Cho biểu thức Q=\(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right):\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)
a) Rút gọn biểu thức
b) Tìm x để Q=2
b) Tính giá trị của Q khi x=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
ai làm giúp mình với :<<< Cảm ơn nhiều lắm
\(M=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right)\div\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\div\dfrac{x-1}{\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+4x\sqrt{x}-4\sqrt{x}}{\left(x-1\right)}\times\dfrac{\sqrt{x}}{x-1}\)
\(=\dfrac{4x^2}{\left(x-1\right)^2}\)
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\(\dfrac{4x^2}{\left(x-1\right)^2}=2\)
\(\Leftrightarrow4x^2=2\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+4x-2=0\)
\(\Leftrightarrow2\left(x+1-\sqrt{2}\right)\left(x+1+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1+\sqrt{2}\\x=-1-\sqrt{2}\end{matrix}\right.\) (nhận)
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\(x=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)\left(4+\sqrt{15}\right)}\)
\(=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(16-15\right)\left(4+\sqrt{15}\right)}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
= 5 - 3 = 2
\(M=\dfrac{4x^2}{\left(x-1\right)^2}=16\)
