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Bài 1:
a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>3 căn x=3
=>căn x=1
hay x=1(loại)
\(a.\dfrac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4\left(x-1\right)}}\left(1-\dfrac{1}{x-1}\right)=\dfrac{\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x^2-4x+4}}.\dfrac{x-2}{x-1}=\dfrac{\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|}{\left|x-2\right|}.\dfrac{x-2}{x-1}\left(x>1\right)\)
Tới đây dễ r , bạn tự chia TH ra làm nhé :D
\(b.\dfrac{1}{\sqrt{x}+\sqrt{x-1}}-\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{\sqrt{x^3}-x}{1-\sqrt{x}}=\dfrac{\sqrt{x}-\sqrt{x-1}-\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}+\dfrac{x\sqrt{x}-x}{\sqrt{x}-1}=-2\sqrt{x-1}+x\left(x\ge1\right)\)
Bạn ơi câu a có vẻ có vấn đề ý. Nếu bạn áp dụng HĐT thì phải là√(x-2)2 chứ nhỉ. Mong bạn giải đáp
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1
=\(\frac{\sqrt{x}+1}{x+1}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)
Em thay vào tính nhé!
c) với x>1
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
Áp dụng bất đẳng thức Cosi
A\(\ge2\sqrt{2}+3\)
Xét dấu bằng xảy ra ....
a: \(A=\dfrac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
b: \(B=\left(\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x+\sqrt{x}+1}{x+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\cdot\dfrac{1}{x+\sqrt{x}+1}=\dfrac{-\sqrt{x}+1}{x+\sqrt{x}+1}\)
a: \(P=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
c: Để \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) là số nguyên thì \(\sqrt{x}+1-2⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\{1;2\right\}\)
=>x=0
a) điều kiện xát định \(x\ge0;x\ne1\)
b) \(\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x-1}{x+\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{\left(2\sqrt{x}+x\right)\left(\sqrt{x}-1\right)-\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}:\dfrac{x-1}{x+\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{2x-2\sqrt{x}+x\sqrt{x}-x-x\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{x-2\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}.\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right)^2}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}.\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\Leftrightarrow\dfrac{x+\sqrt{x}+1}{x^2+x\sqrt{x}-\sqrt{x}-1}\)
a/ đkxđ: \(\left\{{}\begin{matrix}x+1\ge0\\x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ne0\end{matrix}\right.\)
b/ đkxđ: \(\dfrac{1}{1-x}>0\Leftrightarrow1-x>0\Leftrightarrow x< 1\)
( vì 1 - x ≠ 0 mà 1 > 0 nên mk cho cả bt > 0 nhé )
c/ đkxđ: \(\dfrac{1}{1-x^2}\ge0\) và 1 - x2 ≠ 0
mà 1 > 0
=> 1 - x2 > 0 \(\Leftrightarrow\left(1-x\right)\left(1+x\right)>0\)
\(\Leftrightarrow-1< x< 1\)
d/ đkxđ: \(\dfrac{2x-4}{1+x^2}\ge0\) mà 1 + x2 > 0 ∀x
=> 2x - 4 ≥ 0
<=> 2x ≥ 4
<=> x ≥ 2
vậy...............
ĐKXĐ: tự làm.
Rút gọn:
\(A=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(x+\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{0+x-\sqrt{x}}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)