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6 tháng 7 2019
\(a,ĐKXĐ:x\ge0;x\ne4\)
Ta có: \(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}-\frac{5\sqrt{x}+2}{x-4}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2x-4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
Vậy....
\(b,ĐKXĐ:x\ge0;x\ne4\)
\(ĐểP=2\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}=2\)
\(\Leftrightarrow2\left(\sqrt{x}+2\right)=3\sqrt{x}\)
\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)
\(\Leftrightarrow3\sqrt{x}-2\sqrt{x}=4\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\text{(Thỏa mãn ĐKXĐ)}\)
Vậy...
HH
13 tháng 12 2020
a)
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}-\frac{5\sqrt{x}+2}{x-4}\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
b) Thay P = 2 vào , ta được :
\(2=\frac{3\sqrt{x}}{\sqrt{x}+2}\Leftrightarrow2\sqrt{x}+4=3\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)
Vậy x = 16 thì P = 2
VN
21 tháng 3 2017
Ủa mua Vip là giáo viên trả lời hả? :>
\(P=\left(\frac{x+\sqrt{x}-4}{x+\sqrt{x}-3\sqrt{x}-3}+\frac{\sqrt{x}-1}{3-\sqrt{x}}\right):\left(\frac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}\right)\)
\(P=\left(\frac{x+\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{3-\sqrt{x}}\right):\frac{1}{\sqrt{x}-2}\)
\(P=\left(\frac{x+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-1}{\sqrt{x}-3}\right).\left(\sqrt{x}-2\right)\)
\(P=\left(\frac{x+\sqrt{x}-4-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right).\left(\sqrt{x}-2\right)\)
\(P=\frac{x+\sqrt{x}-4-x+\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}.\left(\sqrt{x}-2\right)\)
\(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}.\left(\sqrt{x}-2\right)\)
\(P=\frac{1}{\sqrt{x}+1}.\left(\sqrt{x}-2\right)\)
\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
21 tháng 3 2017
Mình không biết ? Đọc trong THÔNG TIN ghi là GV trả lời cho VIP's member mà bạn.
NV
30 tháng 7 2016
a/ \(P=\left[1-\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left(1-\frac{\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left(\frac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{9-x+x-4\sqrt{x}+4-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\frac{3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{13-4\sqrt{x}-9x}\)
\(=\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}\)
b/ \(P=1\Rightarrow\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}=1\Rightarrow3\sqrt{x}-6=13-4\sqrt{x}-9x\)
\(\Rightarrow9x+7\sqrt{x}-19=0\)
Mình k biết mình sai chỗ nào nữa, bạn xem giúp mình với
HQ
6 tháng 7 2019
Làm sương sương :))
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}-\frac{5\sqrt{x}+2}{x-4}\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{5\sqrt{x}+2}{x-4}\)
\(P=\frac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
Để P = 2 thì \(\frac{3\sqrt{x}}{\sqrt{x}+2}=2\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}-2=0\)
\(\Rightarrow\frac{3\sqrt{x}-2\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=0\Rightarrow\frac{3\sqrt{x}-2\sqrt{x}-4}{\sqrt{x}+2}=0\)
\(\Rightarrow\frac{\sqrt{x}-4}{\sqrt{x}+2}=0\Rightarrow\sqrt{x}-4=0\)
\(\Leftrightarrow x=16\)
PN
31 tháng 7 2017
\(A=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}+2\right)+8\sqrt{x}}{x-4}:\frac{2\left(\sqrt{x}+2\right)-2\sqrt{x}-3}{\sqrt{x}+2}\)
\(A=\frac{2x}{x-4}.\left(\sqrt{x}+2\right)=\frac{2x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{2x}{\sqrt{x}-2}\)
\(\frac{x+12}{x-4}+\frac{1}{\sqrt{x}+2}=\frac{x+12+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+x+10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{\sqrt{x}+x+10}{x-4}\) nhé :/