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\(A=3-x^2+2x=3-\left(x^2-2x\right)=3-\left(x^2-2x+1-1\right)\)
\(=3-\left[\left(x-1\right)^2-1\right]=3-\left(x-1\right)^2+1=4-\left(x-1\right)^2\le4\)
Dấu '=" xảy ra \(< =>\left(x-1\right)^2=0< =>x=1\)
Vậy MaxA=4 khi x=1
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\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\)
vì \(\left|\frac{3}{2}x+\frac{1}{9}\right|\ge0;\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0=>\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0\) (với mọi x,y)
Mà \(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\) (theo đề)
Nên \(\left|\frac{3}{2}x+\frac{1}{9}\right|=0=>\frac{3}{2}x=-\frac{1}{9}=>x=-\frac{2}{27}\)
\(\left|\frac{1}{5}y-\frac{1}{2}\right|=0=>\frac{1}{5}y=\frac{1}{2}=>y=\frac{5}{2}\)
Vậy...........
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