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b, t = \(\sqrt{3- \sqrt{5}}\)(3 +\(\sqrt{5}\)).(\(\sqrt{10}\)-\(\sqrt{2}\))
t = \(\sqrt{3- \sqrt{5}}\)(3 +\(\sqrt{5}\)).\(\sqrt{2}\)(\(\sqrt{5}\) -1)
t = (\(\sqrt{5}\) -1).(\(\sqrt{5}\) -1).(3 +\(\sqrt{5}\))
t = (\(\sqrt{5}\) -1)2.(3 +\(\sqrt{5}\))
t = (5 - \(2\sqrt{5}\)+1).(3 +\(\sqrt{5}\))
t = 15 + \(5\sqrt{5}\) \(-6\sqrt{5}\)-10+1+\(\sqrt{5}\)
t = 6
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
I: Rút gọn
\(A=\sqrt{7-4\sqrt{3}}\\ =\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\\ =\sqrt{\left(2-\sqrt{3}\right)^2}\\ =2-\sqrt{3}\)
\(B=\sqrt{19-8\sqrt{3}}\\ =\sqrt{16-2\cdot4\cdot\sqrt{3}+3}\\ =\sqrt{\left(4-\sqrt{3}\right)^2}\\ =4-\sqrt{3}\)
\(C=\sqrt{21-4\sqrt{5}}\\ =\sqrt{20-2\cdot2\sqrt{5}+1}\\ =\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot1+1}\\ =\sqrt{\left(2\sqrt{5}-1\right)^2}\\ =2\sqrt{5}-1\)
Câu D mình làm chưa ra, sorry :<
b/ ĐKXĐ:...
\(\Leftrightarrow x-19-2\sqrt{x-19}+1+y-7-4\sqrt{y-7}+4+z-1997-6\sqrt{z-1997}+9=0\)
\(\Leftrightarrow\left(\sqrt{x-19}-1\right)^2+\left(\sqrt{y-7}-2\right)^2+\left(\sqrt{z-1997}-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-19}=1\\\sqrt{y-7}=2\\\sqrt{z-1997}=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=20\\y=11\\z=2006\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=3\left(x^2+2\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\) \(\Rightarrow a^2+b^2=x^2+2\)
Pt tương đương:
\(10ab=3\left(a^2+b^2\right)\Leftrightarrow3a^2-10ab+3b^2=0\)
\(\Leftrightarrow\left(3a-b\right)\left(a-3b\right)=0\Rightarrow\left[{}\begin{matrix}3a=b\\a=3b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x+1}=\sqrt{x^2-x+1}\\\sqrt{x+1}=3\sqrt{x^2-x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}9\left(x+1\right)=x^2-x+1\\x+1=9\left(x^2-x+1\right)\end{matrix}\right.\) \(\Leftrightarrow...\)
a/ ĐKXĐ; \(-1\le x\le8\)
Đặt \(\sqrt{1+x}+\sqrt{8-x}=t>0\Rightarrow\sqrt{\left(1+x\right)\left(8-x\right)}=\frac{t^2-9}{2}\)
\(\Rightarrow t+\frac{t^2-9}{2}=3\)
\(\Leftrightarrow t^2+2t-15=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{1+x}+\sqrt{8-x}=3\)
\(\Leftrightarrow9+2\sqrt{\left(1+x\right)\left(8-x\right)}=9\)
\(\Leftrightarrow\left(1+x\right)\left(8-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=8\end{matrix}\right.\)
\(=\left(\sqrt{3}+4\right)\sqrt{\left(4-\sqrt{3}\right)^2}+\left(\sqrt{3}-4\right)\sqrt{\left(4+\sqrt{3}\right)^2}=\left(\sqrt{3}+4\right)\left(4-\sqrt{3}\right)+\left(\sqrt{3}-4\right)\left(4+\sqrt{3}\right)\)
\(=16-3+3-16=0\)