\(E=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x} 1}{x \sqrt{x}}\right) \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right).\left(\frac{\sqrt{x} 1}{\sqrt{x}-1} \frac{\sqrt{x}-1}{\sqrt{x} 1}\right)\)a. Rú...

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

#Toán lớp 9

2

TT

ta thi ngoc anh

6 tháng 10 2019

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

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

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

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

=\(\frac{2\sqrt{x}+2x+2}{\sqrt{x}}\)

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AN

@Nk>↑@

6 tháng 10 2019

ĐKXĐ:\(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

Ta có:

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

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

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

Thay vào E ta được: \(E=2+\frac{2\left(x+1\right)\sqrt{x}}{x}\)

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KV

karry vương

15 tháng 7 2016

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#Toán lớp 3

0

TH

~Tiểu Hoa Hoa~

28 tháng 10 2020

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#Toán lớp 9

2

NN

Nobi Nobita

28 tháng 10 2020

\(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

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

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

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

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

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

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

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

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LD

l҉o҉n҉g҉ d҉z҉

28 tháng 10 2020

Để xem nào ...

Ta có HĐT : \(\hept{\begin{cases}a\sqrt{a}+b\sqrt{b}=\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\\a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\end{cases}\left(a,b\ge0\right)}\)

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

ĐKXĐ : x > 0 ; x khác 1

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

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

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

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

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C

Chung

25 tháng 6 2018

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#Toán lớp 5

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