\(\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}+2}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\)
a/ rút gọn
b/tính gtri của biểu thức tại x=\(4-2\sqrt{3}\)
K
Khách
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Những câu hỏi liên quan
AA
6 tháng 8 2021
a, A= \(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\left(\sqrt{x}\right)^2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{x}{\sqrt{x}+2}\right)\)
A=\(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)}+\frac{x}{\sqrt{x}+2}\right)\)
A=\(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\sqrt{x}+x}{\left(\sqrt{x}+2\right)}\right)\)
A=\(\frac{1}{x+2\sqrt{x}}\)
b, A >= \(\frac{1}{3\sqrt{x}}\)
=> \(\frac{1}{x+2\sqrt{x}}\) >= \(\frac{1}{3\sqrt{x}}\)
=> x <= -1 , x >= 4 (x khác 0)
15 tháng 5 2021
a) Ta có: \(B=\left(\dfrac{x+3\sqrt{x}-3}{x-16}-\dfrac{1}{\sqrt{x}+4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-4}\)
\(=\left(\dfrac{x+3\sqrt{x}-3-\sqrt{x}+4}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-4}\)
\(=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\dfrac{\sqrt{x}-4}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+4}\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 5 2021
ĐKXĐ: \(x>0;x\ne1\)
\(P=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
b.
\(P=x-\sqrt{x}+1=x-\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(P_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{4}\)
MM
3 tháng 5 2021
a) đk: \(\left\{{}\begin{matrix}\sqrt{x}+1>0\\\sqrt{x}-1>0\\x>0\end{matrix}\right.=>\sqrt{x}>\pm1\)
rút gọn pt
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\) \(\dfrac{\left(x^2-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2x+\sqrt{x}\right)\left(\sqrt{x}-1\right)\sqrt{x}.\left(\sqrt{x}+1\right)}{\sqrt{x}.\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2\left(x-1\right)x\left(x+1\right)}{x\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\)
13 tháng 9 2023
ĐKXĐ: x>=0; x<>1
\(B=\dfrac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}}{x-1}:\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-\sqrt{x}+\sqrt{x}}{x-1}\cdot\dfrac{x-1}{x+2\sqrt{x}+1-x+2\sqrt{x}-1}\)
\(=\dfrac{2x+2\sqrt{x}+1}{4\sqrt{x}}\)
Khi \(x=\dfrac{2-\sqrt{3}}{2}=\dfrac{4-2\sqrt{3}}{4}=\left(\dfrac{\sqrt{3}-1}{2}\right)^2\) thì:
\(B=\dfrac{2\cdot\dfrac{2-\sqrt{3}}{2}+2\cdot\dfrac{\sqrt{3}-1}{2}+1}{4\cdot\dfrac{\sqrt{3}-1}{2}}\)
\(=\dfrac{2-\sqrt{3}+\sqrt{3}-1+1}{2\left(\sqrt{3}-1\right)}=\dfrac{2}{2\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{2}\)
Y
1 tháng 7 2023
\(a,A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\left(dk:x\ge0,x\ne1\right)\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\sqrt{x}\left(x-1\right)+\left(x-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{x-1}\right)\)
\(=\dfrac{x-1-2\sqrt{x}+2}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\dfrac{x-1}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(b,x-3\sqrt{x}+2=0\Leftrightarrow x-\sqrt{x}-2\sqrt{x}+2=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
Thay \(x=4\) vào A :
\(A=\dfrac{\sqrt{4}-1}{\sqrt{4}+1}=\dfrac{2-1}{2+1}=\dfrac{1}{3}\)
16 tháng 4 2021
a, \(A=\dfrac{\sqrt{x}-1}{x^2-x}:\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)=\dfrac{\sqrt{x}-1}{x\left(\sqrt{x}\pm1\right)}:\left(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{1}{x\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1}=1\)
16 tháng 4 2021
b, Cho A = 1 rồi còn gì, hay đề lỗi bạn ?
\(x=4+2\sqrt{3}=\sqrt{3}^2+2\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
xem là bài mình làm có sai đâu ko nhé nếu rút gọn ra kq khác thì thay bên trên vào nhé
\(a,\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}+2}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\left(dkxd:x\ge0;x\ne1\right)\)
\(=\left[\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\left(\sqrt{x}-1\right)\)
\(=\dfrac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)\)
\(=\dfrac{\left(x+2\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\)
\(=\sqrt{x}+1\)
\(b,\) Thay \(x=4-2\sqrt{3}\) vào biểu thức trên, ta được:
\(\sqrt{4-2\sqrt{3}}+1\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}+1\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+1\)
\(=\left|\sqrt{3}-1\right|+1\)
\(=\sqrt{3}-1+1\)
\(=\sqrt{3}\)
Vậy: ...
\(\text{#}Toru\)
\(a\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}+2}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\\ =\left(\dfrac{\sqrt{x}-1}{x-1}+\dfrac{x+\sqrt{x}+2}{x-1}\right).\sqrt{x}-1\\ =\dfrac{x+\sqrt{2}+1}{x-1}.\sqrt{x}-1\\ =\sqrt{x}+1\\ b,tacóx=4-2\sqrt{3}=\left(\sqrt{3}-\sqrt{1}\right)^2thãy=\sqrt{3}-\sqrt{1}vàobiểuthức,tađược\\ \sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}-1=\sqrt{3}-1-1=\sqrt{3}-2\)