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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=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. \(VT=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2+\sqrt{3}-2+\sqrt{3}=VP\)
Bài 1.
Ta có : \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3+4\sqrt{3}+4}-\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|\sqrt{3}-2\right|\)
\(=\sqrt{3}+2-\left(2-\sqrt{3}\right)\)
\(=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(P=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)
\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: P=1/4
=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)
=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
=>\(4\sqrt{x}-8-3\sqrt{x}=0\)
=>\(\sqrt{x}=8\)
=>x=64
c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{2-\sqrt{3}}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Khi x = 49 thì:
\(A=\frac{4\sqrt{49}}{\sqrt{49}-1}=\frac{4\cdot7}{7-1}=\frac{28}{6}=\frac{14}{3}\)
2) Ta có:
\(B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}\)
\(B=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
c) \(P=A\div B=\frac{4\sqrt{x}}{\sqrt{x}-1}\div\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{4\sqrt{x}}{\sqrt{x}+1}\)
Ta có: \(P\left(\sqrt{x}+1\right)=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\frac{4\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow4\sqrt{x}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)
Mà \(VT\ge0\left(\forall x\ge0,x\ne1\right)\)
\(\Rightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}=2\\x-4=0\end{cases}}\Rightarrow x=4\)
Vậy x = 4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\)\(\left(\frac{\sqrt{x}+2}{\sqrt{x}}-\frac{\sqrt{x}-4}{\sqrt{x}-2}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)\(:\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{x-4-x+4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-3}{4}\)
\(b,\)Để \(P>0\Rightarrow\frac{\sqrt{x}-3}{4}>0\)
Mà \(4>0\Rightarrow\sqrt{x}-3>0\Rightarrow\sqrt{x}>3\Rightarrow x>9\)
\(c,\sqrt{P}_{min}=0\Rightarrow\frac{\sqrt{x}-3}{4}=0\)
\(\Leftrightarrow\sqrt{x}-3=0\Rightarrow\sqrt{x}=3\Rightarrow x=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(P=\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\)
\(\Leftrightarrow P=\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4x+8\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4\sqrt{x}}{2-\sqrt{x}}\)
2) Để \(P=2\)
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=2\)
\(\Leftrightarrow4\sqrt{x}=4-2\sqrt{x}\)
\(\Leftrightarrow6\sqrt{x}=4\)
\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{4}{9}\)
Vậy để \(P=2\Leftrightarrow x=\frac{4}{9}\)
3) Khi \(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\2\sqrt{x}-1==0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\left(ktm\right)\\x=\frac{1}{4}\left(tm\right)\end{cases}}\)
Thay \(x=\frac{1}{4}\)vào P, ta được :
\(\Leftrightarrow P=\frac{4\sqrt{\frac{1}{4}}}{2-\sqrt{\frac{1}{4}}}=\frac{4\cdot\frac{1}{2}}{2-\frac{1}{2}}=\frac{2}{\frac{3}{2}}=\frac{4}{3}\)
4) Để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)
\(\Leftrightarrow8x-4\sqrt{x}=-x-\sqrt{x}+6\)
\(\Leftrightarrow9x-3\sqrt{x}-6=0\)
\(\Leftrightarrow3x-\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=3x-2\)
\(\Leftrightarrow x=9x^2-12x+4\)
\(\Leftrightarrow9x^2-13x+4=0\)
\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}9x-4=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{9}\\x=1\end{cases}}\)
Thử lại ta được kết quá : \(x=\frac{4}{9}\left(ktm\right)\); \(x=1\left(tm\right)\)
Vậy để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\Leftrightarrow x=1\)
5) Để biểu thức nhận giá trị nguyên
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}\inℤ\)
\(\Leftrightarrow4\sqrt{x}⋮2-\sqrt{x}\)
\(\Leftrightarrow-4\left(2-\sqrt{x}\right)+8⋮2-\sqrt{x}\)
\(\Leftrightarrow8⋮2-\sqrt{x}\)
\(\Leftrightarrow2-\sqrt{x}\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;-2;6;-6;10\right\}\)
Ta loại các giá trị < 0
\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;6;10\right\}\)
\(\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)
\(\)
![](https://rs.olm.vn/images/avt/0.png?1311)
câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
a, ĐK \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(P=\frac{x-1}{\sqrt{x}}:\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
Ta thấy \(P=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}>0\forall x>0,x\ne1\)
b, P=\(\frac{x+2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\frac{2}{2+\sqrt{3}}+2\sqrt{\frac{2}{2+\sqrt{3}}}+1}{\sqrt{\frac{2}{2+\sqrt{3}}}-1}\)
=\(\frac{\frac{4}{\left(\sqrt{3}+1\right)^2}+2.\sqrt{\left(\frac{2}{\left(\sqrt{3}+1\right)^2}\right)}+1}{\sqrt{\left(\frac{2}{2+\sqrt{3}}\right)^2}-1}=\frac{\frac{4}{\left(\sqrt{3}+1\right)^2}+2.\frac{2}{\sqrt{3}+1}+1}{\frac{2}{\sqrt{3}+1}-1}\)
\(=\frac{12+6\sqrt{3}}{1-3}=-6-3\sqrt{3}\)
cậu ơi câu c đâu ạ??