Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(B=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)
\(=\left(2\sqrt{3}-1\right)^2\left(2+\sqrt{3}\right)^2-8\sqrt{20+2\sqrt{\left(4+3\sqrt{3}\right)^2}}\)
\(=\left(3\sqrt{3}+4\right)^2-8\sqrt{20+2\left(4+3\sqrt{3}\right)}\)
\(=\left(3\sqrt{3}+4\right)^2-8\sqrt{28+6\sqrt{3}}\)
\(=\left(3\sqrt{3}+4\right)^2-8\sqrt{\left(3\sqrt{3}+1\right)^2}\)
\(=43+24\sqrt{3}-8\left(3\sqrt{3}+1\right)=35\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-4}+\frac{\sqrt{x}-10}{x-4}\)
\(A=\frac{x+2\sqrt{x}+x-3\sqrt{x}+2+\sqrt{x}-10}{x-4}\)
\(A=\frac{2x-8}{x-4}=\frac{2\left(x-4\right)}{x-4}=2\)
\(B=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{\left(3\sqrt{3}+4\right)^2}}\)
\(B=43+24\sqrt{3}-8\sqrt{20+6\sqrt{3}+8}\)
\(B=43+24\sqrt{3}-8\sqrt{28+6\sqrt{3}}\)
\(B=43+24\sqrt{3}-8\sqrt{\left(3\sqrt{3}+1\right)^2}\)
\(B=43+24\sqrt{3}-24\sqrt{3}-8\)
\(B=35\)
\(A=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\) \(A=\left(6+7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{\left(16+2.4.3\sqrt{3}+27\right)}}\)
\(A=6\left(7+4\sqrt{3}\right)+\left(7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{\left(4+3\sqrt{3}\right)^2}}\)Trong căn là hằng đẳng thức (a+b)^2
\(A=42+24\sqrt{3}+7^2-\left(4\sqrt{3}\right)^2-8\sqrt{20+2\left(4+3\sqrt{3}\right)}\) sử dụng hằng đẳng thức a^2 -b^2\(A=43+24\sqrt{3}-8\sqrt{20+8+2.3\sqrt{3}}\)
\(A=43+24\sqrt{3}-8\sqrt{1+2.3\sqrt{3}+27}\)trong căn tiếp tục là hằng đẳng thức (a+b)^2\(A=43+24\sqrt{3}-8\sqrt{\left(1+3\sqrt{3}\right)^2}\)
\(A=43+24\sqrt{3}-8\left(1+3\sqrt{3}\right)\)
\(A=35\)
chúc bạn thành công nhé
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
\(A=43+24\sqrt{3}-8\sqrt{20+2\sqrt{\left(3\sqrt{3}+4\right)^2}}\)
\(=43+24\sqrt{3}-8\sqrt{20+2\left(3\sqrt{3}+4\right)}\)
\(=43+24\sqrt{3}-8\sqrt{28+6\sqrt{3}}\)
\(=43+24\sqrt{3}-8\sqrt{\left(3\sqrt{3}+1\right)^2}\)
\(=43+24\sqrt{3}-8\left(3\sqrt{3}+1\right)\)
\(=43-8=35\)
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)
\(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-3)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-(\sqrt{20}-3)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}}=\sqrt{\sqrt{5}-(\sqrt{5}-1)}=\sqrt{1}=1\)
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
\(D=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)
\(=\left(2\sqrt{3}-1\right)^2\left(\sqrt{3}+2\right)^2-8\sqrt{20+2\left(3\sqrt{3}+4\right)}\)
\(=\left(4+3\sqrt{3}\right)^2-8\sqrt{28+6\sqrt{3}}\)\(=\left(4+3\sqrt{3}\right)^2-8\left(3\sqrt{3}+1\right)\)
\(=43+24\sqrt{3}-24\sqrt{3}-8=35\)