pt \(\Leftrightarrow\)\(\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}=-y^2+\frac{49}{4}-10\)

\(\Leftrightarrow\)\(\left(x+y+\frac{7}{2}\right)^2=-y^2+\frac{9}{4}\le\frac{9}{4}\)

\(\Leftrightarrow\)\(\frac{-3}{2}\le x+y+\frac{7}{2}\le\frac{3}{2}\)

\(\Leftrightarrow\)\(-4\le x+y+1\le-1\)

Dấu "=" tự xét nhé 

\(\Leftrightarrow\left(x+y\right)^2+7\left(x+y\right)+10=-y^2\)

\(\Leftrightarrow\left(x+y\right)^2+7\left(x+y\right)+10\le0\)

\(\Leftrightarrow\left(x+y+2\right)\left(x+y+5\right)\le0\)

\(\Leftrightarrow-5\le x+y\le-2\)

\(\Leftrightarrow-4\le x+y+1\le-1\)

\(A_{max}=-1\) khi \(\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)

\(A_{min}=-4\) khi \(\left\{{}\begin{matrix}x=-5\\y=0\end{matrix}\right.\)

NHỚ K MK NHA!!!

a)Áp dụng BĐT (x+y)^2>=4xy>>>(3a+5b)^2>=4.3a.5b>>>144>=60ab>>>ab<=12/5

Dấu=xảy ra khi 3a=5b hay khi a=7,5;b=4.5(không nên dùng Cô-si vì không chắc chắn là số dương).

b)Áp dụng BĐT Cô-si>>>(y+10)^2>=40y(do ở đây y>0 nên có thể dùng Cô-si)>>>A<=y/40y=1/40

Dấu= xảy ra khi y=10.

c)A=(x^2+x+1)/x^2+2x+1=1/2(2x^2+2x+1)/x^2+2x+1>>>A/2=(x^2+2x+1)/(x^2+2x+1)+x^2/(x^2+2x+1))>=1+0=1

Dấu= xảy ra khi x=0

\(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0=>x^2+y^2\ge2xy\\\left(x+y\right)^2\ge0=>x^2+y^2\ge-2xy\end{matrix}\right.\)

Ta có:

\(\left\{{}\begin{matrix}2\left(x^2+y^2\right)+xy\ge5xy\\2\left(x^2+y^2\right)+xy\ge-3xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1\ge5xy\\1\ge-3xy\end{matrix}\right.\)

\(\Leftrightarrow-\dfrac{1}{3}\le xy\le\dfrac{1}{5}\)

Ta có:

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

P= \(\dfrac{2\left(1-xy\right)^2}{4}-4\left(xy\right)^2+2+\left(\dfrac{1-xy}{2}+2xy\right)\)

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

Đặt t = xy => \(-\dfrac{1}{3}\le t\le\dfrac{1}{5}\)

Ta có : 

P= \(\dfrac{-7t^2}{2}+\dfrac{t}{2}+3=-\dfrac{7}{2}\left(t-\dfrac{1}{14}\right)^2+\dfrac{169}{56}\)

Ta có: \(-\dfrac{1}{3}-\dfrac{1}{14}\le t-\dfrac{1}{14}\le\dfrac{1}{5}-\dfrac{1}{14}\)

<=>\(-\dfrac{17}{42}\le t-\dfrac{1}{14}\le\dfrac{9}{70}\)

=> 0\(\le\left(t-\dfrac{1}{14}\right)^2\le\left(\dfrac{17}{42}\right)^2\)

\(\dfrac{169}{56}\ge P\ge\dfrac{169}{56}-\dfrac{7}{2}\left(\dfrac{17}{42}\right)^2\)

Max P= \(\dfrac{169}{56}\) => t = 1/14 => \(xy=\dfrac{1}{14}\rightarrow x^2+y^2=\dfrac{13}{14}\) => x,y=...

Min P=\(\dfrac{169}{56}-\dfrac{7}{6}\left(\dfrac{17}{42}\right)^2\) <=> \(t=xy=-\dfrac{1}{3}\)

<=> x=-y=\(\dfrac{1}{\sqrt{3}}\) 

ta có \(\left(x-y\right)^2\le\left(1+x^2\right)\left(1+y^2\right)\)cái này các bạn tự CM

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

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

      \(\Rightarrow\left[\left(x-y\right)\left(1-xy\right)\right]\le\left[\left(1+x^2\right)\left(1+y^2\right)\right]\)cái dấu ngặc vuông là chỉ dấu giá trị tuyệt đối đấy mình ko biết đánh dấu giá trị tuyệt đối

       \(\Rightarrow\left[\frac{\left(x-y\right)\left(1-xy\right)}{\left(1+x^2\right)\left(1+y^2\right)}\right]\le1\)

       \(\Rightarrow-1\le\frac{\left(x-y\right)\left(1-xy\right)}{\left(1+x^2\right)\left(1+y^2\right)}\le1\)\(\Rightarrow-1\le A\le1\)

có z đâu b

cái này làm rôi mà

Bài 3: \(A=\frac{\left(2a+b+c\right)\left(a+2b+c\right)\left(a+b+2c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Đặt a+b=x;b+c=y;c+a=z

\(A=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}\ge\frac{2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}}{xyz}=\frac{8xyz}{xyz}=8\)

Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)

Bài 4: \(A=\frac{9x}{2-x}+\frac{2}{x}=\frac{9x-18}{2-x}+\frac{18}{2-x}+\frac{2}{x}\ge-9+\frac{\left(\sqrt{18}+\sqrt{2}\right)^2}{2-x+x}=-9+\frac{32}{2}=7\)

Dấu = xảy ra khi\(\frac{\sqrt{18}}{2-x}=\frac{\sqrt{2}}{x}\Rightarrow x=\frac{1}{2}\)

\(a)\) Có \(2012=x+y\ge2\sqrt{xy}\)\(\Leftrightarrow\)\(xy\le1006^2\)

\(B=\frac{2x^2+8xy+2y^2}{x^2+2xy+y^2}=\frac{2\left(x^2+2xy+y^2\right)}{x^2+2xy+y^2}+\frac{4xy}{x^2+2xy+y^2}=2+\frac{4xy}{\left(x+y\right)^2}\)

\(\le2+\frac{4.1006^2}{2012^2}=2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1006\)

\(b)\) \(C=\left(1+\frac{2012}{x}\right)^2+\left(1+\frac{2012}{y}\right)^2\ge\left[2+2012\left(\frac{1}{x}+\frac{1}{y}\right)\right]^2\ge\left(2+\frac{2012.4}{x+y}\right)^2\)

\(=\left(2+\frac{2012.4}{2012}\right)^2=36\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1006\)

... 

cảm ơn bạn nhiều

bạn ơi, mik học \(A^2+B^2\ge\left(A+B\right)^2d\text{ấu}"="\) xảy ra <=> \(A.B\ge0\) mà bạn?