`a.`\(A=x^2-3x+5=\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{4}+5=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

Dấu "=" `<=>x=3/2`

`b.`\(A=2x^2-10x+1=2\left(x^2-5x+\dfrac{1}{2}\right)=2\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{23}{2}=\left(x-\dfrac{5}{2}\right)^2-\dfrac{23}{2}\ge-\dfrac{23}{2}\)Dấu "=" `<=>x=5/2`

`c.`\(C=\left(2x-1\right)^2+\left(x+2\right)^2=4x^2-4x+1+x^2+4x+4=5x^2+5\ge5\)

Dấu "=" `<=>x=0`

`d.`\(\left(1-3x\right)^2+\left(5y+2\right)^4-1\ge-1\)

Dấu "=" `<=>` \(\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-\dfrac{2}{5}\end{matrix}\right.\)

`e.`\(E=5x^2-10x+4xy+y^2-10=\left(4x^2+4xy+y^2\right)+\left(x^2-10x+25\right)-35=\left(2x+y\right)^2+\left(x-5\right)^2-35\ge-35\)

Dấu "=" `<=>`\(\left\{{}\begin{matrix}x=5\\y=-10\end{matrix}\right.\)

`f.`\(F=\left(x^2-7x\right)\left(x^2-7x+12\right)\)

Đặt \(x^2-7x=t\)

`=>`\(F=t\left(t+12\right)\)

\(F=t^2+12t=\left(t^2+12t+36\right)-36=\left(t+6\right)^2-36\ge-36\)

Dấu "=" xảy ra \(\Leftrightarrow t=-6\)

\(\Rightarrow x^2-7x=-6\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)

(a+b+c)^2=3(ab+bc+ca)`

`<=>a^2+b^2+c^2+2(ab+bc+ca)=3(ab+bc+ca)`

`<=>a^2+b^2+c^2-(ab+bc+ca)=0`

`<=>a^2+b^2+c^2-ab-bc-ca=0`

`<=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0`

`<=>(a-b)^2+(b-c)^2+(a-c)^2=0`

`=>a=b=c`

`a.(x+y)^2+(x-y)^2`

`=x^2+2xy+y^2+x^2-2xy+y^2`

`=2x^2+2y^2`

`b.2(x-y)(x+y)+(x+y)^2+(x-y)^2`

`=2(x^2-y^2)+2x^2+2y^2`

`=2x^2-2y^2+2x^2+2y^2`

`=4x^2`

`c.(x-y+z)^2+(z-y)^2+2(x-y+z)(y-z)`

`=(x-y+z+y-z)^2`

`=x^2`

`a.`\(I=\dfrac{U}{R}=\dfrac{6}{15}=0,4A\)

`b.`\(\Rightarrow I=0,4+0,3=0,7A\)

\(U=I.R=0,7.15=10,5V\)

`a.`\(\left(3x+1\right)\left(2x-3\right)-\left(x+1\right)\left(6x-1\right)=0\)

`=>`\(6x^2-9x+2x-3-6x^2+x-6x+1=0\)

`=>`\(-12x-2=0\)

`=>-12x=2`

`=>x=-1/6`

`b.`\(\left(2x-1\right)\left(3x-4\right)-\left(x+1\right)\left(x-2\right)+\left(x-1\right)\left(2x-1\right)-7x\left(x-2\right)=0\)

\(\Rightarrow6x^2-8x-3x+4-x^2+2x-x+2+2x^2-x-2x+1-7x^2+14x=0\)

\(\Rightarrow x+7=0\)

`=>x=-7`

\(A=\dfrac{2017-2n}{8n-4}\)

\(A=-\dfrac{\left(2n-1\right)-2016}{4\left(2n-1\right)}\)

\(A=-\dfrac{1}{4}+\dfrac{2016}{4\left(2n-1\right)}\)

\(A=-\dfrac{1}{4}+\dfrac{504}{2n-1}\)

Để A có GTLN thì \(\dfrac{504}{2n-1}\) lớn nhất `=>2n-1` nhỏ nhất

`=>2n-1` là số nguyên dương nhỏ nhất

`=>2n-1=1` `=>n=1`

\(\Rightarrow A=-\dfrac{1}{4}+504=\dfrac{2015}{4}\)

Vậy \(Max_A=\dfrac{2015}{4}\) khi `n=1`

\(\left|x\right|=3\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

`@` Với `x=3`

\(A.B=\left(-2x^2+3x+5\right).\left(x^2-x+3\right)\)

        \(=\left(-2.3^2+3.3+5\right).\left(3^2-3+3\right)\)

        \(=-4.9\)

        `=-36`

`@`Với `x=-3`

\(A.B=\left(-2x^2+3x+5\right).\left(x^2-x+3\right)\)

        \(=\left[-2.\left(-3\right)^2+3.\left(-3\right)+5\right].\left[\left(-3\right)^2-\left(-3\right)+3\right]\)

       \(=-22.15\)

       \(=-330\)

\(P=4x^2-4xy+5y^2-4x+18y-52\)

\(P=\left(4x^2-4xy+y^2\right)+\left(4y^2+16y+16\right)-4x+2y-68\)

\(P=\left(2x-y\right)^2+\left(2y+4\right)^2-2\left(2x-y\right)-68\)

\(P=\left[\left(2x-y\right)^2-2\left(2x-y\right)+1\right]+\left(2y+4\right)^2-69\)

\(P=\left(2x-y+1\right)^2+\left(2y+4\right)^2-69\ge-69\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2x-y+1=0\\2y+4=0\end{matrix}\right.\)

                        \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)

Vậy \(Min_P=-69\) khi \(\left(x;y\right)=\left(-\dfrac{3}{2};-2\right)\)