a) ĐKXĐ : \(x>0;x\ne1\)

b) \(M=\frac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{\sqrt{x}-1}-\frac{x+\sqrt{x}}{\sqrt{x}}\)

\(M=\frac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}-1}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

\(M=\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}-\left(\sqrt{x}+1\right)\)

\(M=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}-\sqrt{x}-1\)

\(M=\sqrt{x}-1-\sqrt{x}-1\)

\(M=-2\)( đpcm )

\(=\left(\frac{\sqrt{x}\left(\sqrt{2}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{2}+2\right)}\right).\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{4\text{x}}}\)

\(=\left(\frac{\sqrt{2\text{x}}+2\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{2}+2\right)}\right).\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{4\text{x}}}\)

\(=\frac{\sqrt{2\text{x}}+x}{\left(\sqrt{x}-2\right)\left(\sqrt{2}+2\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{4\text{x}}}\)

\(=\frac{\sqrt{2\text{x}}+x}{\sqrt{2}+2}.\frac{\sqrt{x}-2}{\sqrt{4\text{x}}}\)

\(=\frac{x\sqrt{2}-2\sqrt{2\text{x}}+x\sqrt{x}-2\text{x}}{2\sqrt{2\text{x}}+4\sqrt{x}}\)

tick cho mình nha

Đk \(x\ge0\) và \(x\ne\pm4\)

\(M=\frac{x+2\sqrt{2}+x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{x-4}{\sqrt{4x}}=\frac{2x\left(x-4\right)}{2\sqrt{2}\left(x-4\right)}=\frac{\sqrt{2}x}{2}\)

c) Để \(M>3\Leftrightarrow\frac{\sqrt{2}x}{2}>3\Leftrightarrow\frac{\sqrt{2}x-6}{2}>0\Leftrightarrow\sqrt{2}x>6\Leftrightarrow x>\frac{6}{\sqrt{2}}\)

a: ĐKXĐ: x=0; x<>1

\(M=\left(2+\sqrt{x}\right)\left(1-2\sqrt{x}-x+1+\sqrt{x}+x\right)\)

\(=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)=4-x\)

b: Sửa đề: P=1/M

P=1/4-x=-1/x-4

Để P nguyên thì x-4 thuộc {1;-1}

=>x thuộc {5;3}

a)\(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

b)\(E=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)

\(=\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)

\(=\frac{x-1}{\sqrt{x}}\)

c)Để E>0 thì \(\frac{x-1}{\sqrt{x}}>0\)

Mà \(\sqrt{x}>0\)

\(\Rightarrow x-1>0\) hay \(x>1\)

a. ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\\x\ne4\end{matrix}\right.\)

\(A=\frac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}=\frac{2\sqrt{x}-4}{3\sqrt{x}+3}\)

b. \(A< 0\Leftrightarrow\frac{2\sqrt{x}-4}{3\sqrt{x}+3}< 0\Leftrightarrow2\sqrt{x}-4< 0\Leftrightarrow x< 4\)\(\Rightarrow\left\{{}\begin{matrix}0\le x< 4\\x\ne1\end{matrix}\right.\)thì \(A< 0\)

a ) \(ĐKXĐ:x\ge0;x\ne1\)

= \(\frac{x+1+\sqrt{x}}{x+1}:\left[\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]-1\)

\(=\frac{x+1+\sqrt{x}}{x+1}:\frac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)

\(=\frac{x+1+\sqrt{x}}{x+1}:\frac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)

\(=\frac{\left(x+1+\sqrt{x}\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)

\(=\frac{x+1+\sqrt{x}}{\sqrt{x}-1}-1=\frac{x+2}{\sqrt{x}-1}\)

B ) Ta có :

 \(Q=P-\sqrt{x}\)

\(=\frac{\sqrt{x}+2}{\sqrt{x}-1}-\sqrt{x}\)

\(=\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}=1+\frac{3}{\sqrt{x}-1}\)

Đế Q nhận giá trị nguyên thì \(1+\frac{3}{\sqrt{x}-1}\in Z\)

\(\Leftrightarrow\frac{3}{\sqrt{x}-1}\in Z\left(vì1\in Z\right)\)

\(\Leftrightarrow\sqrt{x}-1\inƯ\left(3\right)\)

Ta có bảng sau :

Vậy để biểu thức \(Q=P-\sqrt{x}\) nhận giá trị nguyên thì \(x\in\left\{16;4;0\right\}\)


 

a/ ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

b/ Ta có :

\(P=\left(\frac{1}{1-\sqrt{x}}-\frac{1}{1+\sqrt{x}}\right).\frac{x-1}{\sqrt{x}}\)

\(=\left(\frac{1+\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}-\frac{1-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right).\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

\(=\frac{1+\sqrt{x}-1+\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}\)

\(=-2\)

Vậy...

bình phương lên

\(Q^2=\frac{x-2\sqrt{\left(x-1\right)}+x+2\sqrt{\left(x-1\right)}+2\sqrt{\left(x-2\right)^2}}{x^2-4\left(x-1\right)}.\left(\frac{x-2}{x-1}\right)^2\)

\(=\frac{2x+2\left(x-2\right)}{\left(x-2\right)^2}.\frac{\left(x-2\right)^2}{\left(x-1\right)^2}=\frac{2\left(x+x-2\right)}{\left(x-1\right)^2}=\frac{4\left(x-1\right)}{\left(x-1\right)^2}=\frac{4}{x-1}\)

\(\Rightarrow Q=\frac{2}{\sqrt{x-1}}\)