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a: Ta có: \(E=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right):\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right):\left(\dfrac{x-1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{4\sqrt{x}+4\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}}{x-1}\)
\(=\dfrac{4x^2}{\left(x-1\right)^2}\)
b: Để E=2 thì \(4x^2=2\left(x-1\right)^2\)
\(\Leftrightarrow4x^2-2x^2+4x-2=0\)
\(\Leftrightarrow2x^2+4x-2=0\)
\(\Leftrightarrow x^2+2x-1=0\)
\(\Leftrightarrow\left(x+1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{2}-1\\x=\sqrt{2}-1\end{matrix}\right.\)
c: Ta có: \(x=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=2\)
Thay x=2 vào E, ta được:
\(E=\dfrac{4\cdot2^2}{1}=16\)
Cô hướng dẫn nhé :)
a. ĐK: \(x>0;x\ne1\)
Ta có \(E=\frac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)+4\sqrt{x}\left(x-1\right)}{x-1}:\frac{x-1}{\sqrt{x}}\)
\(\Leftrightarrow E=\frac{4x\sqrt{x}}{x-1}.\frac{\sqrt{x}}{x-1}=\frac{4x^2}{\left(x-1\right)^2}\)
b. Để \(E=2\Rightarrow\frac{4x^2}{\left(x-1\right)^2}=2\Leftrightarrow2x^2+4x-2=0\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}-1\\x=-\sqrt{2}-1\left(L\right)\end{cases}}\)
c. \(x=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=2\)
Vậy E = 16.
a)Rút gọn E ta đc:
\(\frac{4x^2+\sqrt{x}\left(2x+2\right)-4x}{x^2-2x+1}\)
b)Với E=2\(\Leftrightarrow\)\(\frac{4x^2+\sqrt{x}\left(2x+2\right)-4x}{x^2-2x+1}=2\)
\(\Leftrightarrow\frac{4x^2}{x^2-2x+1}+\frac{2\sqrt{x^3}}{x^2-2x+1}-\frac{4x}{x^2-2x+1}+\frac{2\sqrt{x}}{x^2-2x+1}-2=0\)
\(\Leftrightarrow\frac{2\left(x^2\sqrt{x^3}+\sqrt{x}-1\right)}{x^2-2x+1}=0\)
\(\Leftrightarrow x^2+\sqrt{x^3}+\sqrt{x}-1=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{-\sqrt{x^3}-\sqrt{x}+1}=0\left(tm\right)\\\sqrt{-\sqrt{x^3}-\sqrt{x}+1}+x=0\left(loai\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\sqrt{5}-3=0\left(loai\right)\\2x+\sqrt{5}-3=0\left(tm\right)\end{cases}}\)
\(\Leftrightarrow x=-\frac{\sqrt{5}-3}{2}\left(tm\right)\)
a, A\(=\left(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{x-1}{\sqrt{x}}\) ĐK x>0 ;\(x\ne1;x\ne-1\)
\(A=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4x\sqrt{x}-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}}{x-1}\)
\(A=\frac{4x\sqrt{x}}{x-1}.\frac{\sqrt{x}}{x-1}\)=\(\frac{4x^2}{\left(x-1\right)^2}\)
b, Để A =2 \(\Rightarrow\frac{4x^2}{\left(x-1\right)^2}=2\Rightarrow4x^2=2\left(x-1\right)^2\)
<=> \(4x^2=2x^2-4x+2\)
<=> \(2x^2+4x-2=0\)
<=> \(x^2+2x-1=0\)
\(\Delta=1^2-1.\left(-1\right)\) = 2
=> \(\orbr{\begin{cases}x_1=-1-\sqrt{2}\left(loại\right)\\x_2=-1+\sqrt{2}\left(nhận\right)\end{cases}}\)
Vậy x=\(-1+\sqrt{2}\)thì A =2
c, Thay x =\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)=2
=>A = \(\frac{4.2^2}{\left(2-1\right)^2}=16\)
Vậy A=16 thì x=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
ĐKXĐ: \(x>0;x\ne1\)
\(E=\left(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right):\left(\frac{x-1}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}}{x-1}+4\sqrt{x}\right):\left(\frac{x-1}{\sqrt{x}}\right)\)
\(=\frac{4x\sqrt{x}}{\left(x-1\right)}.\frac{\sqrt{x}}{\left(x-1\right)}=\frac{4x^2}{\left(x-1\right)^2}\)
Đề có nhầm ko bạn?
\(E=2\Rightarrow\left(\frac{2x}{x-1}\right)^2=2\Rightarrow\left[{}\begin{matrix}\frac{2x}{x-1}=\sqrt{2}\\\frac{2x}{x-1}=-\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\sqrt{2}x-\sqrt{2}\\2x=-\sqrt{2}x+\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-\sqrt{2}}{2-\sqrt{2}}< 0\left(l\right)\\x=\frac{\sqrt{2}}{2+\sqrt{2}}=\sqrt{2}-1\end{matrix}\right.\)
