Cho biểu thức: \(M=\frac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{\sqrt{x}-1}-\frac{x+\sqrt{x}}{\sqrt{x}}\)
a, Tìm điều kiện để biểu thức M có nghĩa
b, C/minh: Biểu thức M không phụ thuộc vào biến.
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TN
17 tháng 6 2019
a/ \(đkxđ\) : \(x\ne0;x\ne1\)
b/
M = \(\frac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{\sqrt{x}-1}-\frac{x+\sqrt{x}}{\sqrt{x}}\)
\(=\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}-\frac{x+\sqrt{x}}{\sqrt{x}}\)
\(=\frac{\left(x-2\sqrt{x}+1\right).\sqrt{x}-\left(x+\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}+x-x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{2\sqrt{x}-2x}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{2\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=-2\)
chúc bn học tốt
TH
7 tháng 10 2015
\(=\left(\frac{\sqrt{x}\left(\sqrt{2}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{2}+2\right)}\right).\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{4\text{x}}}\)
\(=\left(\frac{\sqrt{2\text{x}}+2\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{2}+2\right)}\right).\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{4\text{x}}}\)
\(=\frac{\sqrt{2\text{x}}+x}{\left(\sqrt{x}-2\right)\left(\sqrt{2}+2\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{4\text{x}}}\)
\(=\frac{\sqrt{2\text{x}}+x}{\sqrt{2}+2}.\frac{\sqrt{x}-2}{\sqrt{4\text{x}}}\)
\(=\frac{x\sqrt{2}-2\sqrt{2\text{x}}+x\sqrt{x}-2\text{x}}{2\sqrt{2\text{x}}+4\sqrt{x}}\)
tick cho mình nha
30 tháng 7 2020
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 \(M>3\Leftrightarrow\frac{\sqrt{2}x}{2}>3\Leftrightarrow\frac{\sqrt{2}x-6}{2}>0\Leftrightarrow\sqrt{2}x>6\Leftrightarrow x>\frac{6}{\sqrt{2}}\)
26 tháng 7 2023
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}
AN
6 tháng 10 2019
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 \(x>1\)
29 tháng 11 2019
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 \(A< 0\)
30 tháng 12 2019
a ) \(ĐKXĐ:x\ge0;x\ne1\)
= \(\frac{x+1+\sqrt{x}}{x+1}:\left[\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]-1\)
\(=\frac{x+1+\sqrt{x}}{x+1}:\frac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
\(=\frac{x+1+\sqrt{x}}{x+1}:\frac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
\(=\frac{\left(x+1+\sqrt{x}\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)
\(=\frac{x+1+\sqrt{x}}{\sqrt{x}-1}-1=\frac{x+2}{\sqrt{x}-1}\)
30 tháng 12 2019
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 :
| \(\sqrt{x}-1\) | 3 | -3 | 1 | -1 |
| \(\sqrt{x}\) | 4 | -2 | 2 | 0 |
| \(x\) | 16(t/m) | 4(t/m) | 0(t/m) |
Vậy để biểu thức \(Q=P-\sqrt{x}\) nhận giá trị nguyên thì \(x\in\left\{16;4;0\right\}\)
30 tháng 7 2020
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...
26 tháng 9 2020
bình phương lên
\(Q^2=\frac{x-2\sqrt{\left(x-1\right)}+x+2\sqrt{\left(x-1\right)}+2\sqrt{\left(x-2\right)^2}}{x^2-4\left(x-1\right)}.\left(\frac{x-2}{x-1}\right)^2\)
\(=\frac{2x+2\left(x-2\right)}{\left(x-2\right)^2}.\frac{\left(x-2\right)^2}{\left(x-1\right)^2}=\frac{2\left(x+x-2\right)}{\left(x-1\right)^2}=\frac{4\left(x-1\right)}{\left(x-1\right)^2}=\frac{4}{x-1}\)
\(\Rightarrow Q=\frac{2}{\sqrt{x-1}}\)
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 )