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28 tháng 10 2017

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

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

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

\(B=\frac{3x-12\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)

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

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

vậy \(B=\frac{3\sqrt{x}}{\sqrt{x}+1}\)

8 tháng 3 2022

a, Với x >= 0 ; x khác 16 

\(A=\left(\frac{x+5\sqrt{x}-27+\left(3-\sqrt{x}\right)\left(\sqrt{x}+4\right)}{x-16}\right):\frac{1}{\sqrt{x}+4}\)

\(=\left(\frac{x+5\sqrt{x}-27+3\sqrt{x}+12-x-4\sqrt{x}}{x-16}\right):\frac{1}{\sqrt{x}+4}\)

\(=\left(\frac{4\sqrt{x}-15}{x-16}\right):\frac{1}{\sqrt{x}+4}=\frac{4\sqrt{x}-15}{\sqrt{x}-4}\)

b, Ta có \(B=-2A\Rightarrow\sqrt{x}-4=-\frac{8\sqrt{x}-30}{\sqrt{x}-4}\)

\(\Leftrightarrow x-8\sqrt{x}+16=-8\sqrt{x}+30\Leftrightarrow x-14=0\Leftrightarrow x=14\left(tm\right)\)

31 tháng 7 2017

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

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

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

30 tháng 7 2018

=\(\frac{x\sqrt{x}-2x+28}{x-3\sqrt{x}-4}\)\(\frac{\sqrt{x}-4}{\sqrt{x}+1}\)\(\frac{\sqrt{x}+8}{\sqrt{x}-4}\)

\(\frac{x\sqrt{x}-2x+28-\left(x-16\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)

=\(\frac{x\sqrt{x}-2x+28-x+16-\left(x+9\sqrt{x}+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)

=\(\frac{x\sqrt{x}-3x+44-x-9\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)

=\(\frac{x\sqrt{x}-9\sqrt{x}-4x+36}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)

=\(\frac{\sqrt{x}\left(x-9\right)-4\left(x-9\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)\(\frac{\left(\sqrt{x}-4\right)\left(x-9\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)

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

29 tháng 10 2020

Trả lời nhanh giúp mình với mình cần gấp lắm

2 tháng 3 2018

a) \(P=\left(\frac{x+8}{x\sqrt{x}+8}-\frac{1}{\sqrt{x}+2}\right):\left(1-\frac{x-3\sqrt{x}+6}{x-2\sqrt{x}+4}\right)\)

\(P=\frac{x+8-x+\sqrt{x}-4}{x\sqrt{x}+8}:\frac{x-2\sqrt{x}+4-x+3\sqrt{x}-6}{x-2\sqrt{x}+4}\)

\(P=\frac{\sqrt{x}+4}{x\sqrt{x}+8}:\frac{\sqrt{x}-2}{x-2\sqrt{x}+4}\)

\(P=\frac{\sqrt{x}+4}{\sqrt{x}+2}.\frac{1}{\sqrt{x}-2}\)

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

b) Ta có \(x=6+4\sqrt{2}=2^2+2.2.\sqrt{2}+\left(\sqrt{2}\right)^2=\left(2+\sqrt{2}\right)^2\)

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

Suy ra \(P=\frac{2+\sqrt{2}+4}{6+4\sqrt{2}-4}=\frac{6+\sqrt{2}}{4\sqrt{2}+2}=\frac{11\sqrt{2}-2}{14}\)

2 tháng 3 2018

cô  Hoàng Thị Thu Huyền  ơi e thấy có j đó sai sai ở đây 

chỗ dòng thứ 2 phải là 

\(P=\left[\frac{8}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}-\frac{x-2\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right]\) 

vì theo hằng đẳng thức   A+ B3= (A+B)(A2- AB +B2)

11 tháng 11 2018

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

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

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

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

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

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

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

\(=x-2\sqrt{x}+4\)

=.= hok tốt!!