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S
15 tháng 10 2019
\(\sqrt{9x-9}+1=13\Leftrightarrow3\sqrt{x-1}=12\Leftrightarrow\sqrt{x-1}=4\Leftrightarrow x-1=16\Leftrightarrow x=17\)
\(2.\text{bạn tự tìm đk}\)
\(A=\left(\frac{2}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\frac{2\sqrt{x}-2}{\sqrt{x}-1}\right)\)
\(A=\frac{2\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-2\right)=\sqrt{x}\left(\sqrt{x}-2\right)< 0\Leftrightarrow x-2\sqrt{x}< 0\Leftrightarrow\left(\sqrt{x}-1\right)^2< 1\Leftrightarrow-1< \sqrt{x}-1< 1\)
\(\Leftrightarrow0< x< 4\)
15 tháng 10 2019
Câu 1:
\(\sqrt{9x-9}+1=13\)\(ĐKXĐ:x\ge1\)
\(\Leftrightarrow\sqrt{9\left(x-1\right)}=12\)
\(\Leftrightarrow3\sqrt{x-1}=12\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
\(\Leftrightarrow x=17\)(tm ĐKXĐ)
Câu 2
ĐKXĐ: \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(A=\left(\frac{2}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{x-\sqrt{x}}\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}-\frac{2\sqrt{x}-2}{\sqrt{x}-1}\right)\)
\(=\left(\frac{2}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(=\left(\frac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\sqrt{x}-2\right)\)
\(=\left(\frac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\frac{1}{\sqrt{x}-2}\)
\(=\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\frac{1}{\sqrt{x}-2}\)
\(=\frac{1}{x-2\sqrt{x}}\)
b Để A có giá trị âm \(\Rightarrow\frac{1}{x-2\sqrt{x}}< 0\)
vì 1>0
\(\Rightarrow x-2\sqrt{x}< 0\)
\(\Leftrightarrow0< \sqrt{x}< 2\)
\(\Leftrightarrow0< x< 4\)
kết hợp ĐKXĐ: \(\Rightarrow1< x< 4\)
17 tháng 10 2020
ĐKXĐ : x > 0 ; x ≠ 1 ; x ≠ 4
a) \(A=\left(1-\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{1}{\sqrt{x-1}}\right)\div\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\left(\frac{x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\left(\frac{x-1-4\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}-3}{\sqrt{x}-2}\)
b) Với x = \(11-6\sqrt{2}\)
\(A=\frac{\sqrt{11-6\sqrt{2}}-3}{\sqrt{11-6\sqrt{2}}-2}\)
\(=\frac{\sqrt{2-6\sqrt{2}+9}-3}{\sqrt{2-6\sqrt{2}+9}-2}\)
\(=\frac{\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot3+3^2}-3}{\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot3+3^2}-2}\)
\(=\frac{\sqrt{\left(\sqrt{2}-3\right)^2}-3}{\sqrt{\left(\sqrt{2}-3\right)^2}-2}\)
\(=\frac{\left|\sqrt{2}-3\right|-3}{\left|\sqrt{2}-3\right|-2}\)
\(=\frac{3-\sqrt{2}-3}{3-\sqrt{2}-2}=\frac{-\sqrt{2}}{1-\sqrt{2}}\)
c) Ta có : \(A=\frac{\sqrt{x}-3}{\sqrt{x}-2}=\frac{\sqrt{x}-2-1}{\sqrt{x}-2}=1-\frac{1}{\sqrt{x}-2}\)
Để A nguyên => \(\frac{1}{\sqrt{x}-2}\)nguyên
=> \(1⋮\sqrt{x}-2\)
=> \(\sqrt{x}-2\inƯ\left(1\right)=\left\{\pm1\right\}\)
=> \(\sqrt{x}\in\left\{3;1\right\}\)
=> \(x=9\)( không nhận x = 1 do ĐKXĐ )
d) Để A = -2
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}=-2\)( x > 0 ; x ≠ 1 ; x ≠ 4 )
=> \(\sqrt{x}-3=-2\sqrt{x}+4\)
=> \(\sqrt{x}+2\sqrt{x}=4+3\)
=> \(3\sqrt{x}=7\)
=> \(9x=49\)( bình phương hai vế )
=> \(x=\frac{49}{9}\)( tm )
e) Để A có giá trị âm
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}< 0\)
Xét hai trường hợp :
1.\(\hept{\begin{cases}\sqrt{x}-3>0\\\sqrt{x}-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}>3\\\sqrt{x}< 2\end{cases}}\Leftrightarrow\hept{\begin{cases}x>9\\x< 4\end{cases}}\)( loại )
2. \(\hept{\begin{cases}\sqrt{x}-3< 0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}< 3\\\sqrt{x}>2\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 9\\x>4\end{cases}}\Leftrightarrow4< x< 9\)
Vậy với 4 < x < 9 thì A có giá trị âm
f) Để A < -2
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}< -2\)
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}+2< 0\)
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}+\frac{2\sqrt{x}-4}{\sqrt{x-2}}< 0\)
=> \(\frac{3\sqrt{x}-7}{\sqrt{x}-2}< 0\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}3\sqrt{x}-7< 0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}3\sqrt{x}< 7\\\sqrt{x}>2\end{cases}}\Leftrightarrow\hept{\begin{cases}9x< 49\\x>4\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{49}{9}\\x>4\end{cases}}\Leftrightarrow4< x< \frac{49}{9}\)
2. \(\hept{\begin{cases}3\sqrt{x}-7>0\\\sqrt{x}-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}3\sqrt{x}>7\\\sqrt{x}< 2\end{cases}}\Leftrightarrow\hept{\begin{cases}9x>49\\x< 4\end{cases}}\Leftrightarrow\hept{\begin{cases}x>\frac{49}{9}\\x< 4\end{cases}}\)( loại )
Vậy với 4 < x < 49/9 thì A < -2
g) Để \(A>\sqrt{x}-1\)
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}>\sqrt{x}-1\)
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}-\left(\sqrt{x}-1\right)>0\)
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-2}>0\)
=> \(\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{x-3\sqrt{x}+2}{\sqrt{x}-2}>0\)
=> \(\frac{-x+4\sqrt{x}-5}{\sqrt{x}-2}>0\)
Ta có : \(-x+4\sqrt{x}-5=-\left(x-4\sqrt{x}+4\right)-1=-\left(\sqrt{x}-2\right)^2-1\le-1< 0\left(\forall\ge0\right)\)
Nên để A > 0 thì ta chỉ cần xét \(\sqrt{x}-2< 0\)
\(\sqrt{x}-2< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ => \(\hept{\begin{cases}0< x< 4\\x\ne1\end{cases}}\)thì tm
13 tháng 7 2017
a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne2\end{cases}}\)
\(P=\left(1-\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\right).\left(1+\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\right)\)
\(=\left(1-\sqrt{x}\right).\left(1+\sqrt{x}\right)=1-x\)
b. \(P\ge0\Rightarrow1-x\ge0\Rightarrow x\le1\)
Vậy với \(x\le1\)thì P có giá trị không âm
NV
14 tháng 7 2016
1/
a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)
b/ \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)
\(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)
Vậy x = 9/25 , x = 4
14 tháng 7 2016
1) a) ĐKXĐ : \(0\le x\ne\frac{1}{9}\)
b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)
Đk : x > 0
Để A < 0 thì \(\sqrt{x}.\left(\sqrt{x}-2\right)\)< 0 => \(\sqrt{x}-2\)< 0 ( vì \(\sqrt{x}>0\))
=> \(\sqrt{x}\)< 2 => x < 4 => 0 < x < 4 ( kết hợp đk)
k mk nha