2/

a,Ta có: a+b+c=0

<=>(a+b+c)²=0

<=>a²+b²+c²+2(ab+bc+ca)=0

<=>2+2(ab+bc+ca)=0

<=>ab+bc+ca=\(\frac{-2}{2}=-1\)

<=>(ab+bc+ca)²=1

<=>a²b²+b²c²+c²a²+2abc(a+b+c)=1

<=>a²b²+b²c²+c²a²=1 (vì a+b+c=0)

Lại có: a²+b²+c²=2

<=>(a²+b²+c²)²=4

<=>a⁴+b⁴+c⁴+2(a²b²+b²c²+c²a²)=4

<=>a⁴+b⁴+c⁴+2=4 (vì a²b²+b²c²+c²a²=1)

<=>a⁴+b⁴+c⁴=2

b, tương tự a

1/

b, \(B=9x^2-6x+2=9x^2-6x+1+1=\left(3x-1\right)^2+1\)

Vì \(\left(3x-1\right)^2\ge0\Rightarrow B=\left(3x-1\right)^2+1\ge1\)

Dấu "=" xảy ra khi x=1/3

Vậy Bmin = 1 khi x = 1/3

c,\(C=x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow C=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra khi x=-1/2

Vậy...

d, \(D=2x^2+2x+1=2\left(x^2+x+\frac{1}{2}\right)=2\left(x^2+x+\frac{1}{4}+\frac{1}{4}\right)=2\left(x+\frac{1}{2}\right)^2+\frac{1}{2}\)

Vì \(2\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow D=2\left(x+\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)

Dấu "=" xảy ra khi x=-1/2

Vậy...

\("="\Leftrightarrow x=1;y=3\)

Bạn thêm vào dòng cuối nhé :v Mình quên ghi :v

\(\frac{-7x^2+42x-64}{x^2-6x+10}\)

\(\Rightarrow7+\frac{6}{\left(x-3\right)^2+1}=y^2+2y\)

\(\Rightarrow\frac{6}{\left(x-3\right)^2+1}=\left(y-1\right)^2+6\)

\(\Rightarrow6=\left[\left(y-1\right)^2+6\right]\left[\left(x-3\right)^2+1\right]\)

\(\Rightarrow0=\left(y-1\right)^2\left(x-3\right)^2+6\left(x-3\right)^2+\left(y-1\right)^2\)

ta có 

áp dụng bất đẳng thức Bunhia : 

\(P^2=\left(x+y+2z\right)^2\le\left(1^2+1^2+2^2\right)\left(x^2+y^2+z^2\right)=6\times6=36\)

Do đó \(-6\le P\le6\text{ nên GTNN P= - 6, GTLN P =6}\)

\(\left(1+\frac{b^2+c^2-a^2}{2bc}\right).\frac{1+\frac{a}{b+c}}{1-\frac{a}{b+c}}.\frac{b^2+c^2-\left(b-c\right)^2}{a+b+c}\)

= \(\left(1+\frac{\left(b+c\right)^2-2bc-a^2}{2bc}\right).\frac{\frac{a+b+c}{b+c}}{\frac{b+c-a}{b+c}}.\frac{\left(b+c\right)^2-2bc-\left(b-c\right)^2}{a+b+c}\)

= \(\left(1+\frac{\left(b+c-a\right)\left(b+c+a\right)-2bc}{2bc}\right).\frac{a+b+c}{b+c-a}.\frac{\left(b+c-b+c\right)\left(b+c+b-c\right)-2bc}{a+b+c}\)

= \(\left(1+\frac{\left(b+c-a\right)\left(b+c+a\right)}{2bc}-1\right).\frac{a+b+c}{b+c-a}.\frac{4bc-2bc}{a+b+c}\)

= \(\frac{\left(b+c-a\right)\left(b+c+a\right)}{2bc}.\frac{2bc}{b+c-a}\)

= \(\frac{\left(b+c-a\right)\left(b+c+a\right)}{b+c-a}\)

= \(b+c+a\)