ko biết mới học lớp 6 thui à

\(S=\frac{1}{a^2+b^2}+\frac{1}{ab}+4ab=\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\left(\frac{1}{4ab}+4ab\right)+\frac{1}{4ab}\)

\(\ge\frac{4}{a^2+b^2+2ab}+2.\sqrt{\frac{4ab}{4ab}}+\frac{1}{\left(a+b\right)^2}=4+2+1=7\)

a/ \(\frac{4bc-a^2}{bc+2a^2}.\frac{4ab-c^2}{ab+2c^2}.\frac{4ac-b^2}{ac+2b^2}\)

\(=\frac{4bc-\left(b+c\right)^2}{bc+2\left(b+c\right)^2}.\frac{4\left(-b-c\right)b-c^2}{\left(-b-c\right)b+2c^2}.\frac{4\left(-b-c\right)c-b^2}{\left(-b-c\right)c+2b^2}\)

\(=\frac{-\left(b-c\right)^2}{\left(c+2b\right)\left(b+2c\right)}.\frac{-\left(c+2b\right)^2}{-\left(b-c\right)\left(b+2c\right)}.\frac{-\left(b+2c\right)^2}{\left(b-c\right)\left(c+2b\right)}=1\)

bài 1:

=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)

ta có \(a^2+bc\ge2a\sqrt{bc}\)

=> \(\frac{a}{a^2+bc}\le\frac{a}{2a\sqrt{bc}}=\frac{1}{2\sqrt{bc}}\)

vì \(\frac12\cdot\frac{1}{\sqrt{bc}}\le\frac12\frac{\left(\frac{1}{b}+\frac{1}{c}\right)}{2}=\frac14\left(\frac{1}{b}+\frac{1}{c}\right)\)

=> \(\frac{a}{a^2+bc}\le\frac14\left(\frac{1}{b}+\frac{1}{c}\right)\)

=> \(\frac{a}{a^2+bc}+\frac{b}{b^2+ac}+\frac{c}{c^2+ab}\le\frac14\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right)=\frac12\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac32\)

dấu bằng xảy ra khi a=b=c=1(đpcm)

bài 2:

dự đoán điểm rơi a=3 và b=1

<=> \(S=\left(a^2-2a+9\right)+\left(a+\frac{9}{a}-6\right)+\left(b+\frac{1}{b}-2\right)+4a+2b+8\)

\(S=\left(a-3\right)^2+\frac{a^2-6a+9}{a}+\frac{b^2-2b+1}{b}+2\left(2a+b\right)+8\)

\(S=\left(a-3\right)^2+\frac{\left(a-3\right)^2}{a}+\frac{\left(b-1\right)^2}{b}+2\left(2a+b\right)+8\)

=> \(S\ge0+0+0+17+8=25\)

dấu bằng xảy ra khi và chỉ khi a-3=0=>a=3

b-1=0=>b=1

hình như thiếu đề thì phải đó, thử thay a=b=1 vào:

\(\frac{2}{ab}+\frac{1}{a^2+b^2}+4ab=\frac{2}{1.1}+\frac{1}{1^2+1^2}+4.1.1=\frac{13}{2}< 11\)

a) \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{ab}\)

b) Giống câu a ?

c) \(\left(\sqrt{ab}-\sqrt{\frac{a}{b}}+\frac{1}{a}\sqrt{4ab}+\frac{1}{b}\sqrt{\frac{b}{a}}\right):\left(1+\frac{2}{a}-\frac{1}{b}+\frac{1}{ab}\right)\)

\(=\left(\sqrt{ab}-\sqrt{\frac{a}{b}}+\sqrt{\frac{4b}{a}}+\sqrt{\frac{1}{ab}}\right):\left(\frac{ab+2b-a+1}{ab}\right)\)

\(=\frac{ab-a+2b+1}{\sqrt{ab}}\cdot\frac{ab}{ab+2b-a+1}\)

\(=\sqrt{ab}\)

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Biểu thức không có GTNN nhé.