Lần sau tìm nơi gõ công thức và gõ hẳn ra nhé e <3

Áp dụng BĐT Cauchy-Schwarz ta có:

\(P=x^4+y^4\ge\frac{\left(x^2+y^2\right)^2}{2}\ge\frac{\left(\frac{\left(x+y\right)^2}{2}\right)^2}{2}=\frac{\left(\frac{2^2}{2}\right)^2}{2}=...\text{(tự tính nhé :)}\)

Khi \(x=y=1\)

I spring. Because spring has many beautiful  flowers.

Trả lời:

\(M=\left(x-2020\right)^4+\left(x+y+1\right)^2+5\)

Ta có: \(\left(x-2020\right)^4\ge0\forall x;\left(x+y+1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-2020\right)^4+\left(x+y+1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-2020\right)^4+\left(x+y+1\right)^2+5\ge5\forall x,y\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2020=0\\x+y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2020\\y=-2021\end{cases}}}\)

Vậy GTNN của M = 5 khi x = 2020; y = - 2021

M_min=-1 khi y=3 và x=+-3

Làm như nào vậy. bạn giải rõ ràng ra đi

Ta có: \(\left(x^2-9\right)^2\ge0\forall x\)

\(\left|y-3\right|\ge0\forall y\)

 \(\Rightarrow\left(x^2-9\right)^2+ \left|y-3\right|\ge0\forall x,y\)

 \(\Rightarrow\left(x^2-9\right)^2+\left|y-3\right|-1\ge-1\forall x,y\)

\(\Rightarrow M\ge-1\forall x,y\)

Dấu \("="\) xảy ra khi \(\left(x^2-9\right)^2=0;\left|y-3\right|=0\)

+) \(\left(x^2-9\right)^2=0\Rightarrow x^2-9=0\)

\(\Rightarrow x=+-3\)

+) \(\left|y-3\right|=0\Rightarrow y-3=0\Rightarrow y=3\)

Vậy \(Min_M=-1\) khi \(x=+-3;y=3.\)

Áp dụng Bunyakovsky, ta có :

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)

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

=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)

Mấy cái kia tương tự 

Đạo hàm đi bạn :D Cho nhanh

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

\(\Rightarrow f'\left(x\right)=4x^3-4x\)

\(f'\left(x\right)=0\Leftrightarrow4x^3-4x=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=0\end{matrix}\right.\)

\(f\left(1\right)=-1;f\left(-2\right)=8;f\left(-1\right)=-1;f\left(0\right)=0\)

\(\Rightarrow y_{min}=-1;"="\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

\(y_{max}=8;"="\Leftrightarrow x=-2\)

Đặt \(x^2=t\left(0\le t\le4\right)\)

\(y=f\left(t\right)=t^2-2t\)

\(minf\left(t\right)=min\left\{f\left(0\right);f\left(4\right);f\left(1\right)\right\}=f\left(1\right)=-1\)

\(maxf\left(t\right)=max\left\{f\left(0\right);f\left(4\right);f\left(1\right)\right\}=f\left(4\right)=8\)

\(min=-1\Leftrightarrow x=\pm1\)

\(max=8\Leftrightarrow x=-2\)

\(A=\dfrac{1}{z}\left(\dfrac{x+y}{xy}\right)=\dfrac{1}{z}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge\dfrac{4}{z\left(x+y\right)}\ge\dfrac{16}{\left(x+y+z\right)^2}=16\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\dfrac{1}{4};\dfrac{1}{4};\dfrac{1}{2}\right)\)

Áp dụng BĐT Cô-si cho 2 số thực dương \(\dfrac{xy}{z}\) và \(\dfrac{yz}{x}\) có:

\(\dfrac{xy}{z}+\dfrac{yz}{x}\) \(\ge\) 2\(\sqrt{\dfrac{xy}{z}\cdot\dfrac{yz}{x}}\) = 2\(\sqrt{y^2}\) = 2y (1)

Tương tự: \(\dfrac{yz}{x}+\dfrac{zx}{y}\ge2z\) (2)

\(\dfrac{xy}{z}+\dfrac{zx}{y}\ge2x\) (3)

Từ (1); (2); (3)

\(\Rightarrow\) \(\dfrac{2xy}{z}+\dfrac{2yz}{x}+\dfrac{2zx}{y}\ge2x+2y+2z\)

\(\Leftrightarrow\) 2\(\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\) \(\ge\) 2(x + y + z)

\(\Leftrightarrow\) \(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\ge x+y+z=10\)

Hay P_Min = 10 

Dấu "=" xảy ra \(\Leftrightarrow\) x = y = z = \(\dfrac{10}{3}\)

Vậy ...

Chúc bn học tốt!

Đặt \(\sqrt[3]{x^2+1}=t\left(t\ge1\right)\)

\(y=f\left(t\right)=t^2-t+1\)

\(minf\left(t\right)=f\left(1\right)=1\)

\(minf\left(t\right)=1\Leftrightarrow t=1\Leftrightarrow\sqrt[3]{x^2+1}=1\Leftrightarrow x=0\)