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\(C=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}+2\right)+6\sqrt{x}}{x-4}.\left(x-4\right)=2\sqrt{x}\)
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thay \(x=3-2\sqrt{2}\) vào P ta có:
\(\dfrac{x+8}{\sqrt{x}+1}=\dfrac{3-2\sqrt{2}+8}{\sqrt{3-2\sqrt{2}}+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}-1+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}}\)
\(b,x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\)
Thay vào P, ta được:
\(P=\dfrac{3-2\sqrt{2}+8}{\sqrt{\left(\sqrt{2}-1\right)^2}+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}}=\dfrac{11\sqrt{2}-4}{2}\)
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Đề 1:
Bài 1:
\(a,=\sqrt{\left(\sqrt{7}+1\right)^2}-\left|-1+\sqrt{7}\right|=\sqrt{7}+1-\sqrt{7}+1=2\\ b,=2\sqrt{2}-4\sqrt{2}-5\sqrt{2}+\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{2}-7\sqrt{2}=\dfrac{-13\sqrt{2}}{\sqrt{2}}\)
Bài 2:
\(PT\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=\dfrac{1}{2}\Leftrightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+\dfrac{1}{2}=1\\x=-\dfrac{1}{2}+\dfrac{1}{2}=0\end{matrix}\right.\)
Bài 3:
\(a,M=\dfrac{a-2\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{2\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)}=\dfrac{2}{\sqrt{a}+1}\\ b,M< 1\Leftrightarrow\dfrac{2}{\sqrt{a}+1}-1< 0\Leftrightarrow\dfrac{1-\sqrt{a}}{\sqrt{a}+1}< 0\\ \Leftrightarrow1-\sqrt{a}< 0\left(\sqrt{a}+1>0\right)\\ \Leftrightarrow a>1\)
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ĐKXĐ: x>=0; x<>9
\(B=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
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\(\Delta=b^{^2}-4ac=m^{^2}-4\left(3-m\right)=m^{^2}-12+4m=\left(m+2\right)^{^2}-16\)
Phương trình có hai nghiệm phân biệt khi và chỉ khi:
\(\Delta>0\Leftrightarrow\left(m+2\right)^2-16>0\Leftrightarrow m+2>16\Leftrightarrow m>14\\ Viete:\\ x_1+x_2=-\dfrac{b}{a}=m\\ x_1x_2=\dfrac{c}{a}=3-m\)
x1 là nghiệm phương trình nên:
\(x_1^2=mx_1+m-3=m\left(x_1+1\right)-3\\ \Rightarrow\left[m\left(x_1+1\right)-3+3\right]\left(x_2+1\right)=12\\ m\left(x_1+1\right)\left(x_2+1\right)=12\\ m\left(x_1x_2+x_1+x_2+1\right)=12\\ m\left(3-m+m+1\right)=12\\ 4m=12\\ m=3\left(KTM\right)\)
Vậy không tồn tại m thoả đề bài
Sửa lại: m + 2 > 4 <=> m > 2, m = 3 thoả đề nhé