cho mình hỏi
\(\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\) DK(x\(\ne4\);x\(\ge\)0)
mình chưa hiểu cái chỗ là 4-x khi xét điều kiện thì phải ra là \(x\ne-4\) vì \(4-x\ne0\Rightarrow-X\ne4\Rightarrow x\ne4\)
Những câu hỏi liên quan
1) a) Rut gon \(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\left(x\ge0\right),x\ne4\)
a) Ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
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Cho biểu thức: \(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}-\frac{5\sqrt{x}+2}{x-4}\left(x\ge0,x\ne4\right)\)
a, Rút gọn biểu thức P.
b, Tìm x để P = 2.
\(a,ĐKXĐ:x\ge0;x\ne4\)
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...
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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
Cho biểu thức: \(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}-\frac{5\sqrt{x}+2}{x-4}\left(x\ge0,x\ne4\right)\)
a, Rút gọn biểu thức P.
b, Tìm x để P = 2.
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 \(\frac{3\sqrt{x}}{\sqrt{x}+2}=2\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}-2=0\)
\(\Rightarrow\frac{3\sqrt{x}-2\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=0\Rightarrow\frac{3\sqrt{x}-2\sqrt{x}-4}{\sqrt{x}+2}=0\)
\(\Rightarrow\frac{\sqrt{x}-4}{\sqrt{x}+2}=0\Rightarrow\sqrt{x}-4=0\)
\(\Leftrightarrow x=16\)
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Rút gọn các biểu thức sau:Cleft(frac{sqrt{x}+1}{x-4}-frac{sqrt{x}-1}{x+4sqrt{x}+4}right).frac{xsqrt{x}+2x-4sqrt{x}-8}{sqrt{x}-3}(với xge0,xne4,xne9)Dleft(frac{sqrt{x}+2}{x-9}-frac{sqrt{x}-2}{x+6sqrt{x}+9}right).frac{xsqrt{x}-3x-9sqrt{x}-27}{sqrt{x}-2}(với xge0,xne4,xne9)
Đọc tiếp
Rút gọn các biểu thức sau:
C=\(\left(\frac{\sqrt{x}+1}{x-4}-\frac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right).\frac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}-3}\)(với \(x\ge0\),\(x\ne4,x\ne9\))
D=\(\left(\frac{\sqrt{x}+2}{x-9}-\frac{\sqrt{x}-2}{x+6\sqrt{x}+9}\right).\frac{x\sqrt{x}-3x-9\sqrt{x}-27}{\sqrt{x}-2}\)(với \(x\ge0,x\ne4,x\ne9\))
\(P=\frac{\sqrt{x}+1}{\sqrt{x-2}}+\frac{2\sqrt{x}}{\sqrt{x+2}}+\frac{2+5\sqrt{x}}{4-x}\)
a, Rút gọn nếu \(x\ge0,x\ne4\)
b,Tìm x để P= 2
Rút gọn biểu thức :
a) Afrac{x}{x-4}+frac{1}{sqrt{x}-2}+frac{1}{sqrt{x}+2} đkxđ : xge0;xne4
b) Bleft(frac{sqrt{x}-1}{sqrt{x}+1}-frac{sqrt{x}+1}{sqrt{x}-1}right)left(frac{1}{2sqrt{x}}-frac{sqrt{x}}{2}right)^2
c) Cleft(frac{1}{sqrt{x}}+frac{sqrt{x}}{sqrt{x+1}}right)divfrac{sqrt{x}}{x+sqrt{x}} đkxđ : x 0
Đọc tiếp
Rút gọn biểu thức :
a) \(A=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\) đkxđ : \(x\ge0;x\ne4\)
b) \(B=\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}\right)^2\)
c) \(C=\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x+1}}\right)\div\frac{\sqrt{x}}{x+\sqrt{x}}\) đkxđ : x > 0
A=\(\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
=\(\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x-2}}\)
Vậy A=\(\frac{\sqrt{x}}{\sqrt{x}-2}\)vs x\(\ge0;x\ne4\)
C=\(\left(\frac{1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\times\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{1+x}{\sqrt{x}}\)
Vậy C=\(\frac{1+x}{\sqrt{x}}\)vs x>0
Rút gọn biểu thức:
\(B=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-\sqrt{x}}\)
(với \(x\ge0\)và \(x\ne4\))
Rút gọn biểu thức:
a) Aleft(frac{3x-3sqrt{x}-3}{x+sqrt{x}-2}+frac{1}{sqrt{x}-1}-frac{1}{sqrt{x}+2}right):frac{1}{sqrt{x}+2}left(xge0,xne1right)
b) Bfrac{xsqrt{x}-3}{x-2sqrt{x}-3}-frac{2left(sqrt{x-3}right)}{sqrt{x}+1}+frac{sqrt{x}+3}{3-sqrt{x}}left(x0,xne9right)
c) Cfrac{2sqrt{x}-9}{x-5+6}-frac{sqrt{x}+3}{sqrt{x}-2}-frac{2sqrt{x}+1}{3-sqrt{x}}left(xge0,xne4,xne9right)
Đọc tiếp
Rút gọn biểu thức:
a) \(A=\left(\frac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}+2}\left(x\ge0,x\ne1\right)\)
b) \(B=\frac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\frac{2\left(\sqrt{x-3}\right)}{\sqrt{x}+1}+\frac{\sqrt{x}+3}{3-\sqrt{x}}\left(x>0,x\ne9\right)\)
c) \(C=\frac{2\sqrt{x}-9}{x-5+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4,x\ne9\right)\)
\(ChoC=\left(\frac{x+2\sqrt{x}}{x+4\sqrt{x}+4}\right)\div\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2\sqrt{x}+2}{x+\sqrt{x}}\right)\left(x>0;x\ne4;x\ne9\right)\)
Rút gọn C