K
Khách

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

21 tháng 5 2023

\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+2}\right):\dfrac{x+4}{\sqrt{x}+2}\left(dkxd:x\ne4\right)\)

\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\left(\dfrac{\sqrt{x}+2}{x+4}\right)\)

\(=\dfrac{x+2\sqrt{x}-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x+4}\)

\(=\dfrac{x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x+4}\)

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

Vậy \(B=\dfrac{1}{\sqrt{x}-2}\)

 

21 tháng 5 2023

Giúp em với anh

26 tháng 5 2021

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

\(A=\dfrac{-3\sqrt{x}-x-2-4\sqrt{x}+2x+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

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

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

 

26 tháng 5 2021

Mình sửa đầu bài

2 tháng 10 2017

ĐKXĐ:\(x\ge0,x\ne4\)\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}\)=\(\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)=\(\frac{3\sqrt{x}}{\sqrt{x}+2}\)

2 tháng 10 2017

ko biết

9 tháng 1 2016

Điều kiện : x>=0

\(\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)

\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(2+\sqrt{3}\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)

\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[3]{2+\sqrt{3}}-x}{\sqrt{\sqrt{5}-2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)

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

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

29 tháng 7 2019

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

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

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

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

\(=\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)

\(=\frac{\sqrt{x}+3}{\sqrt{x}-4}\)

a) Ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)

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

\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

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

\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

19 tháng 6 2020

\(\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}=\frac{1+2\sqrt{x}+x-4\sqrt{x}}{1-\sqrt{x}}\) (đk x\(\ge\)0,x\(\ne\)1)

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

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

Vậy \(\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}=1-\sqrt{x}\) (\(x\ge0,x\ne1\))

19 tháng 6 2020

\(\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}\)=\(\frac{1^2+2.1.\sqrt{x}+\left(\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}=\frac{1+2\sqrt{x}+x-4\sqrt{x}}{1-\sqrt{x}}=\frac{1-2\sqrt{x}+x}{1-\sqrt{x}}=\frac{1^2-2.1.\sqrt{x}+\left(\sqrt{x}\right)^2}{1-\sqrt{x}}=\frac{\left(1-\sqrt{x}\right)^2}{1-\sqrt{x}}=1-\sqrt{x}\)