tìm x để
\(x^2-5\sqrt{x}+6\ne0\)
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12 tháng 9 2018
\(B=\frac{2+\sqrt{x}}{x-4\sqrt{x}+4}:\left(\frac{\sqrt{x}+2}{\sqrt{x}}+\frac{1}{\sqrt{x}-2}+\frac{6-x}{x+2\sqrt{x}}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{\sqrt{x}+2}{\sqrt{x}}+\frac{1}{\sqrt{x}-2}+\frac{6-x}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)+\left(6-x\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{x\sqrt{x}-8+x+2\sqrt{x}+6\sqrt{x}-12-x\sqrt{x}+2x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{3x+8\sqrt{x}-20}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{\sqrt{x}\left(2+\sqrt{x}\right)^2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)^2\left(3x+8\sqrt{x}-20\right)}\)
\(B=\frac{\sqrt{x}\left(2+\sqrt{x}\right)^2}{\left(\sqrt{x}-2\right)\left(3x+8\sqrt{x}-20\right)}\)
tới đây mình bí rồi cậu làm giúp mình đi
mại dzo
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 7 2021
\(A=\left(\dfrac{\sqrt{x}+1+x+\sqrt{x}}{\sqrt{x}+1}\right)\left(\dfrac{\sqrt{x}-1-x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}+1}{\sqrt{x}+1}\right)\left(\dfrac{-\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\right)\left(\dfrac{-\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\right)\)
\(=-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)=1-x\)
\(A=-1\Leftrightarrow1-x=-1\Rightarrow x=2\)
11 tháng 7 2021
a) Ta có: \(A=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)
=1-x
b) Để A=-1 thì 1-x=-1
hay x=2
2 tháng 9 2017
a. Ta có:\(\frac{x}{y}\sqrt{\frac{y^2}{x^4}=}\) \(\frac{x}{y}.\frac{\left|y\right|}{x^2}=\frac{x.y}{x^2y}\)\(=\frac{1}{x}\)(Vì \(x\ne0;y>0\))
2 tháng 9 2017
b \(3x^2\sqrt{\frac{8}{x^2}}=3x^2\frac{2\sqrt{2}}{\left|x\right|}=\frac{6x^2\sqrt{2}}{-x}=-6x\sqrt{2}\)( Vì \(x< 0\))
29 tháng 6 2021
b, ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
Ta có : \(B=\dfrac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{x+2}{\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+4}{\sqrt{x}}-\dfrac{x-2\sqrt{x}+4}{\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+4-x+2\sqrt{x}-4+x+2}{\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}+2}{\sqrt{x}}\)
29 tháng 6 2021
b) Ta có: \(B=\dfrac{x\sqrt{x}-8}{x-2\sqrt{x}}-\dfrac{x\sqrt{x}+8}{x+2\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+4}{\sqrt{x}}-\dfrac{x-2\sqrt{x}+4}{\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+x+2}{\sqrt{x}}\)
c) Ta có: \(C=\dfrac{1}{\sqrt{x}+2}-\dfrac{5}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-3-5+\left(x-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
15 tháng 8 2021
1: Ta có: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9-\left(x-9\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
Để \(A=-\dfrac{1}{\sqrt{x}}\) thì \(x+\sqrt{x}=-\sqrt{x}+3\)
\(\Leftrightarrow x+2\sqrt{x}-3=0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow x=1\left(nhận\right)\)
2: Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;2;-2;4;-4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)
\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)
7 tháng 2 2021
a) Ta có: \(P=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)
\(=\left(\dfrac{x+1}{x+1}+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{x+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{x+\sqrt{x}+1}{x+1}\cdot\dfrac{\left(x+1\right)\left(\sqrt{x}-1\right)}{x-2\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1}{1}\cdot\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b) Để \(P=5\) thì \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}=5\)
\(\Leftrightarrow x+\sqrt{x}+1=5\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow x+\sqrt{x}+1=5\sqrt{x}-5\)
\(\Leftrightarrow x+\sqrt{x}+1-5\sqrt{x}+5=0\)
\(\Leftrightarrow x-4\sqrt{x}+6=0\)
\(\Leftrightarrow x-2\cdot\sqrt{x}\cdot2+4+2=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+2=0\)(Vô lý)
Vậy: Không có giá trị nào của x để P=5
NM
Nguyễn Minh Quang
Giáo viên
16 tháng 7 2021
Để M có nghĩa thì \(\hept{\begin{cases}\sqrt{x}-3\ne0\\2-\sqrt{x}\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}}\)
ta có \(M=\frac{2\sqrt{x}-9+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b.\(M=5=\frac{\sqrt{x}+1}{\sqrt{x}-3}\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)