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a. ĐK \(x\ge0\)và \(x\ne1\)
A =\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\frac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\cdot\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+2\sqrt{x}+1+\sqrt{x}-x-1+\sqrt{x}}\)
\(=\frac{x+1}{4\sqrt{x}}\)
b. Thay \(x=\frac{2-\sqrt{3}}{2}\Rightarrow A=\frac{\frac{2-\sqrt{3}}{2}+1}{4\sqrt{\frac{2-\sqrt{3}}{2}}}=\frac{4-\sqrt{3}}{4\left(\sqrt{3}-1\right)}=\frac{4-\sqrt{3}}{4-4\sqrt{3}}=-\frac{1+3\sqrt{3}}{8}\)
c . Ta có \(A-\frac{1}{2}=\frac{x+1}{4\sqrt{x}}-\frac{1}{2}=\frac{x-2\sqrt{x}+1}{4\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}>0\)với \(\forall x>0\)và \(x\ne1\)
Vậy A >1/2
đkxđ là \(x\ne1;x>0\)
\(Q=\frac{\sqrt{x}\left(\left(\sqrt{x}\right)^3-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(Q=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
gtnn \(x-\sqrt{x}+1=x-\frac{1}{2}.2.\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
gtnn 3/4
ý c bạn tự làm nha mk chịu
\(A=\left(\frac{2x+1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(x+\sqrt{x}+1\right)}\right).\left(\frac{\sqrt{x}.\left(3+x\right)}{-2x}-\sqrt{x}\right) \)
\(A=\left(\frac{2x+1-\sqrt{x}.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\right).\left(\frac{3+x}{-2\sqrt{x}}-\sqrt{x}\right)\)
\(A=\left(\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\right).\left(\frac{3+x+2x}{-2\sqrt{x}}\right)\)
\(A=\left(\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\right).\left(\frac{3x+3}{-2\sqrt{x}}\right)\)
\(A=\frac{1}{\sqrt{x}-1}.\frac{3.\left(x+1\right)}{-2\sqrt{x}}\)
\(A=\frac{3x+3}{-2\sqrt{x}.\left(\sqrt{x}+1\right)}\)
P/s: hình như đề sai hay sao á, thường thì người ta không cho mẫu là 2 số trừ được như ( x - 3x ) đâu
a: \(P=\dfrac{\left[\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}-4+2\left(\sqrt{x}+1\right)\right]}{x+4\sqrt{x}+4}\)
\(=\dfrac{x+\sqrt{x}-2\sqrt{x}-4+2\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2}\)
\(=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}+2\right)^2}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
c: Để |P|>P thì P<0
\(\Leftrightarrow\sqrt{x}-1< 0\)
hay 0<x<1
a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne-1\\x\ne1\end{cases}}\)
Ta có \(P=\left(1+\frac{\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)-1\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x-2\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)\(=\frac{x+\sqrt{x}+1}{x+1}.\frac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}-1\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}-1=\frac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}=\frac{x+2}{\sqrt{x}-1}\)
b. Ta có \(P-\sqrt{x}=\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\frac{x+2-x+\sqrt{x}}{\sqrt{x}-1}=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}=1+\frac{3}{\sqrt{x}-1}\)
Để \(P-\sqrt{x}\in Z\Rightarrow\sqrt{x}-1\inƯ\left(3\right)\Rightarrow\sqrt{x}-1\in\left\{-3;-1;1;3\right\}\)
| \(\sqrt{x}-1\) | \(-3\) | \(-1\) | \(1\) | \(3\) |
| \(\sqrt{x}\) | -2 | 0 | 2 | 4 |
| x | 0 | 4 | 16 | |
| (l) | (n) | (n) | (n) |
Vậy \(x\in\left\{0;4;16\right\}\)thì \(P-\sqrt{x}\in Z\)
a) ĐKXĐ: \(x\ge0;x\ne1\)
\(A=\left(1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right).\left(1-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
b) \(A=1-x\le1\) ( vì \(x\ge0\) )
Vậy max A = 1 khi x = 0
sao biểu thức khi rút gọn xấu vậy bạn ? đề có sai khum :vv, thế tìm x dài lắm bạn ạ
a, Với x > 0 ; \(x\ne1\)
\(M=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}-x}\right)\)
\(=\left(\frac{x+\sqrt{x}+x-\sqrt{x}}{x-1}\right):\left(\frac{2\sqrt{x}-2-2+x}{x\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\frac{2x}{x-1}\right):\left(\frac{x+2\sqrt{x}-4}{x\left(\sqrt{x}-1\right)}\right)=\frac{2x^2}{\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}-4\right)}\)
ĐK: x > 0
a) Rút gọn M
M = \(\frac{\sqrt{x}}{x+\sqrt{x}}:\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
= \(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\left(\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
= \(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\left(\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b) \(\frac{1}{M}=\frac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}+1\ge2+1=3\)
=> M \(\le\)1/3
=> GTLN của M =1/ 3 khi \(\sqrt{x}=\frac{1}{\sqrt{x}}\Leftrightarrow x=1\) thỏa mãn
Vậy max M = 1/3 tại x = 1
bn giải thíchcách làm câu b hôk mk vs mk ko hiểu