huhu giai giup minh di

\(3x=2y=>\frac{x}{2}\)\(=\frac{y}{3}\)

Ta có:\(x=2k;y=3k\)

Thay x;y trông phép tính trên ta được 

\(\left(2k+3k\right)^3\)\(-\left(x-y\right)^3\)\(=126\)

\(5k^3\)\(-\left(-1k\right)^3\)\(=126\)

\(5^3\)\(k^3\)\(+1k^3=126\)

\(125k^3\)\(+1k^3=126\)

\(k^3\)\(\left(125+1\right)\)\(=126\)

\(k^3\)\(126=126\)

\(k^3\)\(=126:126=1\)

\(=>k=1\)

\(x=2k=2.1=2\)

\(y=3k=3.1=3\)

\(=>x=2;y=3\)

bằng 1/4 nah

Chúc bạn học tốt !

\(A=\frac{1}{3}+3\left|x-\frac{1}{3}\right|\)

Áp dụng KT \(\left|x\right|\ge0\)\(\forall\)\(x\)

BG :

Ta thấy : \(\left|x-\frac{1}{3}\right|\ge0\)\(\forall\)\(x\); \(3\ge0\)

nên : \(3\left|x-\frac{1}{3}\right|\ge0\)\(\forall\)\(x\)

\(\Rightarrow\)\(\frac{1}{3}+3\left|x-\frac{1}{3}\right|\ge\frac{1}{3}+0\)\(\forall\)\(x\)

hay \(A\ge\frac{1}{3}\)\(\forall\)\(x\)

Dấu "=" xảy ra khi :

\(\Leftrightarrow\)\(\left|x-\frac{1}{3}\right|=0\)

\(\Leftrightarrow\)\(x-\frac{1}{3}=0\)

\(\Leftrightarrow\)\(x=\frac{1}{3}\)

Vậy GTNN của \(A=\frac{1}{3}\)đạt được khi \(x=\frac{1}{3}\)

A=1/3+3x[x-1/3]

=>1/3+3x[x-1/3]=0

            3x[x-1/3]=1/3

                 x-1/3=1/3:3

                       x=1/9+1/3

                       x=4/9         

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