Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)ĐKXĐ:x>=0;x khác 9
A=[\(\frac{\sqrt{x}}{\sqrt{x}-3}\) - \(\frac{3\sqrt{x}+9}{x-9}\)+ \(\frac{2\sqrt{x}}{\sqrt{x}+3}\)] \(\div\) [\(\frac{2\sqrt{x}-2}{\sqrt{x}-3}\)-1]
A=[\(\frac{\sqrt{x}\left(\sqrt{x}-3\right)-3\sqrt{x}-9+2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}\)] \(\div\) [\(\frac{\left(2\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-x+9}{x-9}\)]
A=[\(\frac{3x-12\sqrt{x}-9}{x-9}\)].[\(\frac{x-9}{x-4\sqrt{x}+3}\)]
A=\(\frac{3x-12\sqrt{x}-9}{x-4\sqrt{x}+3}\)
a: ĐKXĐ: x>=0; x<>1
\(A=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b: \(A-\dfrac{2}{3}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{2}{3}\)
\(=\dfrac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}=\dfrac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}< =0\)
Do đó: A<=2/3
ĐK:\(x>0\)
\(C=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}=\frac{\sqrt{x}.\left[\left(\sqrt{x}\right)^3+1\right]}{x-\sqrt{x}+1}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-2\sqrt{x}-1\)
\(=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\)
ĐKXĐ: x>=1 và x<>2
\(A=\dfrac{\sqrt{x-1}+\left|\sqrt{x-1}-1\right|+1}{\left|x-2\right|}\)
Trường hợp 1: \(\sqrt{x-1}>1\Leftrightarrow x>2\)
=>\(A=\dfrac{2\sqrt{x-1}}{\left|x-2\right|}\)
Trường hợp 2: 1<x<2
\(A=\dfrac{2}{\left|x-2\right|}\)
a) ĐKXĐ \(\Leftrightarrow\)\(\begin{cases}\sqrt{a}\ge0\\\sqrt{b}\ge0\\\sqrt{ab}\ge0\\a\ne0\end{cases}\)
\(\Leftrightarrow\)\(\begin{cases}a\ge0\\b\ge0\\a\ne0\end{cases}\)
\(\Leftrightarrow\)\(\begin{cases}a>0\\b\ge0\end{cases}\)
b)\(A=\frac{\sqrt{b}}{\sqrt{a}}-\frac{\sqrt{ab}-\left|a\right|}{a}=\frac{\sqrt{ab}}{a}-\frac{\sqrt{ab}-a}{a}=\frac{\sqrt{ab}-\sqrt{ab}+a}{a}=\frac{a}{a}=1\)
ở câu a, ở dấu tương đương là a>0 và b lớn hơn hoặc = 0 ak??? ở câu b là \(\frac{\sqrt{ab}-\left|a\right|}{a}\) có nghĩa là sao???
a: \(A=\left(x^2+x+1-x\right):\dfrac{1-x^2}{\left(1-x\right)-x^2\left(1-x\right)}\)
\(=\left(x^2+1\right)\cdot\left(1-x\right)\)
b: Để A<0 thì 1-x<0
=>x>1
c: |x-4|=5
=>x-4=5 hoặc x-4=-5
=>x=9(nhận) hoặc x=-1(loại)
Thay x=9 vào A, ta được:
\(A=\left(9^2+1\right)\left(1-9\right)=82\cdot\left(-8\right)=-656\)
\(A=\left(\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right)\left(ĐK:x\ge0;\ne1\right)\)
\(=\left[\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}+2}\right]\)
\(=\frac{3\left(\sqrt{x}+1\right)-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{2\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\)
\(=\frac{2\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}=\frac{2\left(\sqrt{x}+3\right)}{\sqrt{x}+1}\)
a) ĐKXĐ: \(x\ge0;x\ne9\)
\(B=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)
\(B=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{x-9}\)
\(B=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-3+11\sqrt{x}}{x-9}\)
\(B=\frac{2x-6+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)
\(B=\frac{3x-6+15\sqrt{x}}{x-9}\)