1/B=\(-\left(x^2+2y^2+2xy-2y\right)\)

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

     =\(-\left[\left(x+y\right)^2+\left(y-1\right)^2\right]+1\)<=1

Bmax=1 khi x+y=0 và y-1=0=>x=-1;y=1

2/C=\(x^2+x+\frac{1}{4}+y^2+y+\frac{1}{4}+\frac{1}{2}\)

      =\(\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{1}{2}\)>=\(\frac{1}{2}\)

Cmin=\(\frac{1}{2}\)khi \(x+\frac{1}{2}=0\)và \(y+\frac{1}{2}=0\)=>\(x=y=\frac{-1}{2}\)

Thêm đấu ngoặc vô đi 

với x;y>=0 ta có:

\(A^2=\left(\sqrt{2x+1}+\sqrt{2y+1}\right)^2=2x+1+2y+1+2\sqrt{\left(2x+1\right)\left(2y+1\right)}\)

\(=2\left(x+y\right)+2+\sqrt{4xy+2x+2y+1}=2\left(x+y\right)+2+\sqrt{4xy+2\left(x+y\right)+1}\)

\(2=2\left(x^2+y^2\right)=\left(1+1\right)\left(x^2+y^2\right)>=\left(x+y\right)^2\Rightarrow x+y< =\sqrt{2}\)(bđt bunhiacopxki)

\(2xy< =x^2+y^2=1\Rightarrow2\cdot2xy=4xy< =2\cdot1=2\)

\(\Rightarrow A^2=2\left(x+y\right)+2+2\sqrt{4xy+2\left(x+y\right)+1}< =2\sqrt{2}+2+2\sqrt{2+2\sqrt{2}+1}\)

\(=2\sqrt{2}+2+2\sqrt{\left(\sqrt{2}+1\right)^2}=2\sqrt{2}+2+2\left(\sqrt{2}+1\right)4\sqrt{2}+4\)

\(\Rightarrow A< =\sqrt{4\sqrt{2}+4}\)

dấu = xảy ra khi x=y=\(\sqrt{\frac{1}{2}}\)

vậy max A là \(\sqrt{4\sqrt{2}+4}\)khi \(x=y=\sqrt{\frac{1}{2}}\)

bài này ko hay cho lắm, cách làm cụ thể nhất trong cái nhất r` đấy

a)Ta thấy: \(\left|x-5\right|\ge0\)

\(\Rightarrow-\left|x-5\right|\le0\)

\(\Rightarrow1000-\left|x-5\right|\le1000\)

\(\Rightarrow A\le1000\)

Dấu "=" xảy ra khi \(\left|x-5\right|=0\Leftrightarrow x=5\)

Vậy \(Max_A=1000\) khi \(x=5\)

b)Ta thấy: \(\left|y-3\right|\ge0\)

\(\Rightarrow\left|y-3\right|+50\ge50\)

\(\Rightarrow B\ge50\)

Dấu "="xảy ra khi \(\left|y-3\right|=0\Leftrightarrow y=3\)

Vậy \(Min_B=50\) khi \(y=3\)

c)Ta thấy: \(\hept{\begin{cases}\left|x-100\right|\ge0\\\left|y+200\right|\ge0\end{cases}}\)

\(\Rightarrow\left|x-100\right|+\left|y+200\right|\ge0\)

\(\Rightarrow\left|x-100\right|+\left|y+200\right|-1\ge-1\)

\(\Rightarrow C\ge-1\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|x-100\right|=0\\\left|y+200\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=100\\y=-200\end{cases}}\)

Vậy \(Min_C=-1\) khi \(\hept{\begin{cases}x=100\\y=-200\end{cases}}\)

Khó vậy bạn

Mình mới lớp 7

Ai cho mình xin k nhé

Thanks

b: \(B=x^3-8y^3-x^3+4x-4x+8y^3+2021=2021\)

Phân tích đa thức sau thành phân tử 

a, 4x³ - 10x² + 2x

b, x² - 3x + 2

Giúp mk vs m.n

\(A=x^2+y^2+\left(\dfrac{1}{2}\right)^2-2xy+2.\dfrac{1}{2}x-2.\dfrac{1}{2}.y+\dfrac{3}{4}\)

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

\(A_{min}=\dfrac{3}{4}\) khi \(x-y+\dfrac{1}{2}=0\)

Đáp án A

y − 1 x 2 + 2 y − 1 x + 2 y − 1 = 0 1  

Nếu y = 1  thì   x = 1

Nếu y ≠ 1  thì để (1) có nghiệm thì

Δ = 2 y − 1 2 + 4 y − 1 2 y − 1 ≥ 0 ⇔ 2 y − 1 3 − 2 y ≥ 0 ⇔ 1 2 ≤ y ≤ 3 2

⇒ min y = 1 2 ; max y = 3 2 ⇒ min y + max y = 2  

\(1,\\ a,\dfrac{x^2}{x+1}+\dfrac{x}{x+1}=\dfrac{x^2+x}{x+1}=\dfrac{x\left(x+1\right)}{x+1}=x\)

\(b,\left(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}\right):\dfrac{x+y}{2x}=\left(\dfrac{4xy}{2\left(x-y\right)\left(x+y\right)}+\dfrac{\left(x-y\right)^2}{2\left(x-y\right)\left(x+y\right)}\right).\dfrac{2x}{x+y}=\dfrac{4xy+x^2-2xy+y^2}{2\left(x-y\right)\left(x+y\right)}.\dfrac{2x}{x+y}=\dfrac{2x\left(x^2+2xy+y^2\right)}{2\left(x-y\right)\left(x+y\right)^2}=\dfrac{2x\left(x+y\right)^2}{2\left(x-y\right)\left(x+y\right)^2}=\dfrac{x}{x-y}\)

Câu 2: 

a: ĐKXĐ: \(x\ne1\)