\(B=\frac{2\left(x+4\right)}{x-3\sqrt{x}-4}+\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{8}{\sqrt{x}-4}\) \(ĐKXĐ:x\ge0;x\ne16\)
a) rút gọn B
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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 \(B=-2A\Rightarrow\sqrt{x}-4=-\frac{8\sqrt{x}-30}{\sqrt{x}-4}\)
\(\Leftrightarrow x-8\sqrt{x}+16=-8\sqrt{x}+30\Leftrightarrow x-14=0\Leftrightarrow x=14\left(tm\right)\)
PN
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
=\(\frac{x\sqrt{x}-2x+28}{x-3\sqrt{x}-4}\)- \(\frac{\sqrt{x}-4}{\sqrt{x}+1}\)- \(\frac{\sqrt{x}+8}{\sqrt{x}-4}\)
= \(\frac{x\sqrt{x}-2x+28-\left(x-16\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
=\(\frac{x\sqrt{x}-2x+28-x+16-\left(x+9\sqrt{x}+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
=\(\frac{x\sqrt{x}-3x+44-x-9\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
=\(\frac{x\sqrt{x}-9\sqrt{x}-4x+36}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
=\(\frac{\sqrt{x}\left(x-9\right)-4\left(x-9\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)= \(\frac{\left(\sqrt{x}-4\right)\left(x-9\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
=\(\frac{x-9}{\sqrt{x}+1}\)
CH
Cô Hoàng Huyền
Admin
VIP
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 A3 + 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!!
\(B=\frac{2\left(x+4\right)}{x-3\sqrt{x}-4}+\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{8}{\sqrt{x}-4}\)
\(B=\frac{2\left(x+4\right)+\sqrt{x}\left(\sqrt{x}-4\right)-8\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(B=\frac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(B=\frac{3x-12\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(B=\frac{3\sqrt{x}}{\sqrt{x}+1}\)
vậy \(B=\frac{3\sqrt{x}}{\sqrt{x}+1}\)