CMR \(\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}< \sqrt[4]{ab}\)vs a,b>0
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Cho a>0, b>0 CMR \(\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\sqrt[4]{ab}\)
CMR nếu a; b >0 thì ta luôn có
\(\frac{a+2\sqrt{ab}+9b}{\sqrt{a}+3\sqrt{b}-2\cdot\sqrt[4]{ab}}\)\(-2\sqrt{b}=\left(\sqrt[4]{a}+\sqrt[4]{b}\right)^2\)
GIÚP MÌNH LÀM BÀI NÀY NHA !!
\(\left(\sqrt{ab}-\sqrt{\frac{a}{b}}+\frac{1}{4}\sqrt{ab}+\frac{1}{b}\sqrt{\frac{b}{a}}\right):\left(1+\frac{2}{a}-\frac{1}{b}+\frac{1}{ab}\right)\)VS a,b >=0
choa,b,c > 0. Cmr: \(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}\)
Bài 1: CMR:
a, (4+sqrt{3}). (4-sqrt{3})13
b, sqrt{8+2sqrt{7}}-sqrt{8-2sqrt{7}}2
c, frac{sqrt{1}}{2+sqrt{3}}+frac{sqrt{1}}{2-sqrt{3}}4
d, frac{asqrt{b}+bsqrt{a}}{sqrt{ab}}:frac{1}{sqrt{a}-sqrt{b}}a-b(a0, b0, a≠b)
Bài 2: CMR:
a, sqrt{a}+frac{sqrt{1}}{sqrt{a}}ge2left(a0right)
b, a+b+frac{1}{2}gesqrt{a}+sqrt{b}left(a,b0right)
c, frac{1}{x}+frac{1}{y}+frac{1}{z}gefrac{1}{sqrt{xyz}}+frac{1}{sqrt{yz}}+frac{1}{sqrt{zx}}left(x,y,z0right)
d, frac{sqrt{3}+2}{sqrt{3}-2}-frac{sqrt{3}-2}{sqrt{3}+2}-8...
Đọc tiếp
Bài 1: CMR:
a, (4+\(\sqrt{3}\)). (4-\(\sqrt{3}\))=13
b, \(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}=2\)
c, \(\frac{\sqrt{1}}{2+\sqrt{3}}+\frac{\sqrt{1}}{2-\sqrt{3}}=4\)
d, \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=a-b\)(a>0, b>0, a≠b)
Bài 2: CMR:
a, \(\sqrt{a}+\frac{\sqrt{1}}{\sqrt{a}}\ge2\left(a>0\right)\)
b, a+b+\(\frac{1}{2}\ge\sqrt{a}+\sqrt{b}\left(a,b>0\right)\)
c, \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xyz}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}\left(x,y,z>0\right)\)
d, \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=-8\sqrt{3}\)
e, \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}\)=a-b(a>0, b>0, a≠b)
Bài 3: Tìm Min hoặc Max(nếu có):
a, \(\sqrt{x^2+9}\)
b, \(\frac{2}{\sqrt{x^2+1}}\)
c, 1-\(\sqrt{5+2x-x^2}\)
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)
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Xem thêm câu trả lời
Cho a,b,c>0
CMR:
\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}\sqrt{\frac{c+a}{ca}}\)
các bạn giúp mình nha càng nhanh càng tốt
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Cho a,b,c>0 Cmr
\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}\)
Pleft(frac{sqrt{a}-sqrt{b}}{sqrt{a}+sqrt{b}}-frac{sqrt{a}+sqrt{b}}{sqrt{a}-sqrt{b}}right)sqrt{frac{1}{a}-frac{1}{b}}left(frac{left(sqrt{a}-sqrt{b}right)^2}{a-b}-frac{left(sqrt{a}+sqrt{b}right)^2}{a-b}right).sqrt{frac{b-a}{ab}}frac{a-2sqrt{ab}+b-a-2sqrt{ab}-b}{a-b}.sqrt{frac{b-a}{ab}}frac{-4sqrt{ab}}{a-b}.sqrt{frac{b-a}{ab}}frac{-4sqrt{ab}}{2017-2018}.sqrt{frac{2018-2017}{ab}}4sqrt{ab}.sqrt{frac{1}{ab}}sqrt{frac{16ab}{ab}}4
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\(P=\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right)\sqrt{\frac{1}{a}-\frac{1}{b}}\)
\(=\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{a-b}-\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{a-b}\right).\sqrt{\frac{b-a}{ab}}\)
\(=\frac{a-2\sqrt{ab}+b-a-2\sqrt{ab}-b}{a-b}.\sqrt{\frac{b-a}{ab}}\)
\(=\frac{-4\sqrt{ab}}{a-b}.\sqrt{\frac{b-a}{ab}}\)\(=\frac{-4\sqrt{ab}}{2017-2018}.\sqrt{\frac{2018-2017}{ab}}\)
\(=4\sqrt{ab}.\sqrt{\frac{1}{ab}}\)\(=\sqrt{\frac{16ab}{ab}}\)\(=4\)
sao tổng lại lớn hơn hiệu
\(A=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}\) vs a>0, b>0;a#b .
\(A=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\cdot\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)
\(=a-b\)
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