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Bài 1:
Ta có: \(M=4x^2-3x+\dfrac{1}{4x}+2011=4x^2-4x+1+x+\dfrac{1}{4x}+2010\)
\(=\left(4x^2-4x+1\right)+\left(x+\dfrac{1}{4x}\right)+2010\)
\(=\left(2x-1\right)^2+\left(x+\dfrac{1}{4x}\right)+2010\)
Áp dụng BĐT Cô- si cho 2 số không âm, ta có:
\(x+\dfrac{1}{4x}\ge2\sqrt{x.\dfrac{1}{4x}}=2\sqrt{\dfrac{1}{4}}=1\)
Suy ra: \(M=\left(2x-1\right)^2+\left(x+\dfrac{1}{4x}\right)+2010\ge0+1+2010=2011\)
Vậy: \(Min_M=2011\Leftrightarrow x=\dfrac{1}{2}\)
Bài 2: Tham khảo: với hai số thực không âm a, b thỏa a2 + b2 = 4, tìm giá trị lớn nhất của biểu thức M= ab /(a+b+2) | Câu hỏi ôn tập thi vào lớp 10
Đang học Bunyakovsky đúng hong :D
1)
\(S=\sqrt{a^2+4ab+b^2}+\sqrt{b^2+4bc+c^2}+\sqrt{c^2+4ac+a^2}\)
\(S^2=\left(\sqrt{a^2+4ab+b^2}+\sqrt{b^2+4bc+c^2}+\sqrt{c^2+4ac+a^2}\right)^2\)
\(\le\left(1^2+1^2+1^2\right)\left(a^2+4ab+b^2+b^2+4bc+c^2+c^2+4ac+a^2\right)\)
\(=3.2\left(a^2+b^2+c^2+2ab+2bc+2ac\right)=6.\left(a+b+c\right)^2=6.6^2=216\)
\(\Leftrightarrow S\le6\sqrt{6}."="\Leftrightarrow a=b=c=2\)
2) \(M^2=\left(\sqrt{x+1}+\sqrt{y+1}\right)^2\le\left(1^2+1^2\right)\left(x+1+y+1\right)=2.8=16\)
\(M\le4."="\Leftrightarrow x=y=3\)
3)
\(S=ab+2\left(a+b\right)\le\dfrac{\left(a+b\right)^2}{4}+\dfrac{8\left(a+b\right)}{4}\)
\(=\dfrac{\left(a+b\right)^2+8\left(a+b\right)}{4}\)
\(\left(a+b\right)^2\le\left(1^2+1^2\right)\left(a^2+b^2\right)=2\Leftrightarrow a+b\le\sqrt{2}\)
\(\dfrac{\left(a+b\right)^2+8\left(a+b\right)}{4}\le\dfrac{2+8\sqrt{2}}{4}=\dfrac{1+4\sqrt{2}}{2}\)
\(S\le\dfrac{1+4\sqrt{2}}{2}."="\Leftrightarrow a=b=\dfrac{1}{\sqrt{2}}\)
bài 1:
vì \(a+b\ge1\Leftrightarrow b\ge1-a\)
khi đó \(A\ge\dfrac{8a^2+1-a}{4a}+\left(1-a\right)^2=2a+\dfrac{1}{4a}-\dfrac{1}{4}+1-2a+a^2\)
\(=a^2+\dfrac{1}{4a}+\dfrac{3}{4}=a^2+\dfrac{1}{8a}+\dfrac{1}{8a}+\dfrac{3}{4}\)
Áp dụng BĐT cauchy:\(a^2+\dfrac{1}{8a}+\dfrac{1}{8a}\ge3\sqrt[3]{a^2.\dfrac{1}{8a}.\dfrac{1}{8a}}=\dfrac{3}{4}\)
\(\Rightarrow A\ge\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)
Dấu = xảy ra khi \(a^2=\dfrac{1}{8a}\Leftrightarrow a=\dfrac{1}{2}\Rightarrow b=\dfrac{1}{2}\)
Vậy AMIN=\(\dfrac{3}{2}\)khi \(a=b=\dfrac{1}{2}\)
\(1a.\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}=\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}=21-2\sqrt{21}+2\sqrt{21}=21\) \(b.\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}=11+2\sqrt{30}-2\sqrt{30}=11\)
\(2a.\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{b}{a}}=\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{a}{b}.b^2}+\sqrt{\dfrac{a^2}{b^2}.\dfrac{b}{a}}=\sqrt{\dfrac{a}{b}}+b\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{a}{b}}=\left(2+b\right)\sqrt{\dfrac{a}{b}}\) \(b.\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m-8mx+4mx^2}{81}}=\sqrt{\dfrac{m}{\left(x-1\right)^2}}.\sqrt{\dfrac{\left(2\sqrt{m}x-2\sqrt{m}\right)^2}{81}}=\dfrac{\sqrt{m}}{\text{|}x-1\text{|}}.\dfrac{\text{|}2\sqrt{m}x-2\sqrt{m}\text{|}}{9}=\dfrac{\sqrt{m}}{\text{|}x-1\text{|}}.\dfrac{2\sqrt{m}\text{|}x-1\text{|}}{9}=\dfrac{2m}{9}\) \(3a.VP=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2=\left(a+\sqrt{a}+1+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1=VT\)
KL : Vậy đẳng thức được chứng minh.
\(b.VP=\dfrac{a+b}{b^2}.\sqrt{\dfrac{a^2b^4}{a^2+2ab+b^2}}=\dfrac{a+b}{b^2}.\dfrac{b^2\text{|}a\text{|}}{\text{|}a+b\text{|}}=\dfrac{a+b}{b^2}.\dfrac{b^2\text{|}a\text{|}}{a+b}=\text{|}a\text{|}=VT\)
KL : Vậy đẳng thức được chứng minh .
P/s : Dài v ~
