Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(\Leftrightarrow\left|x-3\right|=\sqrt{3}+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\sqrt{3}+4\\x_2=-\sqrt{3}+2\end{matrix}\right.\)
Vậy...
ĐKXĐ : \(x^4+\left(\sqrt{3}-\sqrt{2}\right).x^2-\sqrt{6}\ne0\)
\(\Leftrightarrow x\ne\sqrt[4]{2}\)
\(P=\dfrac{x^2-\sqrt{2}}{x^4+\left(\sqrt{3}-\sqrt{2}\right).x^2-\sqrt{6}}\)
\(=\dfrac{x^2-\sqrt{2}}{\left(x^4-\sqrt{2}x^2\right)+\sqrt{3}\left(x^2-\sqrt{2}\right)}\)
\(=\dfrac{x^2-\sqrt{2}}{\left(x^2+\sqrt{3}\right)\left(x^2-\sqrt{2}\right)}=\dfrac{1}{x^2+\sqrt{3}}\)
ĐK x khác 4 và x không âm
\(=\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{4-x}\\ =\frac{8\sqrt{x}+4x}{4-x}\\ =\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\\ =\frac{4\sqrt{x}}{2-\sqrt{x}}\)
C=\(\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{\left(\sqrt{x}+2\right).\left(x-1\right)-\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{x\sqrt{x}-\sqrt{x}+2x-2-\left(x-1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{x-1+x\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{\left(x-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{1}{\sqrt{x}}=\frac{\sqrt{x}}{x}\)
\(a,\sqrt{\frac{5.\left(38^2-17^2\right)}{8.\left(47^2-19^2\right)}}\)
\(=\sqrt{\frac{5.\left(38-17\right)\left(38+17\right)}{8.\left(47-19\right)\left(47+19\right)}}\)
\(=\sqrt{\frac{5.21.55}{8.28.66}}\)
\(=\sqrt{\frac{5775}{14784}}=\frac{5\sqrt{231}}{2\sqrt{4370}}\)
.bn tính lại \(\sqrt{14784}\)đi sao lạ vậy