đăng hơi nhiều câu hỏi=)

x^2+y^2>=0

=>25*(x^2+y^2)>=0(1)

mà:(12-3x-4y)^2>=0(2)

cộng (1) cho (2)=>25(x^2+y^2) + (12-3x-4y)^2>=0

=>min=0 khi x=y=0

#)Giải :

Thay \(x=\frac{10+4y}{3}\)vào A, ta có :

\(A=\left(\frac{10+4y}{3}\right)^2=\frac{100+80y+16y^2}{9}+y^2\)

\(A=\frac{100+80y+25y^2}{9}=\frac{\left(5y\right)^2+2.5y.8+8^2+36}{9}\)

\(A=\frac{\left(5y+8\right)^2}{9}+4\)

Ta có : \(\frac{\left(5y+8\right)^2}{9}\ge0\)với mọi y \(\Rightarrow A=\frac{\left(5y+8\right)^2}{9}+4\ge4\)

Vậy Min_(A) = 4

Dấu ''='' xảy ra khi y = - 8/5 và x = 6/5

Câu 1:

a, Giả sử \(A=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-\dfrac{a}{b}-\dfrac{b}{a}\ge0\)

\(\Leftrightarrow A=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-2\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge0\)

Mà \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\Leftrightarrow A\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-2\cdot\dfrac{a}{b}-2\cdot\dfrac{b}{a}+2\ge0\)

\(\Leftrightarrow\left(\dfrac{a^2}{b^2}-2\cdot\dfrac{a}{b}+1\right)+\left(\dfrac{b^2}{a^2}-2\cdot\dfrac{b}{a}+1\right)\ge0\\ \Leftrightarrow\left(\dfrac{a}{b}-1\right)^2+\left(\dfrac{b}{a}-1\right)^2\ge0\left(\text{luôn đúng}\right)\)

Dấu \("="\Leftrightarrow a=b\)

b, \(B=\dfrac{a^4}{b^4}+\dfrac{b^4}{a^4}-2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\right)+2+\left(\dfrac{a^2}{b^2}+2+\dfrac{b^2}{a^2}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)-4\)

\(B=\left(\dfrac{a^4}{b^4}-2\cdot\dfrac{a^2}{b^2}+1\right)+\left(\dfrac{b^4}{a^4}-2\cdot\dfrac{b^2}{a^2}+1\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)-2\\ \Leftrightarrow B=\left(\dfrac{a^2}{b^2}-1\right)^2+\left(\dfrac{b^2}{a^2}-1\right)^2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2+\dfrac{a}{b}+\dfrac{b}{a}-4\\ \Leftrightarrow B\ge0+0+0+\dfrac{a^2+b^2}{ab}-4\ge\dfrac{2ab}{ab}-4=2-4=-2\)

Dấu \("="\Leftrightarrow\left(a;b\right)\in\left\{\left(1;-1\right);\left(-1;1\right)\right\}\)

Câu 2:

\(\left(x^2+y^2\right)\left(3^2+4^2\right)\ge\left(3x+4y\right)^2=M^2\\ \Leftrightarrow M^2\le25\cdot25\\ \Leftrightarrow M\le25\)

Dấu \("="\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{25}{25}=1\Leftrightarrow\left\{{}\begin{matrix}x^2=9\\y^2=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)

Vậy \(M_{max}=25\Leftrightarrow\left(x;y\right)=\left(3;4\right)\)

bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được

M=x^2+y^2

Vì x^2 > hoặc bằng 0 Dấu bằng xảy ra khi x=0

y^2>hoặc bằng 0 Dấu bằng xảy ra khi y=0 Vậy min của M=0 khi x=0;y=0