Cho biểu thức A = \(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right)\):\(\dfrac{\sqrt{x}-1}{2}\) (\(x\ge0\); \(x\ne1\)). Chứng minh rằng \(A>0\)
Những câu hỏi liên quan
Cho biểu thức:
\(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
Với \(x\ge0;x\ne1\)
a,Rút gọn biểu thức trên
b,Chứng minh rằng P > 0 với mọi\(x\ge0;x\ne1\)
a,\(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(P=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)
Vậy \(P=\dfrac{2}{x+\sqrt{x}+1}\)
b, Ta có \(x+\sqrt{x}+1=\left(x+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)Suy ra \(\dfrac{2}{x+\sqrt{x}+1}>0\forall x>0,x\ne1\)
hay \(P>0\forall x>0,x\ne1\)(đpcm)
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Cho biểu thức Adfrac{sqrt{x}+1}{sqrt{x}-1} và Pleft(dfrac{x-2}{x+2sqrt{x}}+dfrac{1}{sqrt{x}+2}right)timesdfrac{sqrt{x}+1}{sqrt{x}-1} với x0,xne11) Tính giá trị của biểu thức A khi x 92) Chứng minh rằng Pdfrac{sqrt{x}+1}{sqrt{x}}3) Tìm các giá trị của x để 2P2sqrt{x}+5
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Cho biểu thức \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) và \(P=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\times\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) với \(x>0,x\ne1\)
1) Tính giá trị của biểu thức A khi x = 9
2) Chứng minh rằng \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
3) Tìm các giá trị của x để \(2P=2\sqrt{x}+5\)
1: Khi x=9 thì \(A=\dfrac{3+1}{3-1}=\dfrac{4}{2}=2\)
2: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
3: 2P=2*căn x+5
=>\(\dfrac{2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+5\)
=>\(2x+5\sqrt{x}-2\sqrt{x}-2=0\)
=>\(2x+3\sqrt{x}-4=0\)
=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)
=>\(2\sqrt{x}-1=0\)
=>x=1/4
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Chứng minh các đẳng thức sau:
a) \(\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)=1-x\)
(Với \(x\ge0;x\ne1\))
b) \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}+\dfrac{a-b}{\sqrt{a}-b}=2\sqrt{a}\)
(Với a>0; b>0; \(a\ne b\))
Câu b bạn sửa lại đề
\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)
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a: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
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Cho biểu thức \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\), \(x\ge0,x\ne1\)
a) Rút gọn biểu thức A.
b) Giải phương trình \(\left(\sqrt{x}+1\right).A=x\)
c) Đặt \(B=\dfrac{7A}{3\left(2\sqrt{x}-1\right)};x\ge0,x\ne1,x\ne\dfrac{1}{4}\). Tìm số hữu tỉ x để B có giá trị nguyên.
a: Ta có: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
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b: Ta có: \(\left(\sqrt{x}+1\right)\cdot A=x\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\cdot\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=x\)
\(\Leftrightarrow x-2\sqrt{x}+1=0\)
\(\Leftrightarrow x=1\left(loại\right)\)
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CHo biểu thức:
A=\(\left(\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}-2}-\dfrac{1}{1-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1}{x-1}\). \(\left(x\ge0;x\ne1\right)\)
a) Rút gọn biểu thức A.
b) tìm các giá trị của x để \(\dfrac{1}{A}\) là số tự nhiên.
a: Ta có: \(A=\left(\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}-2}+\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1}{x-1}\)
\(=\dfrac{x+\sqrt{x}+1+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}+1}{1}\)
\(=x+2\sqrt{x}+1\)
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rủ gọn biểu thức
\(C=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right)\) : \(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)vs \(x\ge0,x\ne1\)
\(C=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\sqrt{x}\left(x-1\right)+\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{x-1}\right)\)
\(=\dfrac{x-1-2\sqrt{x}+2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x-1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
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Chứng minh đẳng thức:
a) dfrac{xsqrt{x}+ysqrt{y}}{sqrt{x}+sqrt{y}}-left(sqrt{x}-sqrt{y}right)^2sqrt{xy}left(xge0,yge0,x^2+y^2ne0right)
b) left(dfrac{1}{a-sqrt{a}}+dfrac{1}{sqrt{a}-1}right):dfrac{sqrt{a}+1}{a-2sqrt{a}+1}left(age0,ane1right)
c) sqrt{x+2sqrt{x-2}-1}left(sqrt{x-2}-1right):left(sqrt{x}-sqrt{3}right)sqrt{x}+sqrt{3}left(xge2,xne3right)
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Chứng minh đẳng thức:
a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\sqrt{xy}\left(x\ge0,y\ge0,x^2+y^2\ne0\right)\)
b) \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\left(a\ge0,a\ne1\right)\)
c) \(\sqrt{x+2\sqrt{x-2}-1}\left(\sqrt{x-2}-1\right):\left(\sqrt{x}-\sqrt{3}\right)=\sqrt{x}+\sqrt{3}\left(x\ge2,x\ne3\right)\)
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
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Rút gọn biểu thức:
E=\(\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\right)\times\dfrac{x-1}{2x+\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}vớix\ge0,x\ne1\)
M=\(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right)\div\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2}{a-1}\right)vớia\ge0,a\ne1\)
Lm nhanh giúp mk nhé! Thank!
e) Ta có: \(E=\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\right)\cdot\dfrac{x-1}{2x+\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)-\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(2x-3\sqrt{x}+1\right)-x\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{2x\sqrt{x}-3x+\sqrt{x}-x\sqrt{x}-x-\sqrt{x}}{x+\sqrt{x}+1}\cdot\dfrac{1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}-4x}{x+\sqrt{x}+1}\cdot\dfrac{1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}-4x+\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-4x+x\sqrt{x}+x+\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{2x\sqrt{x}-3x+\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}\)
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m) Ta có: \(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2}{a-1}\right)\)
\(=\left(\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\sqrt{a}-1-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-3}\)
\(=\left(\sqrt{a}-1\right)\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}\left(\sqrt{a}-3\right)}\)
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\(E=\left(\dfrac{\sqrt{x}\left(2\sqrt{x^2}+\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}+1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+1}.\dfrac{\sqrt{x}+1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\)
\(M=\left(\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\sqrt{a}-1-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-3}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)^2}{\sqrt{a}\left(\sqrt{a}-3\right)}\)
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Rút gọn biểu thức:
a) \(\sqrt{\dfrac{x-2\sqrt{x}-1}{x+2\sqrt{x}+1}}\left(x\ge0\right)\)
b) \(\dfrac{x-1}{\sqrt{y}-1}\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\left(x\ne1,y\ne1\right),y\ge0\)
