Áp dụng bđt AM-GM ta có

\(x^4+y^2\ge2x^2y\)

\(x^2+y^4\ge2xy^2\)

\(\Rightarrow M\le\frac{x}{2x^2y}+\frac{y}{2xy^2}=\frac{1}{2xy}+\frac{1}{2xy}=\frac{1}{xy}=1\)

Dấu "=" xảy ra khi \(x=y=1\)

Vậy..........

Bạn tham khảo tại đây:

Câu hỏi của trieu dang - Toán lớp 8 - Học toán với OnlineMath

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Rightarrow\frac{\left(yz+xz+xy\right)}{xyz}=0\)

\(\Rightarrow yz+zx+xy=0\)

Ta có : \(x^2+2yz=x^2+yz+yz\)

                              \(=x^2+yz-zx-xy\)

                              \(=x\left(x-z\right)-y\left(x-z\right)\)

                              \(=\left(x-y\right)\left(x-z\right)\)

Tương tự : \(y^2+2xz=y^2+xz+xz\)

                                    \(=y^2+xz-xy-yz\)

                                    \(=y\left(y-x\right)+z\left(x-y\right)\)

                                    \(=\left(x-y\right)\left(z-y\right)\)

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

\(\Rightarrow M=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(x-y\right)\left(z-y\right)}+\frac{xy}{\left(x-z\right)\left(y-z\right)}\)  \(M=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)

\(M=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{yz\left(y-z\right)-xz\left(x-y+y-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

\(A=\frac{\left(yz-xz\right)\left(y-z\right)+\left(xy-xz\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)

Tính thế làm gì bạn ê

Ta có: \(x+y+z=0\)

\(\Rightarrow x+y=-z\)

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

\(\Leftrightarrow x^2+2xy+y^2=z^2\)

\(\Leftrightarrow x^2+y^2-z^2=-2xy\)

Chứng minh tương tự ta có:

\(x^2+z^2-y^2=-2xz\)

\(y^2+z^2-x^2=-2yz\)

\(\frac{xy}{x^2+y^2-z^2}+\frac{xz}{x^2+z^2-y^2}+\frac{yz}{y^2+z^2-x^2}\)

\(=\frac{xy}{-2xy}+\frac{xz}{-2xz}+\frac{yz}{-2yz}\)

\(=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\)

\(=-\frac{3}{2}\)

Vậy giá trị biểu thức là \(-\frac{3}{2}\)

Ta có (x+y)xy=x²+y²-xy

=> \(\frac{1}{x}+\frac{1}{y}=\frac{1}{x^2}+\frac{1}{y^2}-\frac{1}{xy}\)

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

<=> \(0\le\frac{1}{x}+\frac{1}{y}\le4\)

mà \(A=\frac{1}{x^3+y^3}=\left(\frac{1}{x}+\frac{1}{y}\right)^2\le16\)

Vậy Max A =16 khi \(x=y=\frac{1}{2}\)

Nhận xét :

x² lớn hơn 0 ( với mọi x dương )

y² lớn hơn 0 ( với mọi y dương )

Để A_min => \(\frac{1}{x^2}+\frac{1}{y^2}\) Min => x²  và y²  max 

Nhưng x + y = 2 

=> x = y = 1 

A min = \(\frac{1}{1}+\frac{1}{1}+\frac{3}{1}=5\) 

Vậy A min = 5 <=>  x = y = 1

\(A=\frac{1}{x^2}+\frac{1}{y^2}+\frac{3}{xy}\) và x + y = 2

AM-GM => x + y >= \(2\sqrt{xy}\)

=> \(2\sqrt{xy}\)<= 2

=> xy <= 1

\(\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{1}{xy}\)

=> A >= 1/xy + 3/xy

=> A >= 4/xy

mà xy <= 1

=> A >= 4/1

=> A>= 4 

dấu bằng sảy ra khi x = y = 2/2 = 1

Vậy GTNN của A là 4 khi x = y = 1

Nhầm 1/x^2 + 1/y^2 >= 2/xy

=> A >= 5

khi x = y = 1 nhé

đưa nó vế dạng a^3 + b^3 + c^3 = 3abc

Ta có :

    \(x^3\) + \(y^3\) - xy = \(-\dfrac{1}{27}\)

⇔ \(x^3\) + \(y^3\) - xy + \(\dfrac{1}{27}\) = 0

⇔  \(x^3\) + \(y^3\) + \(\dfrac{1^3}{3^3}\) - 3xy.\(\dfrac{1}{3}\) = 0

⇔ (x + y + \(\dfrac{1}{3}\))(\(x^2\) + \(y^2\) + \(\dfrac{1}{9}\) - xy - \(\dfrac{1}{3}x-\dfrac{1}{3}y\)) = 0

TH1 :

x + y + \(\dfrac{1}{3}\) = 0

⇔ x + y = - \(\dfrac{1}{3}\) (loại vì x>0 ; y>0)

TH2 :

\(x^2+y^2+\dfrac{1}{9}-xy-\dfrac{1}{3}x-\dfrac{1}{3}y=0\)\(\dfrac{1}{3}x-\dfrac{1}{3}y\)

⇔ (\(x-\dfrac{1}{3}\))\(^2\) + (\(y-\dfrac{1}{3}\))\(^2\) + (x - y)\(^2\) = 0

⇒ \(x-\dfrac{1}{3}\) = 0       

    \(y-\dfrac{1}{3}\) = 0

    \(x-y\) = 0

⇔ x = y = \(\dfrac{1}{3}\)

Thay x = y = \(\dfrac{1}{3}\) vào \(\dfrac{x}{y^2}\) ta được :

   \(\dfrac{1}{3}\) : \(\dfrac{1}{9}\)

= \(\dfrac{1}{3}\) . 9

= 3

\(\dfrac{1}{3}\)\(x^2+y^2+\dfrac{1}{9}-xy-\dfrac{1}{3}x-\dfrac{1}{3}y=0\)​

Đặt \(f_{\left(x\right)}=ax^2+bx+c\left(a\ne0\right)\)

\(f_{\left(x\right)}=x\leftrightarrow ax^2+bx+c=x\leftrightarrow ax^2+\left(b-1\right)x+c=0\)

\(\Delta=\left(b-1\right)^2-4ac< 0\)

\(f_{\left(f_{\left(x\right)}\right)}=x\leftrightarrow a\left(ax^2+bx+c\right)^2+b\left(ax^2+bx+c\right)+c=x\)

\(\leftrightarrow\left(a^2x^2+a\left(b+1\right)x+ac+b+1\right)\left(ax^2+\left(b-1\right)x+c\right)=0\)

Do\(\left(ax^2+\left(b-1\right)x+c\right)\ne0\)

\(\leftrightarrow a^2x^2+a\left(b+1\right)x+ac+b+1=0\)

\(\Lambda=\left[a\left(b+1\right)\right]^2-4a^2\left(ac+b+1\right)=a^2\left[\left(b+1\right)^2-4\left(ac+b+1\right)\right]=a^2\left[\left(b-1\right)^2-4ac-4\right]< 0\)

-> đpcm