Đáp án+Giải thích các bước giải:

$a)x^4-2x^3+x^2$

$=x^2(x^2-2x+1)$

$=x^2(x-1)^2$

Do $x^2≥0∀x$

$(x-1)^2≥0∀x$

$⇒ x^2(x-1)^2≥0∀x$

$b)x^2+4x+y^2-2y+5$

$=(x^2+4x+4)+(y^2-2y+1)$

$=(x+2)^2+(y-1)^2$

Do $(x+2)^2≥0∀x$

$(y-1)^2≥0∀x$

$⇒ (x+2)^2+(y-1)^2≥0∀x$

$c) a^2+10>6a$

$⇔ a^2+10-6a>0$

$⇔ a^2-6a+9+1>0$

$⇔ (a-3)^2+1>0$

Do $(a-3)^2≥0∀a$

$⇒ (a-3)^2+1≥1∀a$

$⇒  (a-3)^2+1>0$ (luôn đúng)

$d) a^2+1>a$

$⇔a^2+1-a>0$

$⇔ a^2-2.\dfrac{1}{2}a+\dfrac{1}{4}+\dfrac{3}{4}>0$

$⇔ (a-\dfrac{1}{2})^2+\dfrac{3}{4}>0$

Do $(a-\dfrac{1}{2})^2≥0∀a$

$⇒(a-\dfrac{1}{2})^2+\dfrac{3}{4}≥\dfrac{3}{4}∀a$

$⇒ (a-\dfrac{1}{2})^2+\dfrac{3}{4}>0$ (luôn đúng)

$e)3(a^2+b^2+c^2)≥(a+b+c)^2$
$⇔ 3a^2+3b^2+3c^2-(a+b+c)^2≥0$

$⇔ 3a^2+3b^2+3c^2-(a^2+b^2+c^2+2ab+2bc+2ac)≥0$

$⇔ 3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2bc-2ac≥0$

$⇔ 2a^2+2b^2+2c^2-2ab-2bc-2ac≥0$

$⇔ (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)≥0$

$⇔ (a-b)^2+(b-c)^2+(c-a)^2≥0$

Do $(a-b)^2≥0∀a,b$

$(b-c)^2≥0∀b,c$

$(c-a)^2≥0∀c,a$

$⇒ (a-b)^2+(b-c)^2+(c-a)^2≥0∀a,b,c$

$⇒ (a-b)^2+(b-c)^2+(c-a)^2≥0$ (luôn đúng)

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