Bài 1: Căn bậc hai

Đk x \(\ge0\)

P = \(\dfrac{2\left(\sqrt{x}+1\right)-3}{\sqrt{x}+1}\) = 2 - \(\dfrac{3}{\sqrt{x}+1}\)

Để P_{mi n} <=> 2 - \(\dfrac{3}{\sqrt{x}+1}\) min <=> \(\dfrac{3}{\sqrt{x}+1}\) max <=> \(\sqrt{x}+1\) min

=> \(\sqrt{x}+1\) = 1

<=> x = 0

x = 0 => P_min = -1

Vậy P_min = -1 <=> x = 0

Không phải có cái gì đó sai mà ai rõ rằng

\(\left\{{}\begin{matrix}a=b=0\\c=1\end{matrix}\right.\)

\(VT=0+0+1\le0+0+0=0=VP\)

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1) \(1019x^2+18y^4+1007z^2\)

\(=\left(15x^2+15y^4\right)+\left(3y^4+3z^2\right)+\left(1004x^2+1004z^2\right)\)

\(\ge2\sqrt{15x^2.15y^4}+2\sqrt{3y^4.3z^2}+2\sqrt{1004x^2.1004z^2}=30xy^2+6y^2z+2008xz\left(đpcm\right)\)

ta có:

\(P=\dfrac{\sqrt{\left(x-2016\right).2017}}{\sqrt{2017}\left(x+1\right)}+\dfrac{\sqrt{\left(x-2017\right)2016}}{\sqrt{2016}\left(x-1\right)}\)

Áp dụng BĐT cauchy:\(\sqrt{\left(x-2016\right)2017}\le\dfrac{1}{2}\left(x-2016+2017\right)=\dfrac{1}{2}\left(x+1\right)\)

\(\sqrt{\left(x-2017\right)2016}\le\dfrac{1}{2}\left(x-2017+2016\right)=\dfrac{1}{2}\left(x-1\right)\)

do đó \(P\le\dfrac{x+1}{2\sqrt{2017}\left(x+1\right)}+\dfrac{x-1}{2\sqrt{2016}\left(x-1\right)}=\dfrac{1}{2\sqrt{2017}}+\dfrac{1}{2\sqrt{2016}}\)

đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x-2016=2017\\x-2017=2016\end{matrix}\right.\)\(\Rightarrow x=4033\)

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 A_MIN=\(\dfrac{3}{2}\)khi \(a=b=\dfrac{1}{2}\)

\(Pt\Leftrightarrow x^4-2x^3+6x^2-32x+40=\left(2y-1\right)^2\)

\(\Leftrightarrow\left(x^2+2x+10\right)\left(x-2\right)^2=\left(2y-1\right)^2\)

cách of thím thế này hả

mn cố gắng giúp em với

tìm Max thì bn bình phương lên r bunyakovsky

Min thì Áp dụng \(\sqrt{A}+\sqrt{B}\ge\sqrt{A+B}\)

Ta có:

\(\dfrac{n}{a+b+c}=\dfrac{100a+10b+c}{a+b+c}=1+\dfrac{99a+9b}{a+b+c}\)

\(\ge1+\dfrac{99a+9b}{a+b+9}=10+\dfrac{90a-81}{a+b+9}\ge10+\dfrac{90a-81}{a+18}\)

\(=100+\dfrac{-1701}{a+18}\ge100-\dfrac{1701}{19}=\dfrac{199}{19}\)

Dấu = xảy ra khi:\(\left\{{}\begin{matrix}a=1\\b=c=9\end{matrix}\right.\)