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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)
a) điều kiện : \(x>0;x\ne9\)
ta có : \(A=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}-\dfrac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right)\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\) \(\Leftrightarrow A=\left(\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\right)\) \(\Leftrightarrow A=\dfrac{-3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\) b) \(A< -1\) \(\Leftrightarrow A+1< 0\Leftrightarrow\dfrac{-3\sqrt{x}}{2\sqrt{x}+3}+1< 0\Leftrightarrow\dfrac{-\sqrt{x}+3}{2\sqrt{x}+3}< 0\)\(\Leftrightarrow-\sqrt{x}+3< 0\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
vậy ...
----------------------------(-_-)---------------------------------------------------
A=\(\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\div\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
=\(\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x+3}\right)\left(\sqrt{x}-3\right)}\times\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
=\(\dfrac{-3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
a: ĐKXĐ: x>0; x<>3
b: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\right):\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}-x-9}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
c: Để A=-2 thì \(-3\sqrt{x}=4\sqrt{x}+8\)
=>-7 căn x=8(loại)
a: ĐKXĐ: x>0; x<>9
b: \(A=\dfrac{x-3\sqrt{x}-x-9}{x-9}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=-\dfrac{3\sqrt{x}}{2\sqrt{x}+4}\)
c: Để A=-2 thì \(-3\sqrt{x}=-4\sqrt{x}-8\)
=>căn x=-8(loại)
a/ đkxđ: x > 0; x ≠9
b/ \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{x-9}:\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
c/ \(A=-2\Leftrightarrow\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}=-2\)
\(\Leftrightarrow-3\sqrt{x}=-4\sqrt{x}-8\)
\(\Leftrightarrow\sqrt{x}=-8\) (vô lí)
Vậy không có x nào t/m
E = ( \(\dfrac{\sqrt{x}}{\sqrt{x-1}}\)- \(\dfrac{1}{x-\sqrt{x}}\)) : ( \(\dfrac{1}{\sqrt{x+1}}\)+\(\dfrac{2}{\sqrt{x-1}}\))
a) ta có ĐKXĐ của E là x \(\ne\) 1
x \(\ne\) 0
x \(\ne\) -1
b) ( \(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)- \(\dfrac{\sqrt{1}}{x-\sqrt{x}}\)) : ( \(\dfrac{1}{\sqrt{x}+1}\)+\(\dfrac{2}{\sqrt{x}-1}\))
= (\(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)- \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)) :(\(\dfrac{1}{\sqrt{x}+1}\)+ \(\dfrac{2}{x+1}\))
= ( \(\dfrac{\sqrt{x}.\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)) : (\(\dfrac{1\left(x-1\right)+2\sqrt{x+1}}{\left(\sqrt{x}+1\right)\left(x-1\right)}\))
= ( \(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)) : \(\dfrac{x-1+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}\)
= \(\dfrac{1}{\sqrt{x}}\): \(\dfrac{\left(x-1\right)+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\)
= \(\dfrac{1}{\sqrt{x}}\). \(\dfrac{1}{2}\)
= \(\dfrac{1}{2\sqrt{x}}\)
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: \(=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-2\left(\sqrt{6}-1\right)\)
\(=\sqrt{6}-2\sqrt{6}+2=2-\sqrt{6}\)
a) Để biểu thức E được xác định thì \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\9x-1\ne0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x>0\\x\ne\dfrac{1}{9}\end{matrix}\right.\)
b) \(E=\left(1-\dfrac{2\sqrt{x}}{3\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{9x-1}\right):\left(\dfrac{9\sqrt{x}+6}{3\sqrt{x}+1}-3\right)=\left[\dfrac{3\sqrt{x}+1-2\sqrt{x}}{3\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\dfrac{9\sqrt{x}+6-9\sqrt{x}-3}{3\sqrt{x}+1}\right)=\left[\dfrac{\sqrt{x}+1}{3\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\dfrac{3}{3\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{3\sqrt{x}+1}.\left(1+\dfrac{1}{3\sqrt{x}-1}\right).\dfrac{3\sqrt{x}+1}{3}=\dfrac{\sqrt{x}+1}{3\sqrt{x}+1}.\dfrac{3\sqrt{x}+1}{3}.\dfrac{3\sqrt{x}}{3\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)