Tính \(A=\left(21x^2+6x\right)^{2018}\) biết \(x=\sqrt{\sqrt{3}-\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)
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TA
12 tháng 7 2017
Ta có
\(\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)^2\)
\(=27+10\sqrt{2}+27-10\sqrt{2}-2\sqrt{\left(27+10\sqrt{2}\right)\left(27-10\sqrt{2}\right)}\)
\(=54-2\sqrt{529}=8\)
\(\Rightarrow\) \(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}=\sqrt{8}=2\sqrt{2}\)
Xét tử số
\(\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}\)
\(=\left(\sqrt{27+10\sqrt{2}}.\sqrt{27-10\sqrt{2}}\right)\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)\)
\(=23\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)\)
\(=23.2\sqrt{2}=46\sqrt{2}\)
Lại có \(\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right)^2\)
\(=\sqrt{13}-3+\sqrt{13}+3+2\sqrt{\left(\sqrt{13}-3\right)\left(\sqrt{13}+3\right)}\)
\(=2\sqrt{13}+2\sqrt{4}=2\sqrt{13}+4\)
ta bình phương mẫu số
\(\left(\frac{\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}}{\sqrt{\sqrt{13}+2}}\right)^2=\frac{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right)^2}{\sqrt{13}+2}\)
\(=\frac{2\sqrt{13}+4}{\sqrt{13}+2}=2\)
Vậy mẫu \(=\sqrt{2}\)
Vậy \(x=\frac{46\sqrt{2}}{\sqrt{2}}=46\) thay vào ta đc A = 92880
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=>\(A=\left(4-6x+6x-5\right)^{2009}=\left(-1\right)^{2009}=-1\)
11 tháng 8 2017
ai nay dung kinh nghiem la chinh
cau a)
ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)
\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)
\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)
\(x=\frac{3-1}{1}=2\)
suy ra
x^3-4x+1=1
A=1^2018
A=1
b)
ta thay
\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)
khi do
\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)
\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)
x=2
thay vao
x^3+3x-14=0
B=0^2018
B=0
16 tháng 6 2021
a) ĐK: \(x^2+7x+7\ge0\)
Đặt \(a=\sqrt{x^2+7x+7}\) \(\left(a\ge0\right)\)
PT \(\Rightarrow3a^2-3+2a=2\) \(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=-\dfrac{5}{3}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x^2+7x+7=1\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\) (Thỏa mãn)
Vậy ...
b) ĐK: \(x^2-6x+6\ge0\)
Đặt \(a=\sqrt{x^2-6x+6}\) \(\left(a\ge0\right)\)
PT \(\Rightarrow a^2+3=4a\) \(\Leftrightarrow\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\) (Thỏa mãn)
+) Với \(a=3\) \(\Rightarrow x^2-6x+6=9\) \(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{3}\\x=3-2\sqrt{3}\end{matrix}\right.\) (Thỏa mãn)
+) Với \(a=1\) \(\Rightarrow x^2-6x+6=1\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\) (Thỏa mãn)
Vậy ...
16 tháng 6 2021
c)C1: Áp dụng bđt AM-GM \(\Rightarrow VT\ge2>\dfrac{7}{4}\)
=> Dấu = ko xảy ra hay pt vô nghiệm
C2: Đk:\(x>0\)
Đặt \(a=\sqrt{\dfrac{x^2+x+1}{x}}\left(a>0\right)\) \(\Rightarrow\dfrac{1}{a}=\sqrt{\dfrac{x}{x^2+x+1}}\)
Pttt: \(a+\dfrac{1}{a}=\dfrac{7}{4}\Leftrightarrow4a^2-7a+4=0\)
\(\Delta =-15<0 \) => Pt vô nghiệm
Vậy...
d) Đk: \(x\le-8;x\ge0\)
Đặt \(t=\sqrt{x\left(8+x\right)}\left(t\ge0\right)\)
Pttt: \(t^2-3=2t\Leftrightarrow t^2-2t-3=0\Leftrightarrow\left[{}\begin{matrix}t=3\left(tm\right)\\t=-1\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x\left(8+x\right)}=3\Leftrightarrow x^2+8x-9=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-9\end{matrix}\right.\) (tm)
Vậy...
ta có : \(x=\sqrt{\sqrt{3}-\sqrt{5-\sqrt{13+4\sqrt{3}}}}=\sqrt{\sqrt{3}-\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}\)
\(=\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}}=\sqrt{\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{\sqrt{3}-\sqrt{3}+1}=1\)
\(\Rightarrow A=\left(21x^2+6x\right)^{2018}=\left(21\left(1\right)^2+6.1\right)^{2018}=27^{2018}\)