Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Ta có: \(A=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}\)
\(=2\sqrt{x}-1\)
\(A=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2x\sqrt{x^2-1}}\\ A=\sqrt{\left(\sqrt{x^2-1}+1\right)^2}-\sqrt{\left(\sqrt{x^2-1}-1\right)^2}\\ A=\left|\sqrt{x^2-1}+1\right|-\left|\sqrt{x^2-1}-1\right|\)
\(a,\) A có nghĩa \(\Leftrightarrow x^2-1\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)
\(b,x\ge\sqrt{2}\Leftrightarrow\sqrt{x^2-1}-1\ge\sqrt{\left(\sqrt{2}\right)^2-1}-1=0\\ \Rightarrow A=\sqrt{x^2-1}+1-\left(\sqrt{x^2-1}-1\right)=2\)
a: ĐKXĐ: x>0; x<>1
b: \(A=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2}{x-1}\)
c: A nguyên
=>x-1 thuộc {1;-1;2;-2}
=>x thuộc {2;3}
a: ĐKXĐ: x^2-1>=0
=>x>=1 hoặc x<=-1
\(A=\sqrt{x^2-1+2\sqrt{x^2-1}+1}-\sqrt{x^2-1-2\sqrt{x^2-1}+1}\)
\(=\left|\sqrt{x^2-1}+1\right|-\left|\sqrt{x^2-1}-1\right|\)
x>=căn 2
=>x^2>=2
=>x^2-1>=1
=>căn x^2-1>=1
=>căn(x^2-1)-1>=0
=>\(A=\sqrt{x^2-1}+1-\sqrt{x^2+1}+1=2\)
`a)ĐK:{(x>=0),(sqrtx-1ne0):}`
`<=>{(x>=0),(sqrtxne1):}`
`<=>{(x>=0),(x ne 1):}`
`b)A=(x+1-2sqrtx)/(sqrtx-1)+(x+sqrtx)/(sqrtx+1)`
`=(sqrtx-1)^2/(sqrtx-1)+(sqrtx(sqrtx+1))/(sqrtx+1)`
`=sqrtx-1+sqrtx=2sqrtx-1`
`c)A<-1`
`<=>2sqrtx-1<-1`
`<=>2sqrtx<0`
`<=>sqrtx<0` vô lý vì `sqrtx>=0`
`a)ĐK:` \(\begin{cases}x \ge 0\\x-\sqrt{x} \ne 0\\x-1 \ne 0\\\end{cases}\)
`<=>` \(\begin{cases}x \ge 0\\x \ne 0\\x \ne 1\\\end{cases}\)
`<=>` \(\begin{cases}x>0\\x \ne 1\\\end{cases}\)
`b)A=(sqrtx/(sqrtx-1)-1/(x-sqrtx)):(1/(1+sqrtx)+2/(x-1))`
`=((x-1)/(x-sqrtx)):((sqrtx-1+2)/(x-1))`
`=(x-1)/(x-sqrtx):(sqrtx+1)/(x-1)`
`=(sqrtx+1)/sqrtx:1/(sqrtx-1)`
`=(x-1)/sqrtx`
`c)A>0`
Mà `sqrtx>0AAx>0`
`<=>x-1>0<=>x>1`
a, ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
b, Ta có : \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{1}{\sqrt{x}-1}=\dfrac{x-1}{\sqrt{x}}\)
c, Ta có : \(A>0\)
\(\Leftrightarrow x-1>0\)
\(\Leftrightarrow x>1\)
Vậy ...
a) ĐKXĐ: \(x\ge0,x\ne1\)
b) \(A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
c) \(A=2\sqrt{x}-1< -1\Leftrightarrow2\sqrt{x}< 0\)(vô lý do \(2\sqrt{x}\ge0\forall x\))
Vậy \(S=\varnothing\)
Bài 3:
\(A=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt[]{x}+1}\\ DKXD:x\ne1;x\ge0\\ A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\\ A=\sqrt{x}-1+\sqrt{x}\\ A=2\sqrt{x}+1\)
\(C.A< -1\Leftrightarrow2\sqrt{x}-1< -1\\ \Leftrightarrow2\sqrt{x}< 0\\ \Leftrightarrow x< 0\left(ktmdk\right)\\ =>BPTVN:S=\varnothing\)
a)a>=căn2
b) căn((x^2-1)+2căn(x^2-1)+1)-căn((x^2-1)-2căn(x^2-1)+1)=... tự lm tiếp