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a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)
b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
a: \(=6+2\sqrt{11}-4+\sqrt{11}=2+3\sqrt{11}\)
b: \(=\dfrac{3x+9\sqrt{x}-2x+4\sqrt{x}}{\left(\sqrt{x}+3\right)\left(x-2\sqrt{x}\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}=\dfrac{\sqrt{x}+3}{x-2\sqrt{x}}\)
Ta có: \(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\dfrac{8\sqrt{x}-8x+8x}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(=\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
ta có : \(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
=\(\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)}{4-x}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-x\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
=\(\dfrac{8\sqrt{x}-4x+8x}{4-x}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\dfrac{8\sqrt{x}+4x}{4-x}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x-2}\right)}\) =\(\dfrac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\dfrac{4\sqrt{x}}{2-\sqrt{x}}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\) =\(\dfrac{4x\left(\sqrt{x}-2\right)}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\)
=\(-\dfrac{4x\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\) =\(-\dfrac{4x}{3-\sqrt{x}}\) =\(\dfrac{4x}{\sqrt{x}-3}\)
này mới đúng !!
a: Ta có: \(\frac{8x\cdot\sqrt{x}-1}{2x-\sqrt{x}}-\frac{8x\cdot\sqrt{x}+1}{2x+\sqrt{x}}\)
\(=\frac{\left(2\sqrt{x}\right)^3-1}{\sqrt{x}\left(2\sqrt{x}-1\right)}-\frac{\left(2\sqrt{x}\right)^3+1}{\sqrt{x}\left(2\sqrt{x}+1\right)}\)
\(=\frac{\left(2\sqrt{x}-1\right)\left(4x+2\sqrt{x}+1\right)}{\sqrt{x}\left(2\sqrt{x}-1\right)}-\frac{\left(2\sqrt{x}+1\right)\left(4x-2\sqrt{x}+1\right)}{\sqrt{x}\left(2\sqrt{x}+1\right)}\)
\(=\frac{4x+2\sqrt{x}+1}{\sqrt{x}}-\frac{4x-2\sqrt{x}+1}{\sqrt{x}}=\frac{4\sqrt{x}}{\sqrt{x}}=4\)
Ta có: \(A=\left(\frac{8x\cdot\sqrt{x}-1}{2x-\sqrt{x}}-\frac{8x\cdot\sqrt{x}+1}{2x+\sqrt{x}}\right):\frac{2x+1}{2x-1}\)
\(=4\cdot\frac{2x-1}{2x+1}=\frac{8x-4}{2x+1}\)
b: Để A là số chính phương thì đầu tiên A phải là số tự nhiên
A là số tự nhiên khi \(\begin{cases}8x-4\vdots2x+1\\ \frac{8x-4}{2x+1}\ge0\end{cases}\Rightarrow\begin{cases}8x+4-8\vdots2x+1\\ \frac{2x-1}{2x+1}\ge0\end{cases}\)
=>\(\begin{cases}-8\vdots2x+1\\ \left[\begin{array}{l}x\ge\frac12\\ x<-\frac12\end{array}\right.\end{cases}\Rightarrow\begin{cases}2x+1\in\left\lbrace1;-1;2;-2;4;-4;8;-8\right\rbrace\\ \left[\begin{array}{l}x\ge\frac12\\ x<-\frac12\end{array}\right.\end{cases}\)
=>\(\begin{cases}2x\in\left\lbrace0;-2;1;-3;3;-5;7;-9\right\rbrace\\ \left[\begin{array}{l}x\ge\frac12\\ x<-\frac12\end{array}\right.\end{cases}\)
=>\(\begin{cases}x\in\left\lbrace0;-1;\frac12;-\frac32;\frac32;-\frac52;\frac72;-\frac92\right\rbrace\\ \left[\begin{array}{l}x\ge\frac12\\ x<-\frac12\end{array}\right.\end{cases}\)
=>x∈{-1;1/2;-3/2;3/2;-5/2;7/2;-9/2}
Kết hợp ĐKXĐ, ta được: x\(\in\left\lbrace\frac32;\frac72\right\rbrace\)
TH1: \(x=\frac32\)
=>2x=3
=>2X+1=4; 2x-1=2
\(A=\frac{8x-4}{2x+1}=4\cdot\frac{2x-1}{2x+1}=4\cdot\frac24=2\) không là số chính phương
=>Loại
TH2: \(x=\frac72\)
=>2x=7
=>2x+1=8; 2x-1=6
\(A=4\cdot\frac{2x-1}{2x+1}=4\cdot\frac68=\frac{24}{8}=3\) không là số chính phương
=>Loại
Vậy: x∈∅
Ta có: \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{1}{2\sqrt{x}}\right)\)
\(=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)+8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{2\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)}{2\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{8x-8\sqrt{x}+8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}-2-\sqrt{x}+2}\)
\(=\dfrac{16x-8\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{2\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{2\left(16-8\sqrt{x}\right)}{\sqrt{x}+2}\)
\(=\dfrac{32-16\sqrt{x}}{\sqrt{x}+2}\)
a) \(P=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)
\(=\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\dfrac{2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\dfrac{4}{4-3}\)
\(=4\)
b) \(Q=\left(1+\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\right).\dfrac{1}{\sqrt{x}}vớix>0,x\ne4\)
\(=\left(\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\sqrt{x}-2}\right).\dfrac{1}{\sqrt{x}}\)
\(=\)\(\dfrac{2\sqrt{x}}{\sqrt{x}-2}.\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2}{\sqrt{x}-2}\)
a, \(A=\dfrac{4\left(3-\sqrt{7}\right)}{2}+2\sqrt{7}=\dfrac{12}{2}=6\)
b, \(B=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}}\right):\dfrac{2-\sqrt{x}}{x-1}\)
\(=\left(\dfrac{\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{2-\sqrt{x}}{x-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
nhờ bạn làm rõ vì sao \(\dfrac{\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{2-\sqrt{x}}{x-1}\) lại bằng \(\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
mình xin cảm ơn
a:
Sửa đề: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\right)\cdot\left(\dfrac{\sqrt{x}-7}{\sqrt{x}+1}+1\right)\)
\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right)\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-6}{\sqrt{x}+3}\)
b: P>=1/2
=>P-1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}-\dfrac{1}{2}>=0\)
=>\(\dfrac{-12-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>=0\)
=>\(-\sqrt{x}-15>=0\)
=>\(-\sqrt{x}>=15\)
=>căn x<=-15
=>\(x\in\varnothing\)
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>P>=-2
Dấu = xảy ra khi x=0
\(\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\\ =\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}-\dfrac{2}{\sqrt{x}}\right]\\ =\left[\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]:\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}-\dfrac{2}{\sqrt{x}}\right]\\ =\left[\dfrac{4\sqrt{x}\cdot\left(\sqrt{x}-2\right)-8x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]:\left[\dfrac{\sqrt{x}-1-2\cdot\left(\sqrt{x}-2\right)}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}\right]\\ =\left[\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]:\left[\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}\right]\\ =\left[\dfrac{-4x-8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]:\left[\dfrac{-\sqrt{x}+3}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}\right]\)
\(=\left[\dfrac{-4\sqrt{x}\cdot\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]:\left[\dfrac{-\sqrt{x}+3}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}\right]\\ =\dfrac{-4\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\\ =\dfrac{-4x}{-\sqrt{x}+3}=\dfrac{4x}{\sqrt{x}-3}\)
2
P=\(\frac{4}{\sqrt{x-3}}\)
P\(\rho=\frac{-4x}{x-3\sqrt{x+4}}\)
\[p=\frac{4x}{\sqrt{x}}\]
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