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a)Đkxđ : x#1 , x > 0
Q = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
Q = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
Q=\(\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
Q=\(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
Q=\(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}X\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
Q=\(\dfrac{x-1}{\sqrt{x}}\)
b)Thay x = 2\(\sqrt{2}\)+3 vào phương trình ta được :
Q=\(\dfrac{2\sqrt{2}+3-1}{\sqrt{2\sqrt{2}+3}}\)
Q=\(\dfrac{2\sqrt{2}+2}{\sqrt{\left(\sqrt{2}+1\right)}^2}\)
Q=\(\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
Q= 2
a) Để biểu thức P xác định thì \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
Vậy ĐKXĐ:x\(\ge0\),x\(\ne9\)
\(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}=\left[\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{\left(-3\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\dfrac{-3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\dfrac{-3}{\sqrt{x}+3}\)
b) Ta có \(P< \dfrac{1}{2}\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{1}{2}\Leftrightarrow-6< \sqrt{x}+3\Leftrightarrow\sqrt{x}>-9\)
Vì \(\sqrt{x}\ge0\) và 0>-9
Vậy \(x\ge0\)
Kết hợp với ĐKXĐ, Vậy \(x\ge0\) và \(x\ne9\) thì P<\(\dfrac{1}{2}\)
Bài 3:
a: \(=\left(4\sqrt{2}-6\sqrt{2}\right)\cdot\dfrac{\sqrt{2}}{2}=-2\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}=-2\)
b: \(=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-2\left(\sqrt{6}-1\right)\)
\(=\sqrt{6}-2\sqrt{6}+2=2-\sqrt{6}\)
Câu 2:
a, ĐKXĐ: x\(\ge\)0; x\(\ne\)\(\pm\)1
B=
\(\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{-2.2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =-\dfrac{4}{\sqrt{x}-1}\)
1. \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)
\(=\left(1+\sqrt{2}\right)^2-\sqrt{3}^2\)
\(=1+2\sqrt{2}+2-3\)
\(=2\sqrt{2}\)
3. \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\cdot\left(1+\dfrac{1}{\sqrt{x}}\right)\)(1)
ĐKXĐ \(x>0,x\ne1\)
pt (1) <=> \(\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+1+\sqrt{x}-1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\sqrt{x}\cdot2}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\)
b) Để \(\sqrt{A}>A\Leftrightarrow\sqrt{\dfrac{2}{\sqrt{x}-1}}>\dfrac{2}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}>\dfrac{4}{x-2\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}-\dfrac{4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\cdot\left(\sqrt{x}-1\right)-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-2-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-6}{x-2\sqrt{x}+1}>0\)
Vì \(2\sqrt{2}-6< 0\Rightarrow x-2\sqrt{x}+1< 0\)
mà \(x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\ge0\forall x\)
Vậy không có giá trị nào của x thỏa mãn \(\sqrt{A}>A\)
(P/s Đề câu b bị sai hay sao vậy, chả có số nào mà \(\sqrt{A}>A\) cả, check lại đề giùm với nhé)
\(\text{a) }\dfrac{1}{\sqrt{x-1}}\\ \text{Để biểu thức có nghĩa }\\ thì\Rightarrow\left\{{}\begin{matrix}x-1\ge0\\\sqrt{x-1}\ne0\end{matrix}\right.\\ \Rightarrow x-1>0\\ \Rightarrow x>1\)
\(\text{b) }\dfrac{1}{\sqrt{x-\sqrt{2x-1}}}\\ \text{Để biểu thức có nghĩa }\\ thì\Rightarrow\left\{{}\begin{matrix}x-\sqrt{2x-1}\ge0\\\sqrt{x-\sqrt{2x-1}}\ne0\end{matrix}\right.\\ \Rightarrow x-\sqrt{2x-1}>0\\ \Rightarrow x>\sqrt{2x-1}\\ \Rightarrow x^2>2x-1\\ \Rightarrow x^2-2x+1>0\\ \Rightarrow\left(x-1\right)^2>0\\ \Rightarrow\left|x-1\right|>0\\ \Rightarrow\left[{}\begin{matrix}x-1< 0\\x-1>0\end{matrix}\right.\\ \Rightarrow x-1\ne0\\ \Rightarrow x\ne1\)
\(c\text{) }\sqrt{-\dfrac{1}{x}}\\ \text{Để biểu thức có nghĩa }\\ thì\Rightarrow\left\{{}\begin{matrix}-\dfrac{1}{x}\ge0\\x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}< 0\\x\ne0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x< 0\left(\text{Vì }1>0\right)\\x\ne0\end{matrix}\right.\Rightarrow x< 0\)
\(\text{d) }\sqrt{\dfrac{a+1}{a^2}}\\ \text{Để biểu thức có nghĩa }\\ thì\Rightarrow\left\{{}\begin{matrix}\dfrac{a+1}{a^2}\ge0\\a^2\ne0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a+1\ge0\left(\text{Vì }a^2>0\right)\\a\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a\ge-1\\a\ne0\end{matrix}\right.\)
a) ĐKXĐ : x\(\ne\)1
rút gọn
B =( \(\dfrac{1}{x-\sqrt{x}}\)+\(\dfrac{1}{\sqrt{x-1}}\)) : \(\dfrac{\sqrt{x}-1}{\left(\sqrt{x-1}\right)^2}\)
B=( \(\dfrac{1}{\sqrt{x}\left(\sqrt{x-1}\right)}\)+\(\dfrac{1}{\sqrt{x-1}}\)) : \(\dfrac{1}{\sqrt{x-1}}\)
B= \(\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x-1}\right)}\).\(\dfrac{\sqrt{x-1}}{1}\)
B= \(\dfrac{1+\sqrt{x}}{\sqrt{x}}\)
a. ĐKXĐ:\(\left\{{}\begin{matrix}\sqrt{x-3}\ne0\\x-3\ge0\end{matrix}\right.\)⇌\(\left\{{}\begin{matrix}x\ne3\\x\ge3\end{matrix}\right.\)⇒ x > 3
b.ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x-2}\ge0\\x-2\ne0\end{matrix}\right.\)⇌\(\left\{{}\begin{matrix}x\ge2\\x\ne2\end{matrix}\right.\)⇒x > 2
a) Đk: \(\left\{{}\begin{matrix}x-3\ge0\\x-3\ne0\end{matrix}\right.\Leftrightarrow x-3>0\Leftrightarrow x>3\)
b) \(x-2>0\Leftrightarrow x>2\)