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Con Nít
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✰ɮạċɦ☠ℌổ✰
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Trương quốc trọng
29 tháng 3 2020 lúc 9:26

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✰๖ۣۜŠɦαɗøω✰
29 tháng 3 2020 lúc 10:00

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

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

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

Vậy A = 1

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Cô gái thất thường (Ánh...
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Cao Thành Long
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Nguyễn Linh Chi
29 tháng 6 2019 lúc 17:04

ĐK : x>0, x khác 1

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

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

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

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

đỗ phương anh
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Cù Minh Duy
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Vy Thị Hoàng Lan ( Toán...
24 tháng 7 2019 lúc 15:09

ĐKXĐ: Bạn tự làm nha 

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

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

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

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

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

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

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

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

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

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

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

\(=\frac{\left(\sqrt{a}+1\right)\left(a-1\right)}{\sqrt{a}\left(\sqrt{a}-3\right)}\)

nguyen ha
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Tô Hồng Nhân
27 tháng 9 2015 lúc 20:10

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

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

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

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

tick cho mình nha bạn

nguyen le duy hung
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Không Tên
11 tháng 7 2018 lúc 20:04

Bài 1:

a)  \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)

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

b)   \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)

\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)

c)  ĐK:  \(a\ge0;a\ne1\)

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

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

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

\(=1-a+a=1\)

Love
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mạc trần
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Kiyotaka Ayanokoji
29 tháng 7 2020 lúc 10:29

Trả lời:

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

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

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

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

\(A=\left[\frac{x-1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\right]\div\left[\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\right]\)

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

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

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

Học tốt 

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