Chứng minh : \(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}-24}{x-9}=\frac{\sqrt{x}+8}{\sqrt{x}+3}\left(x\ge0,x\ne9\right)\)
Những câu hỏi liên quan
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)
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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) \(C=\frac{2\sqrt{x}-9}{x-5+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4,x\ne9\right)\)
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 \(x\ge0,x\ne4,x\ne9\))
Cho biểu thức: B = \(\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}+\frac{3x-9}{x-9}\right):\left(\frac{\sqrt{x}-2}{3}+1\right)\)với \(x\ge0;x\ne9\)
Rút gọn B
Cho biểu thức
A = \(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\left(x\ge0\right),x\ne9\)
a) Rứt gọn biểu thức
a) \(A=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{11\sqrt{x}-3}{x-9}=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
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\(A=\left(\frac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}+2}\) với \(x\ge0;x\ne9\)
\(\left(\dfrac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}+\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+2}\right)\) : \(\dfrac{1}{\sqrt{x}+2}\)
=\(\dfrac{3x-3\sqrt{x}-3+\sqrt{x}+2-\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+2)}\) .\(\sqrt{x}+2\)
=\(\dfrac{(3x-3\sqrt{x})(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}\)
=\(\dfrac{3\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}\) =\(3\sqrt{x}\)
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cho biểu thức: P=\(\left[1-\frac{x-3\sqrt{x}}{x-9}\right]:\left[\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9x}{x+\sqrt{x}-6}\right]\) \(\left(x\ge0;x\ne9;x\ne4\right)\)
a) Rút gọn P
b) Tìm giá trị của x để P=1
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/ \(P=1\Rightarrow\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}=1\Rightarrow3\sqrt{x}-6=13-4\sqrt{x}-9x\)
\(\Rightarrow9x+7\sqrt{x}-19=0\)
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rút gọn biểu thức
\(A=\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{3}{\sqrt{x}-3}\right).\frac{\sqrt{x}+3}{x+9}\) với\(x\ge0,x\ne9\)
Rút gọn biểu thức :
\(P=\left(\frac{x+\sqrt{x}-4}{x-2\sqrt{x}-3}+\frac{\sqrt{x}-1}{3-\sqrt{x}}\right):\left(1-\frac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)
( \(x\ge0;x\ne4;x\ne9\)
\(A=\left(\frac{\sqrt{x}-2}{\sqrt{x}-3}+\frac{\sqrt{x}+1}{\sqrt{x}+3}+\frac{x-4\sqrt{x}-9}{9-x}\right):\frac{\sqrt{x}+5}{3-\sqrt{x}}\)với \(x\ge0,x\ne9\)
a, Rút gọn
b, Tìm x để A = 1
c, Tìm x để \(\left|A\right|< \frac{1}{2}\)
P/s: Không chắc lắm nha!