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\(4=2a^2+\dfrac{1}{a^2}+\dfrac{b^2}{4}=\left(a^2+\dfrac{1}{a^2}-2\right)+\left(a^2+\dfrac{b^2}{4}+ab\right)-ab+2\)
\(\Rightarrow4=\left(a-\dfrac{1}{a}\right)^2+\left(a+\dfrac{b}{2}\right)^2-ab+2\)
\(\Rightarrow ab=\left(a-\dfrac{1}{a}\right)^2+\left(a+\dfrac{b}{2}\right)^2-2\ge-2\)
\(M_{min}=-2\) khi \(\left\{{}\begin{matrix}a-\dfrac{1}{a}=0\\a+\dfrac{b}{2}=0\end{matrix}\right.\) \(\Rightarrow\left(a;b\right)=\left(1;-2\right);\left(-1;2\right)\)
Lời giải:
Thay \(a=b+1\) ta có:
\(G=4(b+1)^2+b^2-4b(b+1)+4(b+1)-2b\)
Khai triển thu được:
\(G=b^2+6b+8\)
\(\Leftrightarrow G=(b+3)^2-1\geq -1\)
Do đó \(G_{\min}=-1\). Dấu bằng xảy ra khi \(b=-3\Leftrightarrow a=-2\)
\(G=\left[\left(2a\right)^2-2\left(2a\right).b+b^2\right]+2\left(2a-b\right)\)
\(G=\left(2a-b\right)^2+2\left(2a-b\right)\)
\(G=\left(a+a-b\right)^2+2\left(a+a-b\right)\)
\(G=\left(a+1\right)^2+2\left(a+1\right)\)
\(G=\left(a+1\right)^2+2\left(a+1\right)+1-1\)
\(G=\left(a+1+1\right)^2-1\)
\(G=\left(a+2\right)^2-1\)
\(G\ge-1\)
Đẳng thức khi \(a=-2;b=-3\)
\(\dfrac{\widehat{A}}{1}=\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{3}=\dfrac{\widehat{D}}{4}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}}{1+2+3+4}=\dfrac{360^0}{10}=36^0\\ \Rightarrow\left\{{}\begin{matrix}\widehat{A}=36^0\\\widehat{B}=72^0\\\widehat{C}=108^0\\\widehat{D}=144^0\end{matrix}\right.\)
a.b=25 > a=25/b
G= a+b = a+b
= 25/b + b
= 25 + b^2 >= 25
Vậy giá trị nn của biểu thức là 25 khi b^2=0
a) \(A=x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)=\left(\frac{1}{2}\right)^3-3.\frac{1}{2}=-\frac{11}{8}\)
b) \(B=x^6+\frac{1}{x^6}=\left(x^3+\frac{1}{x^3}\right)^2-2=\frac{-11}{8}-2=-\frac{27}{8}\)
c) \(x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2=\left(\frac{1}{2}\right)^2-2=-\frac{7}{4}\)
\(x^5+y^5=\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)=\frac{-7}{4}.\frac{-11}{8}-\frac{1}{2}=1\frac{29}{32}\)
\(C=x^7+\frac{1}{x^7}=\left(x^6+\frac{1}{x^6}\right)\left(x+\frac{1}{x}\right)-\left(x^5+\frac{1}{x^5}\right)=\frac{-27}{8}.\frac{1}{2}-1\frac{29}{32}=-3\frac{19}{32}\)
Nhầm min not max:v
\(A=\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=1+1+\dfrac{a}{b}+\dfrac{b}{a}\ge1+1+2=4\)
"=" Khi a=b
Áp dụng BĐT Cô - Si dạng Engel , có :
\(\dfrac{1}{a}+\dfrac{1}{b}\) ≥ \(\dfrac{\left(1+1\right)^2}{a+b}=\dfrac{4}{a+b}\)
⇒ \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) ≥ 4
⇒ AMIN = 4 ⇔ a = b
P/S : Tìm GTNN nhé ( sai đề rùi )
\(G=\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{4}{a}+\frac{4}{b}\ge\frac{\left(2+2\right)^2}{a+b}=\frac{16}{4}=4\) ( Cauchy-Schwarz dạng Engel )
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=2\)
Vậy GTNN của \(G\)là \(4\) khi \(a=b=2\)
Chúc bạn học tốt ~
\(G=\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)=1+\frac{a}{b}+\frac{b}{a}+1=2+\left(\frac{a}{b}+\frac{b}{a}\right)\)
Ta có: \(a,b>0\)
Áp dụng BĐT Cauchy ta có:
\(\frac{a}{b}+\frac{b}{a}\ge2.\sqrt{\frac{a}{b}.\frac{b}{a}}=2.1=2\)
Dấu " = " xảy ra \(\Leftrightarrow\frac{a}{b}=\frac{b}{a}\Leftrightarrow a=b\)
\(\Rightarrow a=b=2\)
\(\Rightarrow G\ge2+2=4\)
\(G=4\Leftrightarrow a=b=2\)
Vậy \(G_{min}=4\Leftrightarrow a=b=2\)
Thấy thừa đk a+b=4
Đây là cách khác nhé.