Cho ba số thực dương a,b,c thỏa mãn:\(a^2b^2+b^2c^2+c^2a^2\ge a^2b^2c^2\)

Tìm giá trị nhỏ nhất của:A=\(\dfrac{a^2b^2}{c^3\left(a^2+b^2\right)}+\dfrac{b^2c^2}{a^3\left(b^2+c^2\right)}+\dfrac{c^2a^2}{b^3\left(c^2+a^2\right)}\)

Cho các số thực dương x,y,z thỏa mãn xyz ≤ 1

CMR:\(\dfrac{x\left(1-y^3\right)}{y^3}+\dfrac{y\left(1-z^3\right)}{z^3}+\dfrac{z\left(1-x^3\right)}{x^3}\)≥ 0

Cho a,b,c là các số thực dương thỏa mãn abc=1.CMR:

\(\dfrac{1}{ab+a+2}+\dfrac{1}{bc+b+2}+\dfrac{1}{ca+c+2}\le\dfrac{3}{4}\)

\(\left\{{}\begin{matrix}x+my=2\left(1\right)\\mx-2y=1\left(2\right)\end{matrix}\right.\)

thay pt (1) vào pt (2) ta duoc:\(\left\{{}\begin{matrix}x+my=2\\mx-\left(x+my\right)y=1\left(3\right)\end{matrix}\right.\)

PT (3) tương đương: \(mx-y^2m-yx-1=0\)

<=>\(-y^2m-yx+mx-1=0\)

\(\Delta=b^2-4ac=x^2-4.\left(-m\right).\left(mx-1\right)=x^2+4m^2x-4m\)

theo Vi-ét ta có:\(\left\{{}\begin{matrix}S=\dfrac{-b}{a}=\dfrac{-x}{m}\\P=\dfrac{c}{a}=\dfrac{-mx+1}{m}\end{matrix}\right.\)

Để pt có hai nghiệm lớn hơn 0<=>\(\left\{{}\begin{matrix}\Delta>0\\S>0\\P>0\end{matrix}\right.\)hay \(\left\{{}\begin{matrix}x^2+4m^2x-4m>0\\\dfrac{-x}{m}>0\\\dfrac{-mx+1}{m}>0\end{matrix}\right.\)

tới chỗ này là tìm m được rồi.Chúc bạn học tốt

d/tìm Min:

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

=>Min D=-1.Dấu = xảy ra khi x=-2

TÌM Max:

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

=>Max D=4.Dấu = xảy ra khi x=\(\dfrac{1}{2}\)

các câu kia tương tự nha bạn.chúc bạn học tốt

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