=x+1\)ĐK : \(x\ge-1\)
\(\Leftrightarrow4x+1=\left(x+1\right)^2\)
\(\Leftrightarrow x^2+2x+1-4x-1=0\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)( thỏa )
d) \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}-\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|-\left|\sqrt{x-1}-1\right|=2\)
+) Xét \(x\ge2\)
\(pt\Leftrightarrow\sqrt{x-1}+1-\sqrt{x-1}+1=2\)
\(\Leftrightarrow2=2\)( luôn đúng )
+) Xét \(1\le x< 2\):
\(pt\Leftrightarrow\sqrt{x-1}+1-1+\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)( loại )
Vậy \(x\ge2\)
https://hoc24.vn/hoi-dap/question/407636.html
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}\)
= 9
~ ~ ~ ~ ~
\(M=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{6+2\sqrt{3}-2}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}+1\)
2.1
\(A=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5.1}+1}-\sqrt{5-2\sqrt{5.1}+1}\)
\(=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}=|\sqrt{5}+1|-|\sqrt{5}-1|=2\)
2.2
\(B\sqrt{2}=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{3+2\sqrt{3.5}+5}+\sqrt{3-2\sqrt{3.5}+5}-2\sqrt{5-2\sqrt{5.1}+1}\)
\(=\sqrt{(\sqrt{3}+\sqrt{5})^2}+\sqrt{(\sqrt{3}-\sqrt{5})^2}-2\sqrt{(\sqrt{5}-1)^2}\)
\(=|\sqrt{3}+\sqrt{5}|+|\sqrt{3}-\sqrt{5}|-2|\sqrt{5}-1|=2\)
$\Rightarrow B=\sqrt{2}$
Bài 1:
1. ĐKXĐ: \(\left\{\begin{matrix} 2x-1\geq 0\\ x-3\geq 0\\ 5-x>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ x\geq 3\\ x< 5\end{matrix}\right.\Leftrightarrow 3\leq x< 5\)
2.
ĐKXĐ: \(\left\{\begin{matrix} x-1\geq 0\\ 2-x\geq 0\\ x+1>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x\leq 2\\ x>-1\end{matrix}\right.\Leftrightarrow 1\leq x\leq 2\)
Bài 1:
a)
\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)
b)
\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)
\(=3\sqrt{5}+1\)
c)
\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)
\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)
d)
\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)
\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)
Bài 1:
a)
\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)
b)
\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)
\(=3\sqrt{5}+1\)
c)
\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)
\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)
d)
\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)
\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)
Bài 2:
a)
\(A=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+1+2\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{(\sqrt{3}+1)^2}}}=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}=\sqrt{6+2(\sqrt{3}-1)}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}=\sqrt{(\sqrt{3}+1)^2}=\sqrt{3}+1\)
b)
\(B=\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}\)
\(=\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-\sqrt{9})^2}}=\sqrt{5}-\sqrt{3-(\sqrt{20}-\sqrt{9})}\)
\(=\sqrt{5}-\sqrt{6-2\sqrt{5}}=\sqrt{5}-\sqrt{5+1-2\sqrt{5}}=\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}\)
\(=\sqrt{5}-(\sqrt{5}-1)=1\)
Bài 2:
c)
\(C=\sqrt{3-\sqrt{5}}(\sqrt{10}+\sqrt{2})(3+\sqrt{5})=\sqrt{6-2\sqrt{5}}(\sqrt{5}+1)(3+\sqrt{5})\)
\(=\sqrt{5+1-2\sqrt{5}}(\sqrt{5}+1)(3+\sqrt{5})\)
\(=\sqrt{(\sqrt{5}-1)^2}(\sqrt{5}+1)(3+\sqrt{5})\)
\(=(\sqrt{5}-1)(\sqrt{5}+1)(3+\sqrt{5})=4(3+\sqrt{5})\)
d)
\(D=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(D\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}=\sqrt{7+1+2\sqrt{7.1}}-\sqrt{7+1-2\sqrt{7.1}}\)
\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}=\sqrt{7}+1-(\sqrt{7}-1)=2\)
\(\Rightarrow D=\sqrt{2}\)
Bài 2:
e)
\(E^2=16+2\sqrt{(8+2\sqrt{10+2\sqrt{5}})(8-2\sqrt{10+2\sqrt{5}})}\)
\(=16+2\sqrt{8^2-(2\sqrt{10+2\sqrt{5}})^2}\)
\(=16+2\sqrt{64-4(10+2\sqrt{5})}=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{20+4-2\sqrt{20.4}}\)
\(=16+2\sqrt{(\sqrt{20}-\sqrt{4})^2}=16+2(\sqrt{20}-\sqrt{4})\)
\(=12+2\sqrt{20}=10+2+2\sqrt{10.2}=(\sqrt{10}+\sqrt{2})^2\)
\(\Rightarrow E=\sqrt{10}+\sqrt{2}\) (do $E$ không âm)
g)
\(G=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\sqrt{(x-1)+2\sqrt{x-1}+1}+\sqrt{(x-1)-2\sqrt{x-1}+1}\)
\(=\sqrt{(\sqrt{x-1}+1)^2}+\sqrt{(\sqrt{x-1}-1)^2}\)
\(=|\sqrt{x-1}+1|+|\sqrt{x-1}-1|\)
Bài 2:
h)
\(H=4x-\sqrt{9x^2-12x+4}=4x-\sqrt{(3x)^2-2.3x.2+2^2}\)
\(=4x-\sqrt{(3x-2)^2}=4x-|3x-2|\)
\(=\left\{\begin{matrix} 4x-(3x-2): \text{nếu x}\geq \frac{2}{3}\\ 4x-(2-3x): \text{nếu x}< \frac{2}{3}\end{matrix}\right.\)
\(=\left\{\begin{matrix} x+2:\text{nếu x}\geq \frac{2}{3}\\ 7x-2:\text{nếu x}< \frac{2}{3}\end{matrix}\right.\)
i)
\(\frac{\sqrt{7}-\sqrt{2}}{\sqrt{7}+\sqrt{2}}+\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}}=\frac{(\sqrt{7}-\sqrt{2})^2+(\sqrt{7}+\sqrt{2})^2}{(\sqrt{7}+\sqrt{2})(\sqrt{7}-\sqrt{2})}\)
\(=\frac{7-2\sqrt{14}+2+7+2\sqrt{14}+2}{7-2}=\frac{18}{5}\)