chứng minh a\(\sqrt{b-1}\) + b\(\sqrt{a-1}\) ≤ab
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1/ Cho a,b>0 , thỏa mãn ab = 1. Chứng minh rằng:
\(\dfrac{a}{\sqrt{b+2}}+\dfrac{b}{\sqrt{a+2}}+\dfrac{1}{\sqrt{a+b+ab}}\ge\sqrt{3}\)
2/ Cho a>0. Chứng minh rằng:
a+\(\dfrac{1}{a}\ge\sqrt{\dfrac{1}{a^2+1}}+\sqrt{1+\dfrac{1}{a^2+1}}\)
3/ Cho a, b>0. Chứng minh rằng:
2(a+b)\(\le1+\sqrt{1+4\left(a^3+b^3\right)}\)
Chứng minh các đẳng thức sau:
a) left(1-a^2right):left(left(frac{1-asqrt{a}}{1-sqrt{a}}+sqrt{a}right).left(frac{1+asqrt{a}}{1+sqrt{a}}-sqrt{a}right)right)+1frac{2}{1-a}
b) left(sqrt{a}+frac{b-sqrt{ab}}{sqrt{a}+sqrt{b}}right):left(frac{a}{sqrt{ab}+b}+frac{b}{sqrt{ab}-a}-frac{a+b}{sqrt{ab}}right)sqrt{b}-sqrt{a}
c) frac{sqrt{a}+sqrt{b}-1}{a+sqrt{ab}}+frac{sqrt{a}-sqrt{b}}{2sqrt{ab}}.left(frac{sqrt{b}}{a-sqrt{ab}}+frac{sqrt{b}}{a+sqrt{ab}}right)frac{sqrt{a}}{a}
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Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right)+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
1. chứng minh rằng các hằng đẳng thức sau với điều kiện các biểu thức tồn tại:
a) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}=a-b\)
b)\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\)
a, \(VT=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)=a-b=VP\) đpcm
b,\(VT=1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}-\dfrac{a^2-a}{a-1}=1-\sqrt{a}+\sqrt{a}-a=1-a=VP\) đpcm
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Chứng minh :\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right);\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1\)
chứng minh : \(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}}-\sqrt{ab}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1\)
\(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1\)
\(VT=\dfrac{\left(a-\sqrt{ab}+b-\sqrt{ab}\right)}{a-b}+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{a-b}+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\sqrt{a}-\sqrt{b}+2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1=VP\)
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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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Chứng minh đẳng thức \(\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=a-b\)
ĐK: \(a,b\ge0\); \(a\ne b\)
\(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)\)
\(=a-b=VP\)
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Cho a, b, c > 0 thỏa mãn a + b + c = 1. Chứng minh rằng:
\(\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}\) ≥ 1
\(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{\left(a+\sqrt{bc}\right)^2}=a+\sqrt{bc}\)
Tương tự: \(\sqrt{b+ac}\ge b+\sqrt{ac}\) ; \(\sqrt{c+ab}\ge c+\sqrt{ab}\)
\(\Rightarrow VT\ge a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}\)
\(\Rightarrow VT\ge a+b+c=1\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
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a)Cho a>b>0 chứng minh rằng \(\frac{1}{a+b}\le\frac{1}{2\sqrt{ab}}\)
b) Chứng minh \(\frac{\sqrt{2}-\sqrt{1}}{3}+\frac{\sqrt{3}-\sqrt{2}}{5}+\frac{\sqrt{4}-\sqrt{3}}{7}+...+\frac{\sqrt{2011}-\sqrt{2010}}{4021}< \frac{1}{2}\)
Chứng minh các đẳng thức sau: a) left(1-a^2right):left[left(frac{1-asqrt{a}}{1-sqrt{a}}+sqrt{a}right)left(frac{1
+asqrt{a}}{1+sqrt{a}}-sqrt{a}right)right]+1frac{2}{1-a}b) left(sqrt{a}+frac{b-sqrt{ab}}{sqrt{a}+sqrt{b}}right):left(frac{a}{sqrt{ab}+b}
+frac{b}{sqrt{ab}-a}-frac{a+b}{sqrt{ab}}right)sqrt{b}-sqrt{a}c) frac{sqrt{a}+sqrt{b}-1}{a
+sqrt{ab}}+frac{sqrt{a}-sqrt{b}}{2sqrt{ab}}left(frac{sqrt{b}}{a-sqrt{ab}}+frac{sqrt{b}}{a
+sqrt{ab}}right)frac{sqrt{a}}{a}d) left(frac{asqrt{a}+bsqrt{b}}{sqrt{a}...
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Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left[\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1
+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right]+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}
+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a
+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a
+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
d) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)