Cho biểu thức Q=\(\dfrac{\sqrt{x}+4}{1-7\sqrt{x}}\)+\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)+\(\dfrac{24\sqrt{x}}{7x+6\sqrt{x}-1}\)
1, Tìm đk và rút gọn
2,Tìm x sao cho Q > -6
3,Tìm x thuộc Z sao cho Q thuộc Z
Cho biểu thức: Q = \(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-3\dfrac{\sqrt{x}-1}{x-5\sqrt{x}+6}\).
a) Tìm điều kiện xác định và rút gọn Q.
b) Tìm các giá trị của x để Q < -1.
c) Tìm các giá trị của x \(\in\) Z sao cho 2Q \(\in\) Z.
a, đk: \(x\ge0,x\ne9,x\ne4\)
\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4-x+3\sqrt{x}-\sqrt{x}+3-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2-\sqrt{x}}{-\left(\sqrt{x}-3\right)\left(2-\sqrt{x}\right)}=\dfrac{-1}{\sqrt{x}-3}\)
b,\(Q< -1=>\dfrac{-1}{\sqrt{x}-3}+1< 0< =>\dfrac{-1+\sqrt{x}-3}{\sqrt{x}-3}< 0\)
\(< =>\dfrac{\sqrt{x}-4}{\sqrt{x}-3}< 0\)
\(=>\left\{{}\begin{matrix}\left[{}\begin{matrix}\sqrt{x}-4>0\\\sqrt{x}-3< 0\end{matrix}\right.\\\left[{}\begin{matrix}\sqrt{x}-4< 0\\\sqrt{x}-3>0\end{matrix}\right.\end{matrix}\right.\)\(< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>16\\x< 9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\end{matrix}\right.\)\(< =>9< x< 16\)
c, \(=>2Q=\dfrac{-2}{\sqrt{x}-3}=1+\dfrac{1}{\sqrt{x}-3}\in Z\)
\(< =>\sqrt{x}-3\inƯ\left(1\right)=\left\{\pm1\right\}\)\(=>x\in\left\{16;4\right\}\)(loại 4)
=>x=16
a) \(Q=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-3\dfrac{\sqrt{x}-1}{x-5\sqrt{x}+6}\)
Ta có \(x-5\sqrt{x}+6=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3>0\\\sqrt{x}-2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>9\\x>2\end{matrix}\right.\) \(\Leftrightarrow x>9\)
\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-3\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(x-4\right)-\left(x-2\sqrt{x}-3\right)-\left(3\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\) \(=\dfrac{-\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\) \(=\dfrac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\) \(=\dfrac{-1}{\left(\sqrt{x}-3\right)}=\dfrac{1}{3-\sqrt{x}}\)
b) \(Q< -1\Leftrightarrow\dfrac{1}{3-\sqrt{x}}< -1\) \(\Leftrightarrow\dfrac{1}{3-\sqrt{x}}+1< 0\) \(\Leftrightarrow\dfrac{4-\sqrt{x}}{3-\sqrt{x}}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4-\sqrt{x}>0\\3-\sqrt{x}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}4-\sqrt{x}< 0\\3-\sqrt{x}>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x>16\\x< 9\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow9< x< 16\)
Vậy để \(Q< -1\) thì \(S=\left\{x/9< x< 16\right\}\)
c) \(2Q\in Z\Leftrightarrow\dfrac{2}{3-\sqrt{x}}\in Z\)
\(\Rightarrow3-\sqrt{x}\inƯ\left(2\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}3-\sqrt{x}=2\\3-\sqrt{x}=-2\\3-\sqrt{x}=1\\3-\sqrt{x}=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=25\\x=4\\x=16\end{matrix}\right.\)
Kết hợp với ĐKXĐ,ta có để \(2Q\in Z\) thì \(x\in\left\{16;25\right\}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)
Ta có: \(Q=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{3\sqrt{x}-3}{x-5\sqrt{x}+6}\)
\(=\dfrac{x-4-x+2\sqrt{x}+2-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-1}{\sqrt{x}-3}\)
c) Để 2Q là số nguyên thì \(-2⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1\right\}\)
\(\Leftrightarrow x\in\left\{16;25;1\right\}\)
Câu 6: Cho biểu thức Q = \(\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)\)
a) Tìm x để biểu thức Q xác định và rút gọn biểu thức Q
b) Tìm các giá trị của x để biểu thức Q có giá trị âm
a) ĐKXĐ: \(x>0;x\ne4\)
\(Q=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)
b) Để biểu thức \(Q\) có giá trị âm thì \(\dfrac{3\sqrt{x}}{\sqrt{x}-2}< 0\)
\(\Rightarrow\sqrt{x}-2< 0\) (vì \(3\sqrt{x}>0\forall x>0;x\ne4\))
\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\)
Kết hợp với điều kiện xác định của \(x\), ta được: \(0< x< 4\)
\(\text{#}\mathit{Toru}\)
Cho biểu thức sau: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
1, Rút gọn P
2, Tính giá trị nhỏ nhất của P
3, Tìm \(x\in Z\) sao cho \(Q=\dfrac{2\sqrt{x}}{P}\in Z\)
1: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
Cho biểu thức sau:
\(P=\frac{\sqrt{x}+4}{1-7\sqrt{x}}+\frac{\sqrt{x}-2}{\sqrt{x}+1}+\frac{24\sqrt{x}}{7x+6\sqrt{x}-1}\)
1, Tìm điều keienj để P có nghĩa và rút gọn.
2. Tìm x sao cho \(P\ge-6\)
ta có :
\(P=\frac{\sqrt{x}+4}{1-7\sqrt{x}}+\frac{\sqrt{x}-2}{\sqrt{x}+1}+\frac{24\sqrt{x}}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}\)
\(\frac{-\left(\sqrt{x}+4\right)\left(\sqrt{x}+1\right)+\left(\sqrt{x}-2\right)\left(7\sqrt{x}-1\right)+24\sqrt{x}}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}=\frac{6x+4\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}\)
\(=\frac{6\sqrt{x}+2}{7\sqrt{x}-1}\)
Để \(P\ge-6\Leftrightarrow\frac{6\sqrt{x}+2}{7\sqrt{x}-1}\ge-6\Leftrightarrow\frac{48\sqrt{x}-4}{7\sqrt{x}-1}\ge0\)
\(\Leftrightarrow\orbr{\begin{cases}0\le\sqrt{x}\le\frac{1}{12}\\\sqrt{x}>\frac{1}{7}\end{cases}}\Leftrightarrow\orbr{\begin{cases}0\le x\le\frac{1}{144}\\x>\frac{1}{49}\end{cases}}\)
Cho biểu thức P= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)+\(\dfrac{3}{\sqrt{x}+1}\)-\(\dfrac{6\sqrt{x}-4}{x-1}\) Với x >=0 , x khác 1
a) Rút gọn biểu thức ( câu này mình rút gọn = \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\))
b) Tìm giá trị của x để P =-1
c) Tìm x thuộc z để P thuộc z
d) Só ánh P với 1
e)Tìm giá trị nhỏ nhất của P
mình đag cần gấp ạ!
a) \(P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-1\)
\(\Leftrightarrow-\sqrt{x}-1=\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)
c) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Kết hợp đk:
\(\Rightarrow x\in\left\{0\right\}\)
d) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\)
\(a,P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,P=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\ \Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\\ c,P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}\in Z\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\in Z\\ \Leftrightarrow2⋮\sqrt{x}+1\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}+1\in\left\{1;2\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}\in\left\{0;1\right\}\\ \Leftrightarrow x\in\left\{0;1\right\}\)
\(d,P=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
Có \(\dfrac{2}{\sqrt{x}+1}>0\left(2>0;\sqrt{x}+1>0\right)\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}< 1\Leftrightarrow P< 1\)
\(e,P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
Có \(\sqrt{x}+1\ge1\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\le2\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)
\(P_{min}=-1\Leftrightarrow x=0\)
Cho biểu thức Q = \(\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
a) rút gọn Q
b) Tính giá trị của Q khi x = \(4+2\sqrt{3}\)
c) Tìm các giá trị của x để Q = 3
d) Tìm các giá trị cảu x để Q > \(\dfrac{1}{2}\)
e) Tìm x \(\in\) Z để Q = Z
a: \(Q=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
b: Khi x=4+2căn 3 thì \(Q=\dfrac{\sqrt{3}+1-2}{\sqrt{3}+1+2}=\dfrac{-3+2\sqrt{3}}{3}\)
c: Q=3
=>3căn x+6=căn x-2
=>2căn x=-8(loại)
d: Q>1/2
=>Q-1/2>0
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{1}{2}>0\)
=>2căn x-4-căn x-2>0
=>căn x>6
=>x>36
d: Q nguyên
=>căn x+2-4 chia hết cho căn x+2
=>căn x+2 thuộc Ư(-4)
=>căn x+2 thuộc {2;4}
=>x=0 hoặc x=4(nhận)
Bài 4:
Cho biểu thức \(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
a) Tìm đkxđ của M và rút gọn
b) Tìm x thuộc Z để M thuộc Z
Em cần gấp :<
a) ĐKXĐ: \(x\ge0;x\ne9;x\ne4\)
\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(M=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(M=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) Ta có M ϵ Z thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3}{\sqrt{x}-3}+\dfrac{4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)
Phải thuộc Z vậy:
4 ⋮ \(\sqrt{x}-3\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Mà: \(x\ge0,x\ne4,x\ne9\) nên \(\sqrt{x}-3\in\left\{1;2;-2;4\right\}\)
\(\Rightarrow x\in\left\{16;25;1;49\right\}\)
Cho \(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
a, Rút gọn Q
b, Tính Q biết \(x=6+4\sqrt{2}\)
c, Tìm xϵZ để QϵZ
a) \(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4,9\right)\)
\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) \(\sqrt{x}=\sqrt{6+4\sqrt{2}}=\sqrt{\left(2+\sqrt{2}\right)^2}=2+\sqrt{2}\)
\(\Rightarrow Q=\dfrac{2+\sqrt{2}+1}{2+\sqrt{2}-3}=\dfrac{3+\sqrt{2}}{\sqrt{2}-1}=\dfrac{\left(3+\sqrt{2}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(=4\sqrt{2}+5\)
c) \(Q=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)
Để \(Q\in Z\Rightarrow4⋮\sqrt{x}-3\Rightarrow\sqrt{x}-3\in\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{4;5;7;2;1\right\}\Rightarrow x\in\left\{16;25;49;4;1\right\}\)
a) Ta có: \(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) Thay \(x=6+4\sqrt{2}\) vào Q, ta được:
\(Q=\dfrac{2+\sqrt{2}+1}{2+\sqrt{2}-3}=\dfrac{3+\sqrt{3}}{\sqrt{2}-1}=\left(3+\sqrt{3}\right)\left(\sqrt{2}+1\right)\)
cho biểu thức Q=\(\left(\dfrac{1}{\sqrt{X}-1}-\dfrac{1}{\sqrt{X}}\right):\left(\dfrac{\sqrt{X}+1}{\sqrt{X}-2}-\dfrac{\sqrt{X}+2}{\sqrt{X-1}}\right)\)
a rút gọn Q
b tìm x để Q>0
b) Q > 0
⇔ \(\dfrac{\sqrt{\text{x}}-2}{3\sqrt{\text{x}}}\) > 0
Do \(\text{3}\sqrt{\text{x}}>0\) ∀x⩾0
⇒ \(\sqrt{\text{x}}-2>0\)
⇔ \(\sqrt{\text{x}}>2\)
⇔ x > 4
Vậy x > 4 thì Q > 0