Toán 9 Bất đẳng thức

49
[tex]a\sqrt{b-1}+b\sqrt{a-1}\leq a.\frac{b-1+1}{2}+b.\frac{a-1+1}{2}=\frac{ab}{2}+\frac{ab}{2}=ab[/tex]

chilephuc said:

Gải giùm mk bài bài 47,48 nữa bạn

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tưởng cần 49 thui anh , em test thử kk
Bài 48
với n dương có
[tex]\frac{1}{\sqrt{n-1}+\sqrt{n}}>\frac{1}{\sqrt{n}+\sqrt{n+1}}\\\rightarrow \frac{2}{\sqrt{n-1}+\sqrt{n}}>\frac{1}{\sqrt{n}+\sqrt{n+1}}+\frac{1}{\sqrt{n-1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n-1}[/tex]
Áp dụng
[tex]\frac{2}{\sqrt{1}+\sqrt{2}}>\sqrt{3}-\sqrt{1}\\\frac{2}{\sqrt{3}+\sqrt{4}}>\sqrt{5}-\sqrt{3}\\...\\\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-\sqrt{79}[/tex]
Cộng tất cả trên ta có:
[tex]2S>\sqrt{81}-\sqrt{1}=9-9=8 \rightarrow s>4[/tex]
Bài 47
[tex]ĐB=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{200}}\\=(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}})+(\frac{1}{\sqrt{101}}+\frac{1}{\sqrt{102}}+...+\frac{1}{\sqrt{200}})>(\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}})+(\frac{1}{\sqrt{200}}+\frac{1}{\sqrt{200}}+...+\frac{1}{\sqrt{200}})=\frac{100}{\sqrt{100}}+\frac{100}{\sqrt{200}}=10+\sqrt{50}=10+5\sqrt{2}[/tex]

chilephuc said:

bài khác nè: choa+b=ab Tìm GTNN của:
P=[tex]\frac{1}{a^{2}+2a}+\frac{1}{b^{2}+2b}+\sqrt{(1+a^{2})(1+b^{2})}[/tex]

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a;b dương hay âm ?????

chilephuc said:

bài khác nè: choa+b=ab Tìm GTNN của:
P=[tex]\frac{1}{a^{2}+2a}+\frac{1}{b^{2}+2b}+\sqrt{(1+a^{2})(1+b^{2})}[/tex]

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[tex]P=\frac{1}{a^{2}+2a}+\frac{1}{b^{2}+2b}+\sqrt{(1+a^{2})(1+b^{2})}\\\\ \geq \frac{1}{a^2+\frac{a^2+4}{2}}+\frac{1}{b^2+\frac{b^2+4}{2}}+(1.1+a.b)\\\\ =\frac{2}{3a^2+4}+\frac{2}{3b^2+4}+ab+1\\\\ =\frac{2}{3a^2+4}+\frac{3a^2+4}{8.16}+\frac{2}{3b^2+4}+\frac{3b^2+4}{8.16}-\frac{3a^2+4+3b^2+4}{128}+ab+1\\\\ =\frac{2}{3a^2+4}+\frac{3a^2+4}{8.16}+\frac{2}{3b^2+4}+\frac{3b^2+4}{8.16}-\frac{3a^2+3b^2+8-128ab}{128}+1\\\\ \geq 2\sqrt{\frac{2.(3a^2+4)}{(3a^2+4).8.16}}+2\sqrt{\frac{2.(3b^2+4)}{(3b^2+4).8.16}}-\frac{3.(a^2+b^2)-6ab}{128}+\frac{122ab}{128}-\frac{8}{128}+1[/tex]
mà [tex]a+b\geq 2\sqrt{ab}\\\\ <=> ab\geq 2\sqrt{ab}\\\\ <=> a^2b^2\geq 4ab\\\\ <=> ab\geq 4[/tex]
mặt khác: [tex]a^2+b^2\geq 2ab\\\\ <=> 3.(a^2+b^2)\geq 6ab\\\\ <=> -6ab\leq -3.(a^2+b^2)[/tex]
=> [tex]P\geq 2\sqrt{\frac{2.(3a^2+4)}{(3a^2+4).8.16}}+2\sqrt{\frac{2.(3b^2+4)}{(3b^2+4).8.16}}-\frac{3.(a^2+b^2)-6ab}{128}+\frac{122ab}{128}-\frac{8}{128}+1\\\\ \geq 2.\frac{1}{8}+2.\frac{1}{8}+\frac{0}{128}+\frac{122ab}{128}-\frac{8}{128}+1\\\\ \geq \frac{1}{2}+\frac{122}{128}.4-\frac{1}{16}+1\\\\ =5,25[/tex]
dấu "=" xảy ra <=> a=b=2