K
Khách

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

25 tháng 11 2018

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

Đặt A=\(\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}\Leftrightarrow A^2=\sqrt{2}-1+\sqrt{2}+1-2\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=2\sqrt{2}-2\sqrt{1}=2\sqrt{2}-2\Leftrightarrow A=\pm\sqrt{2\sqrt{2}-2}\)

Ta có \(\sqrt{\sqrt{2}-1}< \sqrt{\sqrt{2}+1}\Leftrightarrow\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}< 0\Leftrightarrow A< 0\)

Vậy A=\(-\sqrt{2\sqrt{2}-2}\)

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

b) \(\sqrt{4-2\sqrt{3}}+\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{27}=\sqrt{3-2.\sqrt{3}.1+1}+\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}-\sqrt{9.3}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\dfrac{4+2\sqrt{3}}{2^2-\left(\sqrt{3}\right)^2}}-3\sqrt{3}=\left|\sqrt{3}-1\right|+\sqrt{4+2\sqrt{3}}-3\sqrt{3}=\sqrt{3}-1-3\sqrt{3}+\sqrt{3+2\sqrt{3}+1}=-2\sqrt{3}-1+\sqrt{\left(\sqrt{3}+1\right)^2}=-2\sqrt{3}-1+\sqrt{3}+1=-\sqrt{3}\)

c) \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{3}{\sqrt{x}+3}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-3x+5\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(3x-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(3x-3\sqrt{x}-2\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left[3\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)\right]}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(\sqrt{x}-1\right)\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(3\sqrt{x}-2\right)}{\sqrt{x}+3}=\dfrac{2-3\sqrt{x}}{\sqrt{x}+3}\)

24 tháng 10 2017

GIÚP MÌNH VỚI MÌNH ĐANG VỘI

a: Ta có: \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

b: Ta có: \(\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)

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

\(=a+\sqrt{a}-2\sqrt{a}+1+1\)

\(=a-\sqrt{a}+2\)

28 tháng 8 2021

a,ĐKXĐ: tự tìm :v

 \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(x+2\sqrt{x}+1\right)-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+1\right)^2-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6+2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

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

\(=\dfrac{\left(9\sqrt{x}-9\right)-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{9\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(10-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\dfrac{10-\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}-2}{\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

4 tháng 9 2023

\(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\) (ĐK: \(x\ne1;x\ge0\))

\(A=\dfrac{15\sqrt{x}-11}{x+3\sqrt{x}-\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(A=\dfrac{15\sqrt{x}-11}{\sqrt{x}\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{\left(15\sqrt{x}-11\right)-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(A=\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(A=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{-\left(5\sqrt{x}-2\right)}{\sqrt{x}+3}\)

\(A=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

\(P=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

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

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

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

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

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

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

b: \(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

c: \(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-3x+8\sqrt{x}+5-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}+8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

29 tháng 10 2023

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)

Sửa đề: \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

=\(\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{2-3\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11+\left(2-3\sqrt{x}\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+3\right)}\)

\(=\dfrac{15\sqrt{x}-11+2\sqrt{x}+6-3x-9\sqrt{x}-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

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

29 tháng 10 2023

\(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\left(x\ge0;x\ne1\right)\)

\(P=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(P=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{2x-2\sqrt{x}+3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(P=\dfrac{15\sqrt{x}-11-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(P=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(P=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(P=\dfrac{-\left(\sqrt{x}-1\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(P=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)

a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)

\(=\dfrac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\)

=2

b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)

\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)

\(=\dfrac{4x-1}{x^2}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Ta có: \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-3x+6\sqrt{x}-11-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+5\sqrt{x}-8}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-1\right)}\)