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\(3,\)Áp dụng bđt Mincopski \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)hai lần có
\(VT\ge\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{yz}+\sqrt{zx}\right)^2}+\sqrt{z+xy}\)
\(\ge\sqrt{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)
\(=\sqrt{x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)
\(=\sqrt{1+2t+t^2}\left(t=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
\(=\sqrt{\left(t+1\right)^2}=t+1=VP\left(Đpcm\right)\)
\(2,\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{2\sqrt{ab}}{2\sqrt{\sqrt{a}.\sqrt{b}}}=\sqrt{\sqrt{ab}}\left(đpcm\right)\)
3/a) \(BĐT\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\)(đúng với mọi x, y không âm)
Đẳng thức xảy ra khi x = y
b) \(BĐT\Leftrightarrow\frac{\left(x-y\right)^2}{xy}\ge0\) (đúng với mọi x, y không âm)
"=" <=> x = y
c) BĐT \(\Leftrightarrow2a+2b+2\ge2\sqrt{ab}+2\sqrt{a}+2\sqrt{b}\)
\(\Leftrightarrow\left(a-2\sqrt{ab}+b\right)+\left(a-2\sqrt{a}+1\right)+\left(b-2\sqrt{b}+1\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{a}-1\right)^2+\left(\sqrt{b}-1\right)^2\ge0\) (đúng)
"=" <=> a = b = 1
1/ \(A=\sqrt{7-2\sqrt{7}.1+1}-\sqrt{7-2\sqrt{7}.\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{7}-1\right|-\left|\sqrt{7}-\sqrt{2}\right|\) (thực ra em nghĩ ko cần thêm trị tuyệt đối đâu nhưng thêm cho chắc:D)
\(=\sqrt{7}-1-\sqrt{7}+\sqrt{2}=\sqrt{2}-1\)
2/Em thấy nó sai sai nên thôi:(
a) \(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}=\frac{1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}=\frac{2\sqrt{x}}{x-1}\)( x > 0 ; x ≠ 1 )
b) \(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)( x > 0 ; x ≠ 4 )
a) Với \(x>0\)và \(x\ne1\)ta có:
\(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}+1\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+1+\sqrt{x}-1+x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) Với \(x>0\)và \(x\ne4\)ta có:
\(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)-2\left(\sqrt{x}+2\right)+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)
2/
a/ \(\sqrt{a}+\frac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\frac{1}{\sqrt{a}}}=2\), dấu "=" khi \(a=1\)
b/ \(a+b+\frac{1}{2}=a+\frac{1}{4}+b+\frac{1}{4}\ge2\sqrt{a.\frac{1}{4}}+2\sqrt{b.\frac{1}{4}}=\sqrt{a}+\sqrt{b}\)
Dấu "=" khi \(a=b=\frac{1}{4}\)
c/ Có lẽ bạn viết đề nhầm, nếu đề đúng thế này thì mình ko biết làm
Còn đề như vậy: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\) thì làm như sau:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\) ; \(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\); \(\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\)
Cộng vế với vế ta được:
\(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\ge\frac{2}{\sqrt{xy}}+\frac{2}{\sqrt{yz}}+\frac{2}{\sqrt{xz}}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\)
Dấu "=" khi \(x=y=z\)
d/ \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\frac{\left(\sqrt{3}-2\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}\)
\(=\frac{7+4\sqrt{3}}{3-4}-\frac{7-4\sqrt{3}}{3-4}=-7-4\sqrt{3}+7-4\sqrt{3}=-8\sqrt{3}\)
e/ \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=\frac{\left(a-b\right)\left(a+b-\sqrt{ab}\right)}{\sqrt{ab}}\)
\(=\frac{a^2-b^2}{\sqrt{ab}}-\left(a-b\right)\) (bạn chép đề sai)
\(x^2-1=\frac{1}{4}\left(a^2+\frac{1}{a^2}+2\right)-1=\frac{1}{4}\left(a-\frac{1}{a}\right)^2\)
\(\Rightarrow\sqrt{x^2-1}=\frac{1}{2}\left(a-\frac{1}{a}\right)\)
Tương tự \(\sqrt{y^2-1}=\frac{1}{2}\left(b-\frac{1}{b}\right)\)
\(A=\frac{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)-\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)+\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}=\frac{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}-ab-\frac{1}{ab}+\frac{a}{b}+\frac{b}{a}}{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}+ab+\frac{1}{ab}-\frac{a}{b}-\frac{b}{a}}\)
\(=\frac{\frac{a}{b}+\frac{b}{a}}{ab+\frac{1}{ab}}=\frac{a^2+b^2}{a^2b^2+1}\)
b/ \(B=\frac{\left(\sqrt{a+bx}+\sqrt{a-bx}\right)^2}{a+bx-\left(a-bx\right)}=\frac{a+\sqrt{a^2-b^2x^2}}{bx}\)
\(a^2-b^2x^2=a^2-\frac{4a^2m^2}{\left(1+m^2\right)^2}=\frac{a^2\left(m^4+2m^2+1\right)-4a^2m^2}{\left(1+m^2\right)^2}=\frac{a^2\left(1-m^2\right)^2}{\left(1+m^2\right)^2}\)
\(\Rightarrow B=\left(a+\frac{a\left(1-m^2\right)}{1+m^2}\right).\left(\frac{1+m^2}{2am}\right)=\frac{a+am^2+a-am^2}{2am}=\frac{1}{m}\)
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1. Ta có : \(\left(\sqrt{a}-\sqrt{b}\right)^2>0\Leftrightarrow a-2\sqrt{ab}+b>0\Leftrightarrow a+b>2\sqrt{ab}\Leftrightarrow\frac{1}{\sqrt{ab}}>\frac{2}{a+b}\)
2. Áp dụng từ câu 1) , ta có :
\(\frac{1}{\sqrt{1.2005}}+\frac{1}{\sqrt{2.2004}}+...+\frac{1}{\sqrt{2005.1}}>\frac{2}{1+2005}+\frac{2}{2+2004}+...+\frac{2}{2005+1}\)
\(\Leftrightarrow\frac{1}{\sqrt{1.2005}}+\frac{1}{\sqrt{2.2004}}+...+\frac{1}{\sqrt{2005.1}}< \frac{2.2005}{2006}=\frac{2005}{1003}\)
3. Ta có : \(\left(\frac{x^2+y^2}{x-y}\right)^2=\frac{x^4+2x^2y^2+y^4}{x^2-2xy+y^2}=\frac{x^4+y^4+2}{x^2+y^2-2}\)
Đặt \(t=x^2+y^2,t\ge0\Rightarrow\frac{x^4+y^4+2}{x^2+y^2-2}=\frac{t^2-2+2}{t-2}=\frac{t^2}{t-2}\)
Xét : \(\frac{t-2}{t^2}=\frac{1}{t}-\frac{2}{t^2}=-2\left(\frac{1}{t^2}-\frac{2}{t.4}+\frac{1}{16}\right)+\frac{1}{8}=-2\left(\frac{1}{t}-\frac{1}{4}\right)^2+\frac{1}{8}\le\frac{1}{8}\)
\(\Rightarrow\frac{t^2}{t-2}\ge8\Rightarrow\left(\frac{x^2+y^2}{x-y}\right)^2\ge8\Leftrightarrow\frac{x^2+y^2}{x-y}\ge2\sqrt{2}\)