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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 \(a=b=\dfrac{1}{2}\)
thi cấp tỉnh mà với có 1 số bài thi vào chuyên đại học với cấp 3 nữa
Bài 2: Ta có:
\(\left(2x+5y+1\right)\left(2020^{\left|x\right|}+y+x^2+x\right)=105\) là số lẻ
\(\Rightarrow\left\{{}\begin{matrix}2x+5y+1\\2020^{\left|x\right|}+y+x^2+x\end{matrix}\right.\) đều lẻ
\(\Rightarrow y⋮2\)\(\Rightarrow2020^{\left|x\right|}⋮̸2\Leftrightarrow\left|x\right|=0\Leftrightarrow x=0\).
Thay vào tìm được y...
\(1,\dfrac{1}{1+x}=1-\dfrac{1}{1+y}+1-\dfrac{1}{1+z}=\dfrac{y}{1+y}+\dfrac{z}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)
Cmtt: \(\dfrac{1}{1+y}\ge2\sqrt{\dfrac{xz}{\left(1+x\right)\left(1+z\right)}};\dfrac{1}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)
Nhân VTV
\(\Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge8\sqrt{\dfrac{x^2y^2z^2}{\left(1+x\right)^2\left(1+y\right)^2\left(1+z\right)^2}}\\ \Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge\dfrac{8xyz}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\\ \Leftrightarrow8xyz\le1\Leftrightarrow xyz\le\dfrac{1}{8}\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{2}\)
\(2,\\ a,2x^2+y^2-2xy=1\\ \Leftrightarrow\left(x-y\right)^2+x^2=1\\ \Leftrightarrow\left(x-y\right)^2=1-x^2\ge0\\ \Leftrightarrow x^2\le1\Leftrightarrow\sqrt{x^2}\le1\Leftrightarrow\left|x\right|\le1\)
a) \(\dfrac{y}{x}\cdot\sqrt{\dfrac{x^2}{y^4}}\)
\(=\dfrac{y}{x}\cdot\dfrac{\sqrt{x^2}}{\sqrt{\left(y^2\right)^2}}\)
\(=\dfrac{y}{x}\cdot\dfrac{x}{y^2}\)
\(=\dfrac{1}{y}\)
b) \(\dfrac{5}{2}x^3y^3\cdot\sqrt{\dfrac{16}{x^4y^8}}\)
\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{\sqrt{16}}{\sqrt{\left(x^2y^4\right)^2}}\)
\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{4}{x^2y^4}\)
\(=\dfrac{20x^3y^3}{2x^2y^4}\)
\(=\dfrac{10x}{y}\)
c) \(ab^2\sqrt{\dfrac{3}{a^2b^4}}\)
\(=ab^2\dfrac{\sqrt{3}}{\sqrt{\left(ab^2\right)^2}}\)
\(=ab^2\cdot\dfrac{\sqrt{3}}{ab^2}\)
\(=\sqrt{3}\)
\(a,\dfrac{y}{x}\cdot\sqrt{\dfrac{x^2}{y^4}}\left(y\ge0;x,y\ne0\right)\) (sửa đề)
\(=\dfrac{y}{x}\cdot\dfrac{\sqrt{x^2}}{\sqrt{y^4}}\)
\(=\dfrac{y}{x}\cdot\dfrac{x}{\sqrt{\left(y^2\right)^2}}\)
\(=\dfrac{y}{x}\cdot\dfrac{x}{y^2}\)
\(=\dfrac{1}{y}\)
\(---\)
\(b,\dfrac{5}{2}x^3y^3\cdot\sqrt{\dfrac{16}{x^4y^8}}\left(x,y\ne0\right)\)
\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{\sqrt{16}}{\sqrt{x^4y^8}}\)
\(=\dfrac{5x^3y^3}{2}\cdot\dfrac{4}{x^2y^4}\)
\(=\dfrac{5x\cdot2}{y}\)
\(=\dfrac{10x}{y}\)
\(---\)
\(c,ab^2\sqrt{\dfrac{3}{a^2b^4}}\left(a>0;b\ne0\right)\) (sửa đề)
\(=ab^2\cdot\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}\)
\(=\dfrac{ab^2\sqrt{3}}{\sqrt{\left(ab^2\right)^2}}\)
\(=\dfrac{ab^2\sqrt{3}}{ab^2}\)
\(=\sqrt{3}\)
#\(Toru\)
