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a: \(P=\dfrac{x+\sqrt{x}+1+11\sqrt{x}-11+34}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+12\sqrt{x}+24}{\sqrt{x}+2}\)
b: Thay \(x=3-2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{3-2\sqrt{2}+12\left(\sqrt{2}-1\right)+24}{\sqrt{2}-1+2}\)
\(=\dfrac{27-2\sqrt{2}+12\sqrt{2}-12}{\sqrt{2}+1}=5+5\sqrt{2}\)
Bài 1
1)
Đkxđ \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có \(4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
Khi đó A=\(\frac{\sqrt{3}-1-1}{\sqrt{3}-1+1}=\frac{\sqrt{3}-2}{\sqrt{3}}\)
2) Đề là \(5-2\sqrt{6}\)sẽ hợp lý hơn nha bn
Đkxđ\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-\sqrt{2}\ne0\end{matrix}\right.\)
Ta có \(5-2\sqrt{6}=\left(1-\sqrt{6}\right)^2\)
Khi đó
B= \(\frac{1-\sqrt{6}}{1-\sqrt{6}-\sqrt{2}}\)
1)
đk: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Rgọn
A=\(\frac{x+12}{x-4}+\frac{1}{\sqrt{x}+2}-\frac{4}{\sqrt{x}-2}\)
= \(\frac{x+12+\sqrt{x}-2-\left(4\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
2)
B=\(\frac{3\sqrt{x}-1}{\sqrt{x}+2}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{10\sqrt{x}}{x-4}\) đk \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
= \(\frac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+10\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
= \(\frac{3x-5\sqrt{x}-2-\left(x+3\sqrt{x}+2\right)+10\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{3x-5\sqrt{x}-2-x-3\sqrt{x}-2+10\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{2x+2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\left(2x+2\sqrt{x}\right)-\left(4\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{2\sqrt{x}\left(\sqrt{x}+2\right)-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\left(\sqrt{x}+2\right)2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=2\)
Chúc bn học tốt
Nhớ tích cho mk nhé