tính \(\left(7+4\sqrt{3}\right)^{2009}.\left(7-4\sqrt{3}\right)^{2009}\)
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QD
1
20 tháng 6 2023
Ta phân tích \(10+6\sqrt{3}=3\sqrt{3}+9+3\sqrt{3}+1\) \(=\left(\sqrt{3}\right)^3+3.\left(\sqrt{3}\right)^2.1+3\sqrt{3}.1^2+1^3\) \(=\left(\sqrt{3}+1\right)^3\).
Vì vậy, \(x=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=\left(\sqrt{3}\right)^2-1=2\)
Vậy \(P=\left(x^3-4x+1\right)^{2009}\)\(=\left(2^3-4.2+1\right)^{2009}\) \(=1\)
31 tháng 5 2021
\(=>x^3=(\sqrt[3]{2\left(\sqrt{3}+1\right)}-\sqrt[3]{2\left(\sqrt{3}-1\right)})^3\)
\(x^3=2\left(\sqrt{3}+1\right)-3.\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]^2.\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\)
+\(3\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]^2\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]-2\left(\sqrt{3}-1\right)\)
\(x^3=\)
\(4-3\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}-\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\)
\(x^3=4-3.\left[\sqrt[3]{4\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\right].\)\(x\)
\(x^3=4-3\left[\sqrt[3]{4\left(3-1\right)}\right].x\)
\(x^3=4-3.2x\)
\(x^3=4-6x\)
thay \(x^3=4-6x\) vào A=>\(A=\left(4-6x+6x-5\right)^{2009}=\left(-1\right)^{2009}=-1\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 4 2020
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:
\(10ab=3\left(a^2+b^2\right)\Leftrightarrow3a^2-10ab+3b^2=0\)
\(\Leftrightarrow\left(3a-b\right)\left(a-3b\right)=0\Rightarrow\left[{}\begin{matrix}3a=b\\a=3b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x+1}=\sqrt{x^2-x+1}\\\sqrt{x+1}=3\sqrt{x^2-x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}9\left(x+1\right)=x^2-x+1\\x+1=9\left(x^2-x+1\right)\end{matrix}\right.\) \(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 4 2020
a/ ĐKXĐ; \(-1\le x\le8\)
Đặt \(\sqrt{1+x}+\sqrt{8-x}=t>0\Rightarrow\sqrt{\left(1+x\right)\left(8-x\right)}=\frac{t^2-9}{2}\)
\(\Rightarrow t+\frac{t^2-9}{2}=3\)
\(\Leftrightarrow t^2+2t-15=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{1+x}+\sqrt{8-x}=3\)
\(\Leftrightarrow9+2\sqrt{\left(1+x\right)\left(8-x\right)}=9\)
\(\Leftrightarrow\left(1+x\right)\left(8-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=8\end{matrix}\right.\)
NQ
1
25 tháng 9 2021
\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)
25 tháng 9 2021
1) \(=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
2) \(=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
3) \(=\sqrt{7}-2+\sqrt{7}+2=2\sqrt[]{7}\)
4) \(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
5) \(=5+6-2\sqrt{5.6}=11-2\sqrt{30}\)
6) \(=3+5-2\sqrt{3.5}=8-4\sqrt{2}\)
7) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)
11 tháng 2 2020
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
Lời giải:
$(7+4\sqrt{3})^{2009}.(7-4\sqrt{3})^{2009}=[(7+4\sqrt{3})(7-4\sqrt{3})]^{2009}$
$=[7^2-(4\sqrt{3})^2]^{2009}=1^{2009}=1$