Đáp án + Giải thích các bước giải:
`1)`
\(a)\\A=4x^2-12x+19=(4x^2-12x+9)+10\\=(2x-3)^2+10 \geq 10 \ >0\\\to đpcm\\b)\\B=9y^2+3y+2=(3y)^2+2. 3y . \dfrac{1}{2}+\dfrac{1}{4}+2-\dfrac{1}{4}\\=\Big(3y+\dfrac{1}{2}\Big)^2+\dfrac{7}{4} \geq \dfrac{7}{4} \ >0\\c)\\C=\dfrac{4}{9}x^2-\dfrac{4}{15}x+\dfrac{26}{25} \\=\dfrac{4}{9}x^2-2. \dfrac{2}{3} . \dfrac{1}{5}+\dfrac{1}{25}+\dfrac{26}{25}-\dfrac{1}{25}\\=\Big(\dfrac{2}{3}x-\dfrac{1}{5}\Big)^2+1 \geq 1 \ >0\\\to đpcm\)
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`2)`
\(a)\\D=-m^2+4m-5=-(m^2-4m+4)-1\\=-(m-2)^2-1 <0 \\ \to đpcm \\b)\\E=-n^2-6n-19=-(n^2+6n+9)-10\\=-(n+3)^2-10 <0 \\\to đpcm\)