Cho bt:
P = \(\left(\frac{1}{\sqrt{x}+1}-\frac{x+2}{x\sqrt{x}+1}\right):\frac{2}{\sqrt{x}}\)
a) Rút gọn
b) CMR bt P luôn luôn âm với mọi x
K
Khách
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Những câu hỏi liên quan
NV
23 tháng 7 2016
1/ ĐKXĐ: \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)
\(A=\left[\frac{x}{\sqrt{x}\left(x-4\right)}-\frac{6}{3\left(\sqrt{x}-2\right)}+\frac{1}{\sqrt{x}-2}\right]:\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(=\left[\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}-2}\right]:\left(\frac{6}{\sqrt{x}+2}\right)\)
\(=\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+2\right)}{6}\)
\(=\frac{-2}{\sqrt{x}-2}.\frac{1}{6}=-\frac{1}{3\left(\sqrt{x}-2\right)}\)
2/ Để \(A>2\Rightarrow\frac{-1}{3\left(\sqrt{x}-2\right)}>2\)\(\Rightarrow6\sqrt{x}-12+1>0\Rightarrow6\sqrt{x}-11>0\Rightarrow\sqrt{x}>\frac{11}{6}\)
\(\Rightarrow x>\frac{121}{36}\)
23 tháng 7 2018
a)\(P=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}-1}-1\right)ĐK:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}.\)
\(=\left(\frac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\frac{\sqrt{x}-1}{\sqrt{x}-\sqrt{x}+1}\right)\)
=\(\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b)P=3/2 <=>\(\frac{2\sqrt{x}+1}{\sqrt{x}+1}=\frac{3}{2}\Leftrightarrow2\sqrt{x}+1=\frac{3}{2}\sqrt{x}+\frac{3}{2}.\)
\(\Leftrightarrow\frac{1}{2}\sqrt{x}=\frac{1}{2}\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
Với x=1 thoả nãm yêu cầu
NT
13 tháng 2 2016
a) ta thấy x-4=(canx-2)(cãnx+2)
2-canx=-(cãnx - 2)
tự học mới giỏi
b)rut gọn roi giai cho
15 tháng 8 2017
a) \(\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\left(\sqrt{x}-\sqrt{y}\right)}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}-x+2\sqrt{xy}-y\)
\(=3\sqrt{xy}\)
b) \(\frac{x-y}{\sqrt{y}-1}.\sqrt{\frac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}=\frac{x-y}{\sqrt{y}-1}.\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\left(x-y\right)\left(\sqrt{y}-1\right)}{\left(x-1\right)^2}\)
15 tháng 8 2017
a) \(=\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=x+\sqrt{xy}+y-x+2\sqrt{xy}-y=3\sqrt{xy}\)
11 tháng 8 2019
các bn ơi đoạn sau mik viết nhầm đấy bỏ phần không có ngặc đi nha
MN
10 tháng 7 2020
a) \(A=\left(\frac{4\sqrt{x}}{\sqrt{x}+2}-\frac{8x}{x-4}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(\Leftrightarrow A=\frac{4\sqrt{x}\left(\sqrt{x}-2\right)-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{-4x-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\frac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{-4\sqrt{x}}{\sqrt{x}-2}\cdot\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\)
\(\Leftrightarrow A=\frac{4x}{\sqrt{x}-3}\)
b) Để \(A=-1\)
\(\Leftrightarrow\frac{4x}{\sqrt{x}-3}=-1\)
\(\Leftrightarrow4x=3-\sqrt{x}\)
\(\Leftrightarrow4x+\sqrt{x}-3=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(4\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+1=0\\4\sqrt{x}-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-1\left(ktm\right)\\\sqrt{x}=\frac{3}{4}\Leftrightarrow x=\frac{9}{16}\left(tm\right)\end{cases}}\)
Vậy để \(A=-1\Leftrightarrow x=\frac{9}{16}\)
c) Khi \(x=36\)
\(\Leftrightarrow A=\frac{4\cdot36}{\sqrt{36}-3}=\frac{144}{3}=48\)
Q
23 tháng 11 2019
a) \(A=\left(\frac{4\sqrt{x}}{\sqrt{x}+2}-\frac{8x}{x-4}\right):\left(\frac{\sqrt{x}-1}{\left(x-2\sqrt{x}\right)}-\frac{2}{\sqrt{x}}\right)\)
\(A=\left(\frac{4\sqrt{x}}{\sqrt{x}+2}-\frac{8x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
\(A=\left(\frac{4\sqrt{x}\left(\sqrt{x}-2\right)-8x}{\left(\sqrt{x}+2\right)\left(x-2\right)}\right):\left(\frac{\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(A=\left(\frac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(A=\left(\frac{-8\sqrt{x}-4x}{\left(\sqrt{x}+2\right)\sqrt{x}}\right):\left(\frac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(A=\left(\frac{-4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\sqrt{x}}\right).\left(\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\right)\)
\(A=\frac{-4\sqrt{x}\left(2-\sqrt{x}\right).\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(3-\sqrt{x}\right)}\)
\(A=\frac{-4\sqrt{x}\left(2-\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(3-\sqrt{x}\right)}\)
.......... Đến đây bạn tự nhân đa thức với đa thức xog rút gọn nha.
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 10 2019
\(P=\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\frac{2}{\sqrt{x}-1}\)
\(=\left(\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\left(\frac{2}{\sqrt{x}-1}\right)\)
\(=\frac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\frac{2}{x+\sqrt{x}+1}\)
Do \(x+\sqrt{x}+1=x+\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\Rightarrow P=\frac{2}{x+\sqrt{x}+1}>0\)
p=\(\frac{-\sqrt{x}}{2\left(x-\sqrt{x}+1\right)}\)
mà \(-\sqrt{x}< 0\) ( vì điều kiện xác định x > 0 ; x \(\ne\) -1 )
mặt khác \(x-\sqrt{x}+1=x-2\sqrt{x}\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
=> \(\frac{-\sqrt{x}}{2\left(x-\sqrt{x}+1\right)}< 0\)
=> p \(< \) 0 => p luôn luôn âm với mọi x
p = \(\frac{1}{\sqrt{x}+1}-\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\frac{\sqrt{x}}{2}\)
p = \(\frac{x-\sqrt{x}+1}{\left(x-\sqrt{x}+1\right)\sqrt{x}+1}-\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\frac{\sqrt{x}}{2}\)
p = \(\frac{x-\sqrt{x}+1-x-2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\frac{\sqrt{x}}{2}\)
p=\(\frac{-\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\frac{\sqrt{x}}{2}\)
p=\(\frac{-1}{\left(x-\sqrt{x}+1\right)}.\frac{\sqrt{x}}{2}\)
p=\(\frac{-\sqrt{x}}{2\left(x-\sqrt{x}+1\right)}\)