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KR
10 tháng 5 2018
pt \(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x-\sqrt{2}-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\x=3+\sqrt{2}\end{cases}}\)
Y
26 tháng 6 2019
1.
\(x+4\sqrt{x}+3=0\left(ĐK:x\ge0\right)\\ \Leftrightarrow x+\sqrt{x}+3\sqrt{x}+3=0\\ \Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)=0\\ \Rightarrow x\in\varnothing\)
2.
\(x^2+3x\sqrt{x}+2x=0\left(ĐK:x\ge0\right)\\ \Leftrightarrow x^2+x\sqrt{x}+2x\sqrt{x}+2x=0\\ \Leftrightarrow x\sqrt{x}\left(\sqrt{x}+1\right)+2x\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x=0\)
3.
\(x+2\sqrt{x}-8=0\\ \Leftrightarrow x-2\sqrt{x}+4\sqrt{x}-8=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)+4\left(\sqrt{x}-2\right)=0\\ \Leftrightarrow\left(\sqrt{x}+4\right)\left(\sqrt{x}-2\right)=0\\ \Leftrightarrow\sqrt{x}-2=0\\ \Leftrightarrow x=4\)
4.
\(x+\sqrt{9x}-\sqrt{100}=0\left(ĐK:x\ge0\right)\\ \Leftrightarrow x+3\sqrt{x}-10=0\\ \Leftrightarrow x+5\sqrt{x}-2\sqrt{x}-10=0\\ \Leftrightarrow\left(\sqrt{x}+5\right)\left(\sqrt{x}-2\right)=0\\ \Leftrightarrow\sqrt{x}-2=0\\ \Leftrightarrow x=4\)
5.
\(x+\sqrt{3x}-\sqrt{2x}-\sqrt{6}=0\left(ĐK:x\ge0\right)\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}+\sqrt{3}\right)-\sqrt{2}\left(\sqrt{x}+\sqrt{3}\right)=0\\ \Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-\sqrt{2}\right)=0\\ \Leftrightarrow\sqrt{x}-\sqrt{2}=0\Leftrightarrow x=2\)
6.
\(\sqrt{5x}-x-\sqrt{15}+\sqrt{3x}=0\left(ĐK:x\ge0\right)\\ \Leftrightarrow\sqrt{x}\left(\sqrt{5}-\sqrt{x}\right)-\sqrt{3}\left(\sqrt{5}-\sqrt{x}\right)=0\\ \Leftrightarrow\left(\sqrt{x}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-\sqrt{3}=0\\\sqrt{5}-\sqrt{x}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)
21 tháng 8 2020
BÀI 1:
a)
PT <=> \(3x-2=7-4\sqrt{3}\)
<=> \(3x=9-4\sqrt{3}\)
<=> \(x=3-\frac{4}{\sqrt{3}}\)
b)
pt => \(x+1=14-6\sqrt{5}\)
<=> \(x=13-6\sqrt{5}\)
BÀI 2:
a)
pt <=> \(\sqrt{x^2-9}=3\sqrt{x-3}\)
<=> \(x^2-9=9\left(x-3\right)\)
<=> \(x^2-9=9x-27\)
<=> \(x^2-9x+18=0\)
<=> \(\orbr{\begin{cases}x=6\\x=3\end{cases}}\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2019
Bài 1:
a/ \(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}\)
\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{4}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{4}}=\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}\)
\(=\frac{10}{2}=5\)
b/ \(=\left(2-\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{\left(3-\sqrt{2}\right)^2}}}\)
\(=\left(2-\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+3-\sqrt{2}}}\)
\(=\left(2-\sqrt{3}\right)\sqrt{2+4\sqrt{6}}\)
Bạn coi lại đề, tới đây ko rút gọn được nữa nên chắc bạn ghi đề nhầm ở chỗ nào đó
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2019
c/ \(=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(5-\sqrt{24}\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-\sqrt{24}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\left(5+2\sqrt{6}\right)\left(5-\sqrt{24}\right)=\left(5+\sqrt{24}\right)\left(5-\sqrt{24}\right)=1\)
d/ Nhân cả tử và mẫu của từng phân số với liên hợp của mẫu, mẫu số sẽ thành 1 hết:
\(=\frac{\sqrt{25}-\sqrt{24}}{\left(\sqrt{25}+\sqrt{24}\right)\left(\sqrt{25}-\sqrt{24}\right)}+\frac{\sqrt{24}-\sqrt{23}}{\left(\sqrt{24}+\sqrt{23}\right)\left(\sqrt{24}-\sqrt{23}\right)}+...+\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)
\(=\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-1\)
\(=\sqrt{25}-1=5-1=4\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 5 2019
ĐKXĐ: \(x\ge0;x\ne1\)
\(\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11+3x+7\sqrt{x}-6-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{3x+19\sqrt{x}-14}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+7\right)\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 6 2019
\(y=\frac{1}{9+4\sqrt{5}}=\frac{1}{\left(\sqrt{5}+2\right)^2}\)
\(\Rightarrow N=\frac{1}{\left(\sqrt{5}-2\right)^2}-\frac{3}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\frac{2}{9+4\sqrt{5}}\)
\(=\frac{1}{9-4\sqrt{5}}+\frac{2}{9+4\sqrt{5}}-3=\frac{9+4\sqrt{5}+18-8\sqrt{5}}{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}-3=24-4\sqrt{5}\)
\(S^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2+y^2+x^2y^2+1+x^2y^2-1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+x^2y^2-1\)
\(=\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2-1\)
\(=2005^2-1\)
\(\Rightarrow S=\pm\sqrt{2005^2-1}\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 6 2019
c/
Giả sử \(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}\)
\(\Leftrightarrow\sqrt[3]{3+\sqrt[3]{3}}-\sqrt[3]{3}< \sqrt[3]{3}-\sqrt[3]{3-\sqrt[3]{3}}\)
\(\Leftrightarrow\frac{\sqrt[3]{3}}{\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}+\sqrt[3]{9}}< \frac{\sqrt[3]{3}}{\sqrt[3]{9}+\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}}\)
\(\Leftrightarrow\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}+\sqrt[3]{9}>\sqrt[3]{9}+\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\)
\(\Leftrightarrow\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}>\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\) (1)
Ta có: \(\left\{{}\begin{matrix}\sqrt[3]{9+3\sqrt[3]{3}}>\sqrt[3]{9-3\sqrt[3]{3}}\\\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}>\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\end{matrix}\right.\)
Nên (1) đúng
Vậy BĐT ban đầu đúng
Đặt \(\hept{\begin{cases}\sqrt[3]{x+2}=a\\\sqrt[3]{5-x}=b\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a+b=1\\a^3+b^3=7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a+b=1\\\left(a+b\right)\left(a^2-ab+b^2\right)=7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a+b=1\\a^2-ab+b^2=7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a+b=1\\\left(a+b\right)^2-3ab=7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a+b=1\\1-3ab=7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a+b=1\\ab=-2\end{cases}}\)
Làm nốt
Ta có: \(\sqrt[3]{x+2}+\sqrt[3]{5-x}=1\)
<=> \(x+2+5-x+3\sqrt[3]{\left(x+2\right)\left(5-x\right)}\left[\sqrt[3]{x+2}+\sqrt[3]{5-x}\right]=1\)
<=> \(7+3\sqrt[3]{\left(x+2\right)\left(5-x\right)}.1=1\)
<=> \(3\sqrt[3]{\left(x+2\right)\left(5-x\right)}=-6\)
<=> \(\sqrt[3]{\left(x+2\right)\left(5-x\right)}=-2\)
<=>\(3x+10-x^2=-8\)
<=> \(x^2-3x-18=0\)
<=> \(x^2-6x+3x-18=0\)
<=> \(\left(x-6\right)\left(x+3\right)=0\)
<=> \(\orbr{\begin{cases}x=6\\x=-3\end{cases}}\)