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THN
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TXTpro
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Mysterious Person
7 tháng 10 2018 lúc 10:27

sữa đề chút

a) đkxđ : \(x>2;x\ne3\)

b) ta có : \(A=\dfrac{\sqrt{x-1-2\sqrt{x-2}}}{\sqrt{x-2}-1}=\dfrac{\sqrt{\left(\sqrt{x-2}-1\right)^2}}{\sqrt{x-2}-1}=1\)

Thuy Duong Nguyen
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Nguyễn Công Tỉnh
8 tháng 7 2019 lúc 14:56

\(ĐKXĐ:x\ge0\)

Đề sai??? 

Sửa lại

\(a,P=\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}-1}{x-\sqrt{x}+1}\)

\(=\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\frac{x+2-x+\sqrt{x}-1+x-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\frac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)

erffsdaseefd
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Trần Hồ Tú Loan
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tranhuuphuoc
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phamnam
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Bùi Thế Hào
12 tháng 5 2017 lúc 11:25

a/ \(P=\left(\frac{3}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)

\(P=\left(\frac{3}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2}{\sqrt{x^2}-1+\sqrt{x}-1}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)

\(P=\left(\frac{3\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{1}{\left(\sqrt{x}-1\right)}\right)\)

\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{\sqrt{x}-1}{1}\right)\)

=> \(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)}\)

b/ \(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)}=\sqrt{x}-1\)

<=> \(4\sqrt{x}=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)

<=> \(4\sqrt{x}=x-1\). Bình phương 2 vế, ta được:

<=> 16x=(x-1)2

<=> 16x=x2-2x+1

<=> x2-18x+1=0

\(\Delta'=81-1=80=>\sqrt{\Delta'}=4\sqrt{5}\)

=> \(x_1=9-4\sqrt{5}\)

\(x_2=9+4\sqrt{5}\)

Kathy Nguyễn
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DƯƠNG PHAN KHÁNH DƯƠNG
19 tháng 1 2019 lúc 8:44

a ) ĐK : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)\(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^{^2}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+3}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x-2\sqrt{x}+1}{x+4\sqrt{x}+3}\)

nguyen dinh son
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Vy Thị Hoàng Lan ( Toán...
26 tháng 7 2019 lúc 10:26

ĐKXĐ: \(x\ge0;x\ne1\)

\(P=\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}}{\sqrt{x}-1}\right):\frac{2}{\sqrt{x}+1}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{2}{\sqrt{x}+1}\)

\(=\frac{x-\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{2}{\sqrt{x}+1}\)

\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+1}{2}\)

\(=\frac{-\sqrt{x}}{\sqrt{x}-1}\)

Để p = -2 \(\Rightarrow\frac{-\sqrt{x}}{\sqrt{x}-1}=-2\)

\(\frac{-\sqrt{x}}{\sqrt{x}-1}=-2\)

\(\Rightarrow-\sqrt{x}=-2\left(\sqrt{x}-1\right)\)

\(\Rightarrow-\sqrt{x}=-2\sqrt{x}+2\)

\(\Rightarrow-\sqrt{x}+2\sqrt{x}=2\)

\(\Rightarrow\sqrt{x}=2\)

\(\Rightarrow x=4\)