Cho x = \(\sqrt{2+\sqrt{2+\sqrt{2}}}-\sqrt{6-3\sqrt{2+\sqrt{3}}}\) .Tính x4-16x
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KM
1
14 tháng 8 2020
<=> \(x^2=2+\sqrt{2+\sqrt{3}}+6-3\sqrt{2+\sqrt{3}}-2\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(6-3\sqrt{2+\sqrt{3}}\right)}\)
<=> \(x^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{12-6\sqrt{2+\sqrt{3}}+6\sqrt{2+\sqrt{3}}-3\left(2+\sqrt{3}\right)}\)
<=> \(x^2=8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-2\sqrt{12-6-3\sqrt{3}}\)
<=> \(x^2=8-\sqrt{2}.\sqrt{\left(\sqrt{3}+1\right)^2}-2\sqrt{6-3\sqrt{3}}\)
<=> \(x^2=8-\sqrt{2}\left(\sqrt{3}+1\right)-\sqrt{2}.\sqrt{12-6\sqrt{3}}\)
<=> \(x^2=8-\sqrt{6}-\sqrt{2}-\sqrt{2}.\sqrt{\left(3-\sqrt{3}\right)^2}\)
<=> \(x^2=8-\sqrt{6}-\sqrt{2}-\sqrt{2}\left(3-\sqrt{3}\right)\)
<=> \(x^2=8-\sqrt{6}-\sqrt{2}-3\sqrt{2}+\sqrt{6}\)
<=> \(x^2=8-4\sqrt{2}\)
<=> \(8-x^2=4\sqrt{2}\)
<=> \(\left(8-x^2\right)^2=\left(4\sqrt{2}\right)^2\)
<=> \(x^4-16x^2+64=32\)
<=> \(x^4-16x^2=-32\)
VẬY \(x^4-16x^2=-32\)
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26 tháng 6 2023
Giải
Ta có:
\(x=\sqrt{2+\sqrt{2+\sqrt{3}}-\sqrt{6-3\sqrt{2+\sqrt{3}}}}\)
Khi đó:
\(x^2=\left(\sqrt{2+\sqrt{2+\sqrt{3}}-\sqrt{6-3\sqrt{2+\sqrt{3}}}}\right)^2\\ =2+\sqrt{2+\sqrt{3}}+6-3\sqrt{2+\sqrt{3}}-2\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(6-3\sqrt{2+\sqrt{3}}\right)}\\ =8-2\sqrt{2+\sqrt{3}}-2\sqrt{12-3\left(2+\sqrt{3}\right)}\\ =8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-2\sqrt{6-3\sqrt{3}}\\ =8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-\sqrt{2}.\sqrt{12-6\sqrt{3}}\\ =8-\sqrt{2}.\left(\sqrt{4+2\sqrt{3}}+\sqrt{12-6\sqrt{3}}\right)\\ =8-\sqrt{2}.\left(\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}+1}+\sqrt{9-2.3\sqrt{3}+\left(\sqrt{3}\right)^2}\right)\\ 8-\sqrt{2}.\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(3-\sqrt{3}\right)^2}\right)\\ =8-\sqrt{2}.\left(\sqrt{3}+1+3-\sqrt{3}\right)\\ =8-4\sqrt{2}\\ \Rightarrow x^4-16x^2=\left(8-4\sqrt{2}\right)^2-16.\left(8-4\sqrt{2}\right)\\ =96-64\sqrt{2}-128+64\sqrt{2}=-32\)
Vậy \(S=-32\)
AT
1
VT
17 tháng 8 2018
Đặt \(\sqrt{2+\sqrt{3}}=a\left(a>0\right)\)
Ta có x=\(\sqrt{2+a}-\sqrt{3\left(2-a\right)}\Rightarrow x^2=2+a+3\left(2-a\right)-2\sqrt{3\left(2+a\right)\left(2-a\right)}\)\(=8-2a-2\sqrt{3\left(4-a^2\right)}=8-2a-2\sqrt{3\left(4-2-\sqrt{3}\right)}=8-2a-\sqrt{6}\sqrt{4-2\sqrt{3}}\)
\(=8-2\sqrt{2+\sqrt{3}}-\sqrt{6}\left(\sqrt{3}-1\right)=8-\sqrt{2}\sqrt{4+2\sqrt{3}}-3\sqrt{2}+\sqrt{6}\)
\(=8-\sqrt{2}\left(\sqrt{3}+1\right)-3\sqrt{2}+\sqrt{6}=8-\sqrt{6}-\sqrt{2}-3\sqrt{2}+\sqrt{6}=8-4\sqrt{2}\)
\(\Rightarrow x^2-8=-4\sqrt{2}\Rightarrow\left(x^2-8\right)^2=32\Rightarrow x^4-16x^2+64=32\Rightarrow x^4-16x^2+32=0\left(ĐPCM\right)\)
24 tháng 9 2021
1) \(ĐK:x\in R\)
2) \(ĐK:x< 0\)
3) \(ĐK:x\in\varnothing\)
4) \(=\sqrt{\left(x+1\right)^2+2}\)
\(ĐK:x\in R\)
5) \(=\sqrt{-\left(a-4\right)^2}\)
\(ĐK:x\in\varnothing\)
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
a.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$
$\Leftrightarrow \sqrt{2x}=3$
$\Leftrightarrow 2x=9$
$\Leftrightarrow x=\frac{9}{2}$ (tm)
b.
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$
$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$
$\Leftrightarrow 3\sqrt{x+2}=15$
$\Leftrightarrow \sqrt{x+2}=5$
$\Leftrightarrow x+2=25$
$\Leftrightarrow x=23$ (tm)
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
c.
$\sqrt{(x-2)^2}=12$
$\Leftrightarrow |x-2|=12$
$\Leftrightarrow x-2=12$ hoặc $x-2=-12$
$\Leftrightarrow x=14$ hoặc $x=-10$
e.
PT $\Leftrightarrow |2x-1|-x=3$
Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$
$\Leftrightarrow x=4$ (tm)
Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$
$\Leftrightarrow x=\frac{-2}{3}$ (tm)
10 tháng 9 2018
a) ta có : \(A+B=\sqrt[3]{7-5\sqrt{2}}+\sqrt[3]{20+14\sqrt{2}}\)
\(=\sqrt[3]{\left(1-\sqrt{2}\right)^3}+\sqrt{\left(\sqrt{2}+2\right)^3}=1-\sqrt{2}+\sqrt{2}+2=3\)
b) ở đây : https://hoc24.vn/hoi-dap/question/650070.html
DN
11 tháng 9 2018
a, Ta có: \(A+B=\sqrt[3]{7-5\sqrt{2}}+\sqrt[3]{20+14\sqrt{2}}\)
\(=\sqrt[3]{\left(1-\sqrt{2}\right)^3+\sqrt[3]{\left(\sqrt{2}+2\right)^3}}\)
\(=1-\sqrt{2}+\sqrt{2}+2=1+2=3\)
Vậy ...
\(b,x=\sqrt{6-3\sqrt{2+\sqrt{3}}}-\sqrt{2+\sqrt{2+\sqrt{3}}}\)
Đặt \(\sqrt{2+\sqrt{3}=t}\) , ta có:
\(x=\sqrt{6-3.t}-\sqrt{2+t}\)
\(\Rightarrow x^2=2+t+3.\left(2-t\right)-2\sqrt{3}\left(2+t\right)\left(2-t\right)\)
\(=8-2t-2\sqrt{3\left(4-t^2\right)}\)
\(=8-2t-2\sqrt{3\left(4-2-\sqrt{3}\right)}\)
\(=8-2t-\sqrt{6}.\sqrt{4-2\sqrt{3}}\)
\(=8-2\sqrt{2+\sqrt{3}}-\sqrt{6}\left(\sqrt{3}-1\right)\)
\(=8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-3\sqrt{2}+\sqrt{6}\)
\(=8-\sqrt{2}\left(\sqrt{3}+1\right)-3\sqrt{2}+\sqrt{6}\)
\(=8-\sqrt{6}-\sqrt{2}-3\sqrt{2}+\sqrt{6}\)
\(=8-4\sqrt{2}\)
\(\Rightarrow x^2-8=-4\sqrt{2}\)
\(\Rightarrow\left(x^2-8\right)^2=32\)
\(\Rightarrow x^4-16x^2+64=32\)
\(\Rightarrow x^4-16x^2+64-32=0\)
\(\Rightarrow x^4-16x^2+32=0\) (đpcm)
Chúc bạn hok tốt!!!