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Em tách nhỏ ra rồi hỏi nhe!! VD như 1 bài hỏi 1 lần á
3) Ta có: \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}-\dfrac{3\sqrt{5}}{4}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}-\dfrac{3\sqrt{5}}{4}\)
\(=\dfrac{2\sqrt{5}-3\sqrt{5}}{4}\)
\(=\dfrac{-\sqrt{5}}{4}\)
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a: Ta có: \(A=\dfrac{2x-3\sqrt{x}-14}{x-7\sqrt{x}+12}-\dfrac{\sqrt{x}+4}{\sqrt{x}-3}-\dfrac{\sqrt{x}-1}{\sqrt{x}-4}\)
\(=\dfrac{2x-3\sqrt{x}-14-x+16-x+4\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}\)
Ta có: \(B=\dfrac{x-2\sqrt{x}+1}{x-4\sqrt{x}+3}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)
b: Ta có: M=A:B
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
\(=\dfrac{1}{\sqrt{x}-4}\)
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1.
\(\sqrt{x}-2+x\sqrt{x}-2x=(\sqrt{x}-2)+(x\sqrt{x}-2x)=(\sqrt{x}-2)+x(\sqrt{x}-2)\)
\(=(\sqrt{x}-2)(1+x)\)
2.
\(x-10\sqrt{x}+25=(\sqrt{x})^2-2.5.\sqrt{x}+5^2=(\sqrt{x}-5)^2\)
3.
\(4x+4\sqrt{x}+1=(2\sqrt{x})^2+2.2\sqrt{x}+1=(2\sqrt{x}+1)^2\)
4.
\(9x-6\sqrt{x}+1=(3\sqrt{x})^2-2.3\sqrt{x}+1=(3\sqrt{x}-1)^2\)
5.
\(\sqrt{x-1}-5x+5=\sqrt{x-1}-5(x-1)=\sqrt{x-1}(1-5\sqrt{x-1})\)
6.
\(\sqrt{x-3}-2x+6=\sqrt{x-3}-2(x-3)=\sqrt{x-3}(1-2\sqrt{x-3})\)
7.
\(x\sqrt{x}-1=(\sqrt{x})^3-1^3=(\sqrt{x}-1)(x+\sqrt{x}+1)\)
8.
\(x-10\sqrt{x}+21=x-3\sqrt{x}-(7\sqrt{x}-21)\)
\(=\sqrt{x}(\sqrt{x}-3)-7(\sqrt{x}-3)=(\sqrt{x}-7)(\sqrt{x}-3)\)
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Bài 8:
a: Thay x=16 vào P, ta được:
\(P=\dfrac{4+2}{4-1}=\dfrac{6}{3}=2\)
b: Thay \(x=3+2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{\sqrt{2}+1+2}{\sqrt{2}+1-1}=\dfrac{3+\sqrt{2}}{\sqrt{2}}=3\sqrt{2}+2\)
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\(ĐK:x\ge0;x\ne1\\ 1,P=\dfrac{x-2\sqrt{x}+1-x-\sqrt{x}+5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2}{\sqrt{x}-1}\\ 2,P< 0\Leftrightarrow\sqrt{x}-1< 0\left(2>0\right)\\ \Leftrightarrow\sqrt{x}< 1\Leftrightarrow0\le x< 1\)
1) Ta có: \(\dfrac{1}{\sqrt{3}-1}+\dfrac{1}{4+2\sqrt{3}}-\dfrac{2}{\sqrt{3}}-\dfrac{3}{2}\)
\(=\dfrac{\sqrt{3}+1}{2}+\dfrac{2-\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{3}-\dfrac{3}{2}\)
\(=\dfrac{\sqrt{3}+1+2-\sqrt{3}-3}{2}-\dfrac{2\sqrt{3}}{3}\)
\(=-\dfrac{2\sqrt{3}}{3}\)
3) Ta có: \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}-\dfrac{3\sqrt{5}}{4}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}-\dfrac{3\sqrt{5}}{4}\)
\(=\dfrac{2\sqrt{5}-3\sqrt{5}}{4}\)
\(=\dfrac{-\sqrt{5}}{4}\)