Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
a: ĐKXĐ: x>=0; x<>1
\(B=\dfrac{\sqrt{x}\left(1-x\right)^2}{x+1}:\left[\left(x-2\sqrt{x}+1\right)\left(x+2\sqrt{x}+1\right)\right]\)
\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{x+1}\cdot\dfrac{1}{\left(x-1\right)^2}=\dfrac{\sqrt{x}}{x+1}\)
b: Để B=2/5 thì \(\dfrac{\sqrt{x}}{x+1}=\dfrac{2}{5}\)
\(\Leftrightarrow2x-5\sqrt{x}+2=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
=>x=1/4 hoặc x=4
c: Thay \(x=12-6\sqrt{3}=\left(3-\sqrt{3}\right)^2\) vào A, ta được:
\(A=\dfrac{3-\sqrt{3}}{12-6\sqrt{3}+1}=\dfrac{3-\sqrt{3}}{13-6\sqrt{3}}=\dfrac{21+5\sqrt{3}}{61}\)
a) D (ĐKXĐ: x\(\ge0,x\ne1\))
=\(\left(\dfrac{2x-\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(1-\sqrt{x}+x-\sqrt{x}\right)\)
=\(\dfrac{2x-x\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)^2\)
\(=\dfrac{\left(x-x\sqrt{x}-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)}\)
=\(\dfrac{\sqrt{x}\left(\sqrt{x}-x-1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)}\)
=\(-\sqrt{x}\left(\sqrt{x}-1\right)=\sqrt{x}-x\)
b) \(\sqrt{x}-x=3\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)=3\)
=\(\sqrt{x}-x-3=0\Leftrightarrow\left(x-2.\dfrac{1}{2}x+\dfrac{1}{4}\right)-\dfrac{13}{4}=0\)
\(\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{13}{4}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-\dfrac{1}{2}=\dfrac{13}{4}\\\sqrt{x}-\dfrac{1}{2}=-\dfrac{13}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{225}{16}\\x=\dfrac{121}{16}\end{matrix}\right.\)
mk nghỉ ở giữa 2 ngoặc là dấu chia mới đúng chứ :
đk : \(x\ge0;x\ne9\)
\(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\right)\)
\(D=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-\left(3x+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(D=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(D=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(D=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
Bài 1:
a: \(A=\dfrac{\sqrt{x}+2}{2\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4+x-4\sqrt{x}+4}{2\left(x-4\right)}\)
\(=\dfrac{2x+8}{2\left(x-4\right)}=\dfrac{x+4}{x-4}\)
b: Để A=8 thì x+4=8(x-4)
=>x+4=8x-32
=>-7x=-36
hay x=36/7(nhận)
Bài 1:
\(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)
\(=\left(\dfrac{x}{\left(x-7\right)\left(x+7\right)}-\dfrac{x-7}{x\cdot\left(x+7\right)}\right)\cdot\dfrac{x^2+7x}{2x-7}+\dfrac{x}{-\left(x-7\right)}\)
\(=\dfrac{x^2-\left(x-7\right)^2}{x\cdot\left(x-7\right)\left(x+7\right)}\cdot\dfrac{x\cdot\left(x+7\right)}{2x-7}-\dfrac{x}{x-7}\)
\(=\dfrac{\left(x-\left(x-7\right)\right)\cdot\left(x+x-7\right)}{x-7}\cdot\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)
\(=\dfrac{\left(x-x+7\right)\cdot\left(2x-7\right)}{x-7}\cdot\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)
\(=\dfrac{7}{x-7}-\dfrac{x}{x-7}\)
\(=\dfrac{7-x}{x-7}\)
\(=\dfrac{-\left(x-7\right)}{x-7}\)
\(=-1\)
A = \(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)
A = \(\left(\dfrac{x}{\left(x+7\right)\left(x-7\right)}-\dfrac{x-7}{x\left(x+7\right)}\right):\dfrac{2x-7}{x\left(x+7\right)}+\dfrac{x}{7-x}\)
A = \(\left(\dfrac{x^2-\left(x-7\right)^2}{\left(x+7\right)\left(x-7\right)x}\right):\dfrac{2x-7}{x\left(x+7\right)}-\dfrac{x}{x-7}\)
A = \(\left(\dfrac{x^2-\left(x^2-14x+49\right)}{\left(x+7\right)\left(x-7\right)x}\right):\dfrac{\left(2x-7\right)\left(x-7\right)-\left(x^3+7x^2\right)}{\left(x+7\right)\left(x-7\right)x}\)
A = \(\dfrac{14x-49}{\left(x+7\right)\left(x-7\right)x}:\dfrac{-x^3-5x^2-21x+49}{\left(x+7\right)\left(x-7\right)x}\)
A = \(\dfrac{14x-49}{\left(x+7\right)\left(x-7\right)x}.\dfrac{\left(x+7\right)\left(x-7\right)x}{-x^3-5x^2-21x+49}\)
A = \(\dfrac{14x-49}{-x^3-5x^2-21x+49}\)
b: \(P=\left(\sqrt{x}+\sqrt{x-1}-\left(\sqrt{x-1}+\sqrt{2}\right)\right)\cdot\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
\(=-\dfrac{\sqrt{x}-\sqrt{2}}{\sqrt{x}}\)
c: Thay \(x=3-2\sqrt{2}\) vào P, ta được:
\(P=-\dfrac{\sqrt{2}-1-\sqrt{2}}{\sqrt{2}-1}=\dfrac{1}{\sqrt{2}-1}=\sqrt{2}+1\)
a, ĐKXĐ: \(x\ge0;x\ne9\)
b, rút gọn
A=\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x}{x-9}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}-3}-1\right)\)
\(=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}+1}{x-3}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{x+1}\\ =\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ =\dfrac{-3}{\sqrt{x}+3}\)
c,Cho \(A\le-\dfrac{1}{3}\)
\(< =>\dfrac{3}{\sqrt{x}+3}\le-\dfrac{1}{3}\\ < =>\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{3}\le0\\ < =>\dfrac{-9+\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}\le0\\ < =>\dfrac{\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}\le0\\ < =>\sqrt{x}-6\le0\\ < =>\sqrt{x}\le36\\ < =>0\le x\le36\)
Vậy để \(A\le-\dfrac{1}{3}\) thì \(0\le x\le36\)và\(x\ne9\)
d, \(A=\dfrac{-3}{\sqrt{x}+3}\)
Ta có: \(\sqrt{x}+3\ge3\\ =>\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\\ =>\dfrac{-3}{\sqrt{x}+3}\ge\dfrac{-3}{3}\\ =-1\)
Vậy GTNN của A=-1
Xấu ''='' xảy ra khi \(\sqrt{x}=0\\ \Leftrightarrow x=0\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne1\\x>0\end{matrix}\right.\)
b)
\(D=\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\dfrac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(1-\sqrt{x}+x-\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)^2\)
\(=\sqrt{x}-1\)
c)
Giả sử \(D>\dfrac{-2}{\sqrt{x}}\)
\(\Rightarrow\sqrt{x}-1>-\dfrac{2}{\sqrt{x}}\Leftrightarrow\sqrt{x}-1+\dfrac{2}{\sqrt{x}}>0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right)\sqrt{x}+2}{\sqrt{x}}>0\Leftrightarrow x-\sqrt{x}+2>0\Leftrightarrow\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{7}{4}>0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\)(luôn đúng)