Cho \(P=\frac{2x}{\sqrt{x}-2}\)
Với \(x\ge0;x\ne4\) tìm \(x\in Z\) để \(P\in Z\)
Những câu hỏi liên quan
Rút gọn: \(B=\frac{\sqrt{x}}{x+1}-\frac{4\sqrt{x}+2}{x\sqrt{x}-2x+\sqrt{x}-2}\) với x\(\ge0,x\ne4\)
Ta có: \(B=\frac{\sqrt{x}}{x+1}-\frac{4\sqrt{x}+2}{x\sqrt{x}-2x+\sqrt{x}-2}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(x+1\right)\left(\sqrt{x}-2\right)}-\frac{4\sqrt{x}+2}{\left(x+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-2\sqrt{x}-4\sqrt{x}-2}{\left(x+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-6\sqrt{x}-2}{\left(x+1\right)\left(\sqrt{x}-2\right)}\)
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Rút gọn:
\(\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\) với \(x\ge0;x\ne1\)
\(\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
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Cho biểu thức: B = \(\left(\frac{8x\sqrt{x}-1}{2x-\sqrt{x}}-\frac{8x\sqrt{x}+1}{2x+\sqrt{x}}\right):\frac{2x+1}{2x-1}\) \(\left(x\ge0;x\ne\pm\frac{1}{2}\right)\)
Rút gọn B
A frac{x-4sqrt{x}+2}{sqrt{x}-2} (xge0;xne4)
B frac{xsqrt{x}-1}{x-1} (xge0;xne1)
C frac{xsqrt{x}-1}{x-sqrt{x}}-frac{xsqrt{x}+1}{x+sqrt{x}}+frac{x+1}{sqrt{x}} ( x0;xne1)
D sqrt{x+2sqrt{2x-4}}+sqrt{x-2sqrt{2x-4}} (xge2)
E frac{x+sqrt{x^2-2x}}{x-sqrt{x^2}-2x}-frac{x-sqrt{x^2-2x}}{x+sqrt{x^2}-2x}
Đọc tiếp
A = \(\frac{x-4\sqrt{x}+2}{\sqrt{x}-2}\) (\(x\ge0;x\ne4\))
B = \(\frac{x\sqrt{x}-1}{x-1}\) (\(x\ge0;x\ne1\))
C = \(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\) ( \(x>0;x\ne1\))
D = \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) (\(x\ge2\))
E = \(\frac{x+\sqrt{x^2-2x}}{x-\sqrt{x^2}-2x}-\frac{x-\sqrt{x^2-2x}}{x+\sqrt{x^2}-2x}\)
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: Q = \(\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\frac{x+\sqrt{x}}{x-1}\right)\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)với \(x\ge0,x\ne\frac{1}{4}v\text{à}x\ge1\)
1) Rút gon Q
2) Với giá trị nào của x thì biểu thức Q đạt giá trị nhỏ nhất. Tìm giá trị nhỏ nhất đó.
Giúp mik vs
A frac{x-4sqrt{x}+2}{sqrt{x}-2} left(xge0;xne4right)
B frac{xsqrt{x}-1}{x-1} left(xge0;xne1right)
C frac{xsqrt{x}-1}{x-sqrt{x}}-frac{xsqrt{x}+1}{x+sqrt{x}}+frac{x+1}{sqrt{x}} left(x0;xne1right)
D sqrt{x+2sqrt{2x-4}}+sqrt{x-2sqrt{2x-4}} left(xge2right)
E frac{x+sqrt{x^2}-2x}{x-sqrt{x^2-2x}}-frac{x-sqrt{x^2-2x}}{x+sqrt{x^2-2x}}
Đọc tiếp
A= \(\frac{x-4\sqrt{x}+2}{\sqrt{x}-2}\) \(\left(x\ge0;x\ne4\right)\)
B= \(\frac{x\sqrt{x}-1}{x-1}\) \(\left(x\ge0;x\ne1\right)\)
C= \(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\) \(\left(x>0;x\ne1\right)\)
D= \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) \(\left(x\ge2\right)\)
E= \(\frac{x+\sqrt{x^2}-2x}{x-\sqrt{x^2-2x}}-\frac{x-\sqrt{x^2-2x}}{x+\sqrt{x^2-2x}}\)
B=\(\frac{x\sqrt{x}-1}{x-1}\)(x>0,x≠1)
=\(\frac{\sqrt{x^3}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
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A frac{x-4sqrt{x}+2}{sqrt{x}-2} left(xge0;xne1right)B frac{xsqrt{x}-1}{x-1} left(xge0;xne1right)C frac{xsqrt{x}-1}{x-sqrt{x}}-frac{xsqrt{x}+1}{x+sqrt{x}}+frac{x+1}{sqrt{x}}left(xge2right)D sqrt{x+2sqrt{2x-4}}+sqrt{x-2sqrt{2x-4}}left(xge2right)E frac{x+sqrt{x^2-2x}}{x-sqrt{x^2}-2x}-frac{x-sqrt{x^2}-2x}{x+sqrt{x^2}-2x}
Đọc tiếp
A = \(\frac{x-4\sqrt{x}+2}{\sqrt{x}-2}\) \(\left(x\ge0;x\ne1\right)\)
B = \(\frac{x\sqrt{x}-1}{x-1}\) \(\left(x\ge0;x\ne1\right)\)
C = \(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)\(\left(x\ge2\right)\)
D = \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)\(\left(x\ge2\right)\)
E = \(\frac{x+\sqrt{x^2-2x}}{x-\sqrt{x^2}-2x}-\frac{x-\sqrt{x^2}-2x}{x+\sqrt{x^2}-2x}\)
C = \(\left(\frac{2x+1}{\sqrt{x3}-1}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\frac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)với \(x\ge0;x\ne1\). So sánh C với \(-\frac{2}{\sqrt{x}}\)
\(C=\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)^2\)
\(=\sqrt{x}-1\)
Ta co:
\(\sqrt{x}-1+\frac{2}{\sqrt{x}}=\frac{x-\sqrt{x}+2}{\sqrt{x}}=\frac{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}}{\sqrt{x}}>0\)
\(\Rightarrow\sqrt{x}-1>-\frac{2}{\sqrt{x}}\)
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Cho \(B=1+\left(\frac{2x+\sqrt{x}-1}{1-x}-\frac{2x\sqrt{x}-\sqrt{x}+x}{1-x\sqrt{x}}\right).\frac{x-\sqrt{x}}{2\sqrt{x}-1}\)
a) Rút gọn B.
b) Tìm x để \(B=\frac{6-\sqrt{6}}{5}\)
c) CMR: \(B>\frac{2}{3}\) với mọi x thỏa mãn \(x\ge0,x\ne1,x\ne\frac{1}{4}\)
a/ \(B=\frac{1+x}{1+\sqrt{x}+x}\)
b/ Giải phương trình bậc 2 thì dễ rồi ha
c/ \(\frac{1+x}{1+\sqrt{x}+x}>\frac{2}{3}\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2>0\)đung vì x khac 1
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Phương trình bậc hai là\(x-\sqrt{6x}+1=0\) thì giải làm sao bạn ơi??
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