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9 tháng 12 2023
a: \(\dfrac{6}{2-\sqrt{10}}-\dfrac{2\sqrt{5}-5\sqrt{2}}{\sqrt{2}-\sqrt{5}}+\sqrt{49+4\sqrt{10}}\)
\(=\dfrac{6\left(2+\sqrt{10}\right)}{4-10}-\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}+\sqrt{49+2\cdot2\sqrt{10}}\)
\(=\dfrac{6\left(2+\sqrt{10}\right)}{-6}-\sqrt{10}+\sqrt{49+2\cdot\sqrt{40}}\)
\(=-2-\sqrt{10}-\sqrt{10}+\sqrt{49+4\sqrt{10}}\)
\(=-2-2\sqrt{10}+\sqrt{49+4\sqrt{10}}\)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)
\(\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x}{\sqrt{x}+1}\)
\(=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{x}{\sqrt{x}+1}\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\cdot\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\cdot\left(\sqrt{x}+1\right)}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)\)
ND
0
NN
2 tháng 7 2017
a)\(A=\sqrt{2}-\sqrt{12-8\sqrt{2}}\)
\(A=\sqrt{2}-\sqrt{\left(2\sqrt{2}-2\right)^2}\)
\(A=\sqrt{2}-2\sqrt{2}+2\)
\(A=2-\sqrt{2}\)
c)\(C=\dfrac{2\sqrt{3-\sqrt{5}}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{3-\sqrt{5}}}{\sqrt{2}\left(\sqrt{5}-1\right)}=\dfrac{\sqrt{2}\sqrt{3-\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{5}-1}=\dfrac{\sqrt{5}-1}{\sqrt{5}-1}=1\)
d)với x,y,x>0 xyz=100 =>\(\sqrt{xyz}=\sqrt{100}=10\)
\(D=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+10}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{10\sqrt{z}}{\sqrt{xz}+10\sqrt{z}+10}\)
\(D=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+\sqrt{xyz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{xyz^2}}{\sqrt{xz}+\sqrt{xyz^2}+\sqrt{xyz}}\)
\(D=\dfrac{1}{\sqrt{y}+1+\sqrt{yz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{yz}}{1+\sqrt{yz}+\sqrt{y}}\)
\(D=\dfrac{1+\sqrt{y}+\sqrt{yz}}{\sqrt{yz}+\sqrt{y}+1}=1\)
mình chỉ giải được câu a,c,d còn câu b mình nghĩ sai đề
18 tháng 9 2021
\(a,ĐK:\dfrac{3}{x+7}\ge0\Leftrightarrow x+7>0\left(3>0;x+7\ne0\right)\Leftrightarrow x>-7\\ b,ĐK:\dfrac{-2}{5-x}\ge0\Leftrightarrow5-x< 0\left(2-< 0;5-x\ne0\right)\Leftrightarrow x>5\\ c,ĐK:x^2-7x+10\ge0\Leftrightarrow\left(x-5\right)\left(x-2\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5\ge0\\x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5\le0\\x-2\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge5\\x\le2\end{matrix}\right.\)
\(d,ĐK:x^2-8x+10\ge0\Leftrightarrow\left(x-4-\sqrt{6}\right)\left(x-4+\sqrt{6}\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-4-\sqrt{6}\ge0\\x-4+\sqrt{6}\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-4-\sqrt{6}\le0\\x-4+\sqrt{6}\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge4+\sqrt{6}\\x\ge4-\sqrt{6}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le4+\sqrt{6}\\x\le4-\sqrt{6}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4+\sqrt{6}\\x\le4-\sqrt{6}\end{matrix}\right.\)
\(e,ĐK:9x^2+1\ge0\Leftrightarrow x\in R\left(9x^2+1\ge1>0\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 6 2021
\(3T=\left(\sqrt{x^2-6x+19}-\sqrt{x^2-6x+10}\right)\left(\sqrt{x^2-6x+19}+\sqrt{x^2-6x+10}\right)\)
\(=x^2-6x+19-\left(x^2-6x+10\right)=9\)
\(\Rightarrow T=3\)
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
26 tháng 8 2023
2:
ĐKXĐ: x>=3
\(\Leftrightarrow\sqrt{x-3+2\cdot\sqrt{x-3}\cdot\sqrt{3}+3}+\sqrt{x-3-2\cdot\sqrt{x-3}\cdot\sqrt{3}+3}=2\sqrt{3}\)
=>\(\left|\sqrt{x-3}+\sqrt{3}\right|+\left|\sqrt{x-3}-\sqrt{3}\right|=2\sqrt{3}\)
\(\Leftrightarrow\sqrt{x-3}+\sqrt{3}+\left|\sqrt{x-3}-\sqrt{3}\right|=2\sqrt{3}\)
\(\Leftrightarrow\sqrt{x-3}+\left|\sqrt{x-3}-\sqrt{3}\right|=\sqrt{3}\)(1)
TH1: x>=6
(1) trở thành \(\sqrt{x-3}+\sqrt{x-3}-\sqrt{3}=\sqrt{3}\)
=>\(2\sqrt{x-3}=2\sqrt{3}\)
=>x-3=3
=>x=6(nhận)
TH2: 3<=x<6
Phương trình (1) sẽ là;
\(\sqrt{x-3}+\sqrt{3}-\sqrt{x-3}=\sqrt{3}\)
=>\(\sqrt{3}=\sqrt{3}\)(luôn đúng)
1:
\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}\)
\(=16+2\cdot\sqrt{64-4\cdot\left(10+2\sqrt{5}\right)}\)
\(=16+2\cdot\sqrt{24-8\sqrt{5}}\)
\(=16+2\cdot\sqrt{20-2\cdot2\sqrt{5}\cdot2+4}\)
\(=16+2\cdot\sqrt{\left(2\sqrt{5}-2\right)^2}\)
\(=16+2\cdot\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)
\(=10+2\cdot\sqrt{10}\cdot\sqrt{2}+2\)
\(=\left(\sqrt{10}+\sqrt{2}\right)^2\)
=>\(A=\sqrt{10}+\sqrt{2}\)
\(\sqrt{x-10}=-2\)
Ta có: \(\sqrt{x-10}\ge0\)
`=>` Đề bài vô lý
Vậy pt vô nghiệm