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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:
\(E=\dfrac{4\cdot2^2}{1}=16\)
\(1,\\ a,E=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ b,E>0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}>0\Leftrightarrow\sqrt{x}-1>0\left(\sqrt{x}>0\right)\\ \Leftrightarrow x>1\\ 2,\\ a,B=\dfrac{x-\sqrt{x}+\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\left(\sqrt{x}+1\right)\\ B=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,B=2\Leftrightarrow\sqrt{x}-1=2\left(\sqrt{x}+1\right)\\ \Leftrightarrow\sqrt{x}-1=2\sqrt{x}+2\\ \Leftrightarrow\sqrt{x}=-3\Leftrightarrow x\in\varnothing\)
a: Sửa đề: \(E=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right):\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}+4\sqrt{x}\right):\dfrac{x-1}{\sqrt{x}}\)
\(=\left(\dfrac{4\sqrt{x}+4\sqrt{x}\left(x-1\right)}{x-1}\right)\cdot\dfrac{\sqrt{x}}{x-1}\)
\(=\dfrac{4\sqrt{x}\left(1+x-1\right)}{x-1}\cdot\dfrac{\sqrt{x}}{x-1}=\dfrac{4x^2}{\left(x-1\right)^2}\)
b: Để E=2 thì \(4x^2=2\left(x-1\right)^2\)
\(\Leftrightarrow4x^2-2\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow4x^2-2x^2+4x-2=0\)
\(\Leftrightarrow2x^2+4x-2=0\)
\(\Leftrightarrow x^2+2x-1=0\)
\(\Leftrightarrow\left(x+1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{2}\\x+1=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}-1\left(nhận\right)\\x=-\sqrt{2}-1\left(loại\right)\end{matrix}\right.\)
c: \(x=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
Thay x=2 vào E, ta được:
\(E=\dfrac{4\cdot2^2}{\left(2-1\right)^2}=16\)
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\(\Leftrightarrow\)\(\frac{4x^2+\sqrt{x}\left(2x+2\right)-4x}{x^2-2x+1}=2\)
\(\Leftrightarrow\frac{4x^2}{x^2-2x+1}+\frac{2\sqrt{x^3}}{x^2-2x+1}-\frac{4x}{x^2-2x+1}+\frac{2\sqrt{x}}{x^2-2x+1}-2=0\)
\(\Leftrightarrow\frac{2\left(x^2\sqrt{x^3}+\sqrt{x}-1\right)}{x^2-2x+1}=0\)
\(\Leftrightarrow x^2+\sqrt{x^3}+\sqrt{x}-1=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{-\sqrt{x^3}-\sqrt{x}+1}=0\left(tm\right)\\\sqrt{-\sqrt{x^3}-\sqrt{x}+1}+x=0\left(loai\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\sqrt{5}-3=0\left(loai\right)\\2x+\sqrt{5}-3=0\left(tm\right)\end{cases}}\)
\(\Leftrightarrow x=-\frac{\sqrt{5}-3}{2}\left(tm\right)\)
\(a)E=\left(\dfrac{x-2\sqrt{x}}{x-4}-1\right):\left(\dfrac{4-x}{x-\sqrt{x}-6}+\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\right)\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-1\right):\left(\dfrac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+2}\right):\dfrac{4-x+x-4-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}+2}:\dfrac{9-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{-2}{\sqrt{x}+2}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{9-x}\\ =\dfrac{-2\left(\sqrt{x}-3\right)}{9-x}=\dfrac{2\left(\sqrt{x}-3\right)}{x-9}\\ =\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{2}{\sqrt{x}+3}\)
\(b)\)E dương
\(\Leftrightarrow E>0\\ \Leftrightarrow\dfrac{2}{\sqrt{x}+3}>0\\ \Leftrightarrow\sqrt{x}+3>0\left(Vì.2>0\right)\\ \Leftrightarrow\sqrt{x}>-3\forall x\in R\\ \Rightarrow x\ge0\)
Kết hợp đk
\(x\ge0;x\ne4;x\ne9\)
Vậy \(x\ge0;x\ne4;x\ne9\) thì E dương
\(a,E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(x>0;x\ne1\right)\\ E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\dfrac{x}{\sqrt{x}-1}\\ b,E>1\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\\ \Leftrightarrow\sqrt{x}-1>0\left[x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\right]\\ \Leftrightarrow x>1\left(tm\right)\)
\(c,E=\dfrac{x}{\sqrt{x}-1}=\dfrac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\\ E=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\ge2\sqrt{\dfrac{\sqrt{x}-1}{\sqrt{x}-1}}+2=2+2=4\\ E_{min}=4\Leftrightarrow\sqrt{x}-1=1\Leftrightarrow x=4\)
a) Điều kiện: \(x\ge0;x\ne1;x\ne\dfrac{1}{4}\)\(E=\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt[]{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(E=\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right).\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(E=\dfrac{2x\sqrt{x}+x-\sqrt{x}-x\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(E=\dfrac{x\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(E=\dfrac{x\sqrt{x}-2\sqrt{x}}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(E=\dfrac{x\sqrt{x}-2\sqrt{x}+x\sqrt{x}+x+\sqrt{x}}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)
\(E=\dfrac{2x\sqrt{x}-\sqrt{x}+x}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)
\(E=\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)
\(E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)
\(E=\dfrac{x+\sqrt{x}}{x+\sqrt{x}+1}\)
b)Vì \(x\ge0\) nên \(x+\sqrt{x}\ge0\) và \(x+\sqrt{x}+1>0\)
Do đó: \(E\ge0\). Dấu "=" xảy ra \(\Leftrightarrow x=0\)
c)\(E\ge\dfrac{6}{7}\Leftrightarrow\dfrac{x+\sqrt{x}}{x+\sqrt{x}+1}\ge\dfrac{6}{7}\Leftrightarrow7x+7\sqrt{x}\ge6x+6\sqrt{x}+6\)
\(\Leftrightarrow x+\sqrt{x}-6\ge0\Leftrightarrow x-2\sqrt{x}+3\sqrt{x}-6\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ge0\)
\(\Leftrightarrow\sqrt{x}-2\ge0\Leftrightarrow\sqrt{x}\ge2\Leftrightarrow x\ge4\)
@Akai Haruma giup mk vs
a: ĐKXĐ: x>0; x<>1
b: \(E=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4\sqrt{x}\left(x-1\right)}{x-1}:\dfrac{x-1}{\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+4x\sqrt{x}-4\sqrt{x}}{x-1}\cdot\dfrac{\sqrt{x}}{x-1}\)
\(=\dfrac{4x^2}{\left(x-1\right)^2}\)
c: Để E=2 thì \(4x^2=2x^2-4x+2\)
\(\Leftrightarrow2x^2+4x-2=0\)
hay \(x\in\left\{-1+\sqrt{2};-1-\sqrt{2}\right\}\)