A = (x-2)² + (y -1/2)² +3 -4 -1/4

GTNN A = -5/4

\(A=x^2-4x+y^2-y+3\)

\(=\left(x^2-4x+4\right)+\left(y^2-y+\frac{1}{4}\right)+3-4-\frac{1}{4}\)

\(=\left(x^2-4x+4\right)+\left(y^2-y+\frac{1}{4}\right)-\frac{5}{4}\)

\(=\left(x-2\right)^2+\left(y-\frac{1}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

Dấu "=" khi \(\begin{cases}\left(x-2\right)^2=0\\\left(y-\frac{1}{2}\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=2\\y=\frac{1}{2}\end{cases}\)

Vậy Min_A=\(-\frac{5}{4}\) khi \(\begin{cases}x=2\\y=\frac{1}{2}\end{cases}\)

bài này mà lop6 thi khó wa, cj nhẩm:

gtnn = -2 em thử làm xem, k dc cj tip

a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)

\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)

\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)

Vậy Max_Q=10 khi x=2, y=-2

b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)

\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)

Vậy Max_A=14 khi x=-3

+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)

Vậy Max_B=5 khi x=-1/2, y=1/3

c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)

Vậy Min_P=2 khi x=1, y=-3

\(y=x+\dfrac{1}{x}-5\ge2\sqrt{\dfrac{x}{x}}-5=-3\)

\(y_{min}=-3\) khi \(x=1\)

\(y=4x^2+\dfrac{1}{2x}+\dfrac{1}{2x}-4\ge3\sqrt[3]{\dfrac{4x^2}{2x.2x}}-4=-1\)

\(y_{min}=-1\) khi \(x=\dfrac{1}{2}\)

\(y=x+\dfrac{4}{x}\Rightarrow y'=1-\dfrac{4}{x^2}=0\Rightarrow x=-2\)

\(y\left(-2\right)=-4\Rightarrow\max\limits_{x>0}y=-4\) khi \(x=-2\)

\(A=x^2-4x+4+y^2-y+\dfrac{1}{4}-\dfrac{5}{4}\)

\(=\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\forall x\)

Dấu '=' xảy ra khi x=2 và y=1/2

\(\left\{{}\begin{matrix}4x^2+9y^2=9\\A=x-2y+3\end{matrix}\right.\)

Áp dụng bất đẳng thức Bunhiacopxki cho các cặp số \(\left(\dfrac{1}{2};2x\right);\left(-\dfrac{2}{3};3y\right)\)

\(x-2y=\dfrac{1}{2}.x+\left(-\dfrac{2}{3}\right).3y\)

\(\Rightarrow\left[\dfrac{1}{2}.2x+\left(-\dfrac{2}{3}\right).3y\right]^2\le\left(\dfrac{1}{4}+\dfrac{4}{9}\right)\left(4x^2+9y^2\right)=\dfrac{25}{36}.9\)

\(\Rightarrow x-2y\le\dfrac{5}{6}.3=\dfrac{5}{2}\)

\(\Rightarrow A=x-2y+3\le\dfrac{5}{2}+3\)

\(\Rightarrow A=x-2y+3\le\dfrac{11}{2}\)

Dấu "=" xảy ra khi và chỉ khi

\(\dfrac{\dfrac{1}{2}}{2x}=\dfrac{-\dfrac{2}{3}}{3y}\)

\(\Rightarrow\dfrac{2x}{\dfrac{1}{2}}=\dfrac{3y}{-\dfrac{2}{3}}\)

\(\Rightarrow\dfrac{4x^2}{\dfrac{1}{4}}=\dfrac{9y^2}{\dfrac{4}{9}}=\dfrac{4x^2+9y^2}{\dfrac{1}{4}+\dfrac{4}{9}}=\dfrac{9}{\dfrac{25}{36}}=\dfrac{9.36}{25}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{9.36}{25}.\dfrac{1}{16}\\y^2=\dfrac{9.36}{25}.\dfrac{4}{36}=\dfrac{9.4}{25}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3.6}{5}.\dfrac{1}{4}=\dfrac{9}{10}\\y=\dfrac{3.2}{5}=\dfrac{6}{5}\end{matrix}\right.\)

Vậy \(GTLN\left(A\right)=\dfrac{11}{2}\left(tạix=\dfrac{9}{10};y=\dfrac{6}{5}\right)\)

2.

$y=\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x$

$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x$

Vì: $0\leq \sin ^22x\leq 1$

$\Rightarrow 1\geq 1-\frac{1}{2}\sin ^22x\geq \frac{1}{2}$

Vậy $y_{\max}=1; y_{\min}=\frac{1}{2}$

3.

$0\leq |\sin x|\leq 1$

$\Rightarrow 3\geq 3-2|\sin x|\geq 1$

Vậy $y_{\min}=1; y_{\max}=3$

1.

\(y=\cos x+\cos (x-\frac{\pi}{3})=\cos x+\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)

\(=\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)

\(y^2=(\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x)^2\leq (\cos ^2x+\sin ^2x)(\frac{9}{4}+\frac{3}{4})\)

\(\Leftrightarrow y^2\leq 3\Rightarrow -\sqrt{3}\leq y\leq \sqrt{3}\)

Vậy $y_{\min}=-\sqrt{3}; y_{max}=\sqrt{3}$