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1/ Ta có: \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{2^2}{4}=\frac{4}{4}=1\)
Dấu "=" xảy ra khi x=y=1
Máy mình bị lỗi nên ko nhìn được các bài tiếp theo
Chúc bạn học tốt :)
Ta có : x+y=2 => x=2-y. Thay vào bt ta đc : xy= (2-y).y = 2y -y^2
Vì y^2 >= 0 =>2y-y^2 nhỏ hơn hoặc bằng 0
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{4}\right|\)
\(=\left|x+\dfrac{1}{2}\right|+\left|-x-\dfrac{1}{4}\right|+\left|x+\dfrac{1}{3}\right|\)
\(\ge\left|x+\dfrac{1}{2}-x-\dfrac{1}{4}\right|+0=\dfrac{1}{4}\)
Dấu = xảy ra khi \(x=-\dfrac{1}{3}\)
\(A=\frac{3}{\left(x+2\right)^2+4};\left(x+2\right)^2\in N\)
\(\Rightarrow A_{max}\Leftrightarrow\left(x+2\right)^2=0\Leftrightarrow\left(x+2\right)^2+4=4\)
\(\Rightarrow A_{max}=\frac{3}{4}\)
b, \(B=\left(x+1\right)^2+\left(y+3\right)^2+1\)
Mặt khác: \(\left(x+1\right)^2;\left(y+3\right)^2\in N\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2\ge0\)
\(\Rightarrow B_{min}\Leftrightarrow\left(x+1\right)^2+\left(y+3\right)^2=0\Rightarrow B_{min}=1\)
\(A=\frac{3}{\left(x+2\right)^2+4}\)
Để A max
=>(x+2)^2+4 min
Mà\(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+4\ge4\)
Vậy Min = 4 <=>x=-2
Vậy Max A = 3/4 <=> x=-2
\(b,B=\left(x+1\right)^2+\left(y+3\right)^2+1\)
Có \(\left(x+1\right)^2\ge0;\left(y+3\right)^2\ge0\)
\(\Rightarrow B\ge0+0+1=1\)
Vậy MinB = 1<=>x=-1;y=-3
\(A=4x\left(x+y-2\right)^2+\left|2y-3\right|+1,5\)
Ta có:
\(4x\left(x+y-2\right)^2\ge0\)
\(\left|2y-3\right|\ge0\)
\(\Leftrightarrow4x\left(x+y-2\right)^2+\left|2y-3\right|\ge0\)
\(\Leftrightarrow4x\left(x+y-2\right)^2+\left|2y-3\right|+1,5\ge1,5\)
Dấu = xảy ra khi : \(x+y-2=0\Leftrightarrow x+y=2\)
\(2y-3=0\Leftrightarrow y=\frac{3}{2}\Leftrightarrow x=\frac{1}{2}\)
Vậy .....................
a) Vì (x+2)2 >/ 0
=> \(A\le\frac{3}{0+4}=\frac{3}{4}\Rightarrow Amax=\frac{3}{4}\Leftrightarrow x+2=0\Rightarrow x=-2\)
b) Vì \(\left(x+1\right)^2\ge0;\left(y+3\right)^2\ge0\)
\(B\ge0+0+1=1\Rightarrow Bmin=1\Leftrightarrow\int^{x+1=0}_{y+3=0}\Rightarrow\int^{x=-1}_{y=-3}\)
1 )Vì \(\left(x+2\right)^2\ge0;\left(y-3\right)^2\ge0\)
\(\Rightarrow\left(x+2\right)^2+\left(y-3\right)^2\ge0\)
\(\Rightarrow\left(x+2\right)^2+\left(y-3\right)^2+1\ge1\)
Dấu "=: xảy ra <=> \(\orbr{\begin{cases}\left(x+2\right)^2=0\\\left(y-3\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\y=3\end{cases}}}\)
Vậy ........
2 ) \(\frac{1}{\left(x-2\right)^2+2}\ge\frac{1}{2}\)
Dấu "=" xảy ra <=> x = 2
Vậy ..........
Vì \(2x^2\ge0\Rightarrow2x^2+1\ge1\Rightarrow\left(2x^2+1\right)^4\ge1\Rightarrow\left(2x^2+1\right)^4-3\ge-2\)
\(\Rightarrow B\ge-2\)
\(\Rightarrow MIN_B=-2\Leftrightarrow\left(2x^2+1\right)^4=1\Leftrightarrow2x^2+1=1\Leftrightarrow2x^2=0\Leftrightarrow x=0\)