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a/ Đkxđ: x\(\ge\)0 x\(\ne\)4
=\(\frac{3\left(\sqrt{x}+2\right)+2\left(\sqrt{x}-2\right)+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{5\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{5}{\sqrt{x}-2}\)
b/ Với x\(\ge\)0 vã\(\ne\)4
Để M\(\in\)Z \(\Leftrightarrow\) \(\frac{5}{\sqrt{x}-2}\in Z\)
\(\Rightarrow\) \(\sqrt{x}-2\inƯ\left(5\right)\)
\(\begin{cases}\sqrt{x}-2=5\\\sqrt{x}-2=-5\\\sqrt{x}-2=1\\\sqrt{x}-2=-1\end{cases}\Rightarrow\begin{cases}x=49\left(tmĐKXĐ\right)\\KhongcogiatriTm\\x=9\left(tmĐKXĐ\right)\\x=1\left(tmĐKXĐ\right)\end{cases}\)
Vậy để M\(\in\)Z thì x=.....
c/ Với...
Để M<2 thì \(\frac{5}{\sqrt{x}-2}< 2\Rightarrow\frac{5-2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}< 0\)
\(\left[\begin{array}{nghiempt}\hept{\begin{cases}9-2\sqrt{x}>0\\\sqrt{x}-2< 0\end{array}\right.\\\hept{\begin{cases}9-2\sqrt{x}< 0\\\sqrt{x}-2>0\end{array}\right.\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}\hept{\begin{cases}x< \frac{81}{4}\\x< 4\end{array}\right.\\\hept{\begin{cases}x>\frac{81}{4}\\x>4\end{array}\right.\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x< 4\\x>\frac{81}{4}\end{array}\right.}\)
a, Mk làm hơi tắt chút bạn thông cảm nha . mk vội ý mà 
\(A=\left(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right).\left(x-3\sqrt{x}+2\right)\)
\(A=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Câu c : \(A\in Z\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}\in Z\Leftrightarrow1-\dfrac{1}{\sqrt{x}}\in Z\)
Để : \(1-\dfrac{1}{\sqrt{x}}\in Z\) thì \(\sqrt{x}\inƯ\left(1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=1\)
Ta có A=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) với x≥ 9, x ∈ R
Để A > 0 \(\Leftrightarrow\) \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) > 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-2< 0\\\sqrt{x}+1>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}< -1\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 2\\\sqrt{x}>-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 4\\x>1\end{matrix}\right.\end{matrix}\right.\)
Kết hợp với ĐKXĐ\(\Rightarrow\) x ∈ ∅
ĐKXĐ: x≥9, x∈R
Ta có:
A= \(\left[\dfrac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right]\)
= \(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\dfrac{1}{1+\sqrt{x}}\):\(\dfrac{1}{\sqrt{x}-2}\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
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) Rut gon H
\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\)
\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}-\dfrac{1}{\sqrt{a}-2}\)
DKXD : \(\left\{{}\begin{matrix}\sqrt{a}+3\ne0\\\sqrt{a}-2\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a\ne9\\a\ne4\end{matrix}\right.\)
Ta co : \(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{a-\sqrt{a}-6}{a+\sqrt{a}-6}\)
a) đk:\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+1\ne0\end{matrix}\right.\\x-1\ne0\\\Rightarrow x\ne1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}y=\sqrt{x}\Rightarrow y>0;y\ne1\\A=\dfrac{\left(y+1\right)^{^2}-\left(y-1\right)^2}{y^2-1}-\dfrac{3y+1}{y^2-1}\end{matrix}\right.\)
\(A=\dfrac{+4y-\left(3y+1\right)}{y^2-1}=\dfrac{y-1}{\left(y-1\right)\left(y+1\right)}=\dfrac{1}{y+1}=\dfrac{1}{\sqrt{x}+1}\)
\(A\left(3-\sqrt{2}\right)=\dfrac{1}{\sqrt{3-\sqrt{2}}+1}\)
c) \(\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{2}\Rightarrow\sqrt{x}=1\Rightarrow x=1\left(loai\right)\) vo nghiệm
d) \(\dfrac{1}{\sqrt{x}+1}< \dfrac{3}{4}\Rightarrow3\sqrt{x}>1\Rightarrow\sqrt{x}>\dfrac{1}{3}\Rightarrow x>\dfrac{1}{9}\)
a) * Đk: \(x\ne\pm1\)
* \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}-1}{x-1} \)
\(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)-\left(3\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x+\sqrt{x}+\sqrt{x}+1-x+\sqrt{x}+\sqrt{x}-1-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{1}{\sqrt{x}-1}\)
a , thu gọn
\(A=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3x+3}{x-9}\right]:\left[\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}\right]\)
\(A=\left(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(A=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(A=\dfrac{-3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(A=-\dfrac{3}{\sqrt{x}+3}\)
b , tự làm
\(a\text{) Để biểu thức xác định }\\ \text{thì }\Rightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3\ne0\\x-9\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
\(\text{b) }A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\\ =\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}\right)\\ =\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ =\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ =\dfrac{-3}{\sqrt{x}+3}\)
\(c\text{) Để }A\le-\dfrac{1}{3}\\ \text{thì }\Rightarrow\dfrac{-3}{\sqrt{x}+3}\le-\dfrac{1}{3}\\ \Rightarrow\dfrac{3}{\sqrt{x}+3}\ge\dfrac{1}{3}\\ \Rightarrow\dfrac{3}{\sqrt{x}+3}-\dfrac{1}{3}\ge0\\ \Rightarrow\dfrac{9}{3\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}\ge0\\ \Rightarrow\dfrac{9-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}\ge0\\ \Rightarrow\dfrac{\sqrt{x}-6}{\sqrt{x}+3}\le0\\ \Leftrightarrow\sqrt{x}-6\ge0\left(\text{Vì }\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}\ge6\\ \Leftrightarrow x\ge36\)
\(d\text{) Do }\sqrt{x}\ge0\\ \Rightarrow\sqrt{x}+3\ge3\\ \Rightarrow\dfrac{-3}{\sqrt{x}+3}\ge-1\\ \text{Dấu }"="\text{ }xảy\text{ }ra\text{ }khi:\text{ }x=0\)
Vậy..............
a: \(A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{x-1}\)
\(=\dfrac{2x-3\sqrt{x}+1}{x-1}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b: Thay \(x=4-2\sqrt{3}\) vào A, ta được:
\(A=\dfrac{2\left(\sqrt{3}-1\right)-1}{\sqrt{3}-1+1}=\dfrac{2\sqrt{3}-3}{\sqrt{3}}=2-\sqrt{3}\)
c: Để A=1/2 thì \(4\sqrt{x}-2=\sqrt{x}+1\)
=>x=1(loại)
1) +) ta có : \(C-\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}=\dfrac{3\sqrt{x}-x+\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{-\left(x-4\sqrt{x}+4\right)+3}{3\left(x+\sqrt{x}+1\right)}=\dfrac{-\left(\sqrt{x}-2\right)^2+3}{3\left(x+\sqrt{x}+1\right)}\)
không thể cm được đâu bn --> xem lại đề
2) +) ta có : \(D=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)
--> để \(D\in Z\Leftrightarrow\sqrt{x}+2\) là ước của 3 \(\Leftrightarrow\sqrt{x}+2\in\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow x=1\) vậy \(x=1\)
3) +) tương tự 2)
4) a) +) điều kiện xác định : \(x>0;x\ne4\)
ta có : \(A=\left(\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}}\right):\dfrac{\sqrt{x}-2}{x+3\sqrt{x}}\)
\(\Leftrightarrow A=\left(\dfrac{2\sqrt{x}-\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}}{\sqrt{x}-2}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)
b) ta có : \(A=3\Leftrightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=3\Leftrightarrow\sqrt{x}-3=3\sqrt{x}-6\)
\(\Leftrightarrow2\sqrt{x}=3\Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\) vậy \(x=\dfrac{9}{4}\)
c) ta có : \(B=A.\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}.\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{x-9}{x-4}=1-\dfrac{5}{x-4}\)
tương tự 2 )
\(\)
a: ĐKXĐ: a>0; a<>1
\(M=\dfrac{\sqrt{a}-\sqrt{a}+1+a-1}{a-\sqrt{a}}=\dfrac{a}{a-\sqrt{a}}=\dfrac{\sqrt{a}}{\sqrt{a}-1}\)
b: Để M là số nguyên thì \(\sqrt{a}-1+1⋮\sqrt{a}-1\)
\(\Leftrightarrow\sqrt{a}-1\in\left\{1;-1\right\}\)
=>\(a=4\)
c: Để M<1 thì M-1<0
=>\(\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}-1}< 0\)
=>căn a-1<0
=>0<a<1
d: Thay a=3+2 căn 2 vào M, ta được:
\(M=\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{2+\sqrt{2}}{2}\)