K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

29 tháng 5 2019

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

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

\(=\left(\frac{4}{4-4a}-\frac{a^2+1}{\left(1-a\right).\left(1+a\right)}\right).\left(\frac{a+1}{a}\right)=\frac{\left(1+a\right)}{\left(1-a\right).\left(1+a\right)}\cdot\frac{a+1}{a}=\frac{1+a}{\left(1-a\right).a}=\frac{a+1}{a-a^2}\)

2 tháng 10 2019

\(K=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}}}\)

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

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

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

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

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\)

Chúc bạn học tốt !!!

16 tháng 3 2018

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

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

\(A=\frac{\left(a-1\right)^2}{4a}.\)\(\frac{\left(\sqrt{a}-1-\sqrt{a}-1\right)\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\)

\(A=\frac{a-1}{4a}.\left(-2\right).2\sqrt{a}\)

\(A=\frac{\left(a-1\right).\left(-4\sqrt{a}\right)}{4a}\)

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

16 tháng 3 2018

\(ĐKXĐ:a\ne0,1\)

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

\(A=\left(\frac{a}{4}-\frac{1}{2}+\frac{1}{4a}\right)\left(\frac{a-2\sqrt{a}+1}{a-1}-\frac{a+2\sqrt{a}+1}{a-1}\right)\)

\(A=\left(\frac{a^2-2a+1}{4a}\right)\left(\frac{-4\sqrt{a}}{a-1}\right)\)

\(A=\frac{\left(a-1\right)^2}{4a}\times\frac{-4\sqrt{a}}{a-1}\)

\(A=\frac{-4\sqrt{a}\left(a-1\right)^2}{4a\left(a-1\right)}\)

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

Vậy........

k mk nha

30 tháng 7 2020

- Các ĐKXĐ tự tìm dùm mình hen :)

Ta có : \(D=\left(\frac{5}{x-\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}-3}\)

=> \(D=\left(\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{1}{\sqrt{x}+2}\right)\left(\sqrt{x}-3\right)\)

=> \(D=\left(\frac{5+\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\right)\left(\sqrt{x}-3\right)\)

=> \(D=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\right)\left(\sqrt{x}-3\right)\)

=> \(D=\left(\frac{1}{\sqrt{x}-3}\right)\left(\sqrt{x}-3\right)=1\)

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

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

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

=> \(E=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{\sqrt{a}-1}{\sqrt{a}}\)

( làm đến đây thôi câu còn lại bạn tự làm hen )

Ghét nhất mấy câu viết sai đề b, c sai rất nhiều bạn ới

30 tháng 7 2020

đấy là mình đánh máy tính nên kéo dài hơi nhầm bạn ơi chứ không phải sai đề :))

15 tháng 1 2020

Cảm ơn chú đã kb giờ thì t sẽ làm hộ chú :V

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

\(P=\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}+\frac{\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}-\frac{ab-1}{ab-1}\right]\)

\(:\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab-1}\right)}\right]-\frac{\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}+\frac{ab-1}{ab-1}\)

\(P=\frac{\left(a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1\right)+\left(ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}\right)-\left(ab-1\right)}{ab-1}\)

\(:\frac{\left(a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1\right)-\left(ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}\right)+\left(ab-1\right)}{ab-1}\)

\(P=\frac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1+ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}-ab+1}{ab-1}\)

\(:\frac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1-ab-\sqrt{ab}-a\sqrt{b}-\sqrt{a}+ab-1}{ab-1}\)

\(P=\frac{2a\sqrt{b}+2\sqrt{ab}}{ab-1}:\frac{-2\sqrt{a}-2}{ab-1}\)

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

P/s: :V bài này tính toán kĩ nhưng chưa chắc đúng :VVVV

Mr.Fuff cảm ơn ạ >_<!!!!

30 tháng 4 2020

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

\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}-\frac{\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\left[\frac{x+2\sqrt{x}}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x+2}\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\frac{\sqrt{x}}{x-4}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)

\(A=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(x-4\right)}\)

\(A=\frac{\sqrt{x}-2}{x-4}\)

7 tháng 7 2018

mk làm luôn

a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right).\)

=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)

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

=\(\frac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)

=

6 tháng 7 2018

a/ \(A=\frac{\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}}\)

\(A=\frac{\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)-\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}+1-3}{3\sqrt{x}+1}}\)

\(A=\frac{\frac{3x-4\sqrt{x}+1-3\sqrt{x}-1}{\left(3\sqrt{x}\right)^2-1}-\frac{8\sqrt{x}}{9x-1}}{1-1-\frac{3}{3\sqrt{x}+1}}\)

\(A=\frac{\frac{3x-7\sqrt{x}}{9x-1}-\frac{8\sqrt{x}}{9x-1}}{-\frac{3}{3\sqrt{x}+1}}\)

\(A=\frac{3x-7\sqrt{x}-8\sqrt{x}}{9x-1}\left(\frac{-\left(3\sqrt{x}+1\right)}{3}\right)\)

\(A=\frac{3x-15\sqrt{x}}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)

\(A=\frac{3\left(x-3\sqrt{x}\right)}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)

\(A=\frac{\left(x-3\sqrt{x}\right)\left(-3\sqrt{x}-1\right)}{9x-1}\)

\(A=\frac{3x\sqrt{x}+8x+3\sqrt{x}}{9x-1}\)

\(A=\frac{3x\sqrt{x}}{9x-1}+\frac{8x}{9x-1}+\frac{3\sqrt{x}}{9x-1}\)

\(A=\frac{x\sqrt{x}}{x-\frac{1}{3}}+\frac{8x}{9x-1}+\frac{\sqrt{x}}{x-\frac{1}{3}}\)

\(A=\frac{\sqrt{x}\left(x-1\right)}{x-\frac{1}{3}}+\frac{\frac{8}{3}x}{x-\frac{1}{3}}\)

\(A=\frac{\sqrt{x}\left(x-1\right)+\frac{8}{3}x}{x-\frac{1}{3}}\)

7 tháng 7 2018

bạn huy hoàng làm sai rồi