a, \(A=2x^2-8x-10=2\left(x^2-4x+4\right)-18=2\left(x-2\right)^2-18\ge-18\)

Dấu "=" xảy ra <=> x-2=0 <=> x=2

Vậy MinA = -18 khi x=2

b, \(B=x-x^2=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Dấu "=" xảy ra <=> x-1/2=0 <=> x=1/2

Vậy MaxB = 1/4 khi x=1/2

a) \(A=2x^2-8x-10\)
\(=2\left(x^2-4x-5\right)\)

\(=2\left(x^2-2.x.2+2^2-2^2-5\right)\)
\(=2\left[\left(x-2\right)^2-9\right]\)
\(=2\left(x-2\right)^2-18\)

Vì \(2\left(x-2\right)^2\ge0\forall x\)

Nên \(2\left(x-2\right)^2\ge-18\)

Hay \(A\ge-18\)

Vậy gtnn của A là -18 khi \(2\left(x-2\right)^2=0\)

\(x-2=0\)

\(x=2\)

b) \(B=x-x^2\)

\(=-x^2-x\)

\(=-\left(x^2-x\right)\)

\(=-\text{[}x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\text{]}\)

\(=-\text{[}\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\text{]}\)

\(=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\)

Vì \(-\left(x-\frac{1}{2}\right)^2\le0\forall x\)
Nên \(-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x \)
Vậy gtln của B là \(\frac{1}{4}\)khi \(x-\frac{1}{2}=0\)

\(x=\frac{1}{2}\)

Sửa đề cho x/y-z + y/z-x + z/x-y =0,tính Q=x/(y-z)^2 + y/(z-x)^2 + z/(x-y)^2

Ta có: \(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}=0\Rightarrow\frac{x}{y-z}=-\left(\frac{y}{z-x}+\frac{z}{x-y}\right)\)

\(\Rightarrow\frac{x}{y-z}=\frac{y}{x-z}+\frac{z}{y-x}=\frac{y^2-xy+xz-z^2}{\left(x-y\right)\left(z-x\right)}\)

\(\Rightarrow\frac{x}{\left(y-z\right)^2}=\frac{y^2-xy+xz-z^2}{\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)

Tương tự ta có: \(\frac{y}{\left(z-x\right)^2}=\frac{z^2-yz+yx-x^2}{\left(y-z\right)\left(z-x\right)\left(x-y\right)};\frac{z}{\left(x-y\right)^2}=\frac{x^2-zx+zy-y^2}{\left(z-x\right)\left(x-y\right)\left(y-z\right)}\)

Cộng ba đẳng thức trên vế theo vế, ta được:

\(\frac{x}{\left(y-z\right)^2}+\frac{y}{\left(z-x\right)^2}+\frac{z}{\left(x-y\right)^2}=\frac{y^2-xy+xz-z^2+z^2-yz+yx-x^2+x^2-zx+zy-y^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)

Vậy Q = 0

(x-1)(x+1) +x(x-9)=2x² -4

=>x²-1+x²-9x=2x²-4

=>2x²-2x²-9x=3

=>9x=3

=>x=\(\frac{1}{3}\)

Vậy \(x=\frac{1}{3}\)

\(\left(x-1\right)\left(x+1\right)+x\left(x-9\right)=2x^2-4.\)

\(x^2-1+x^2-9x=2x^2-4\)

\(\left(2x^2-2x^2\right)-1-9x=-4\)

\(-1-9x=-4\)

\(-9x=-3\)

\(\Rightarrow\)\(x=\frac{1}{3}\)

17 + 9 + 2005 = 2031

 Chúc bạn học tốt ^^

\(17+9+2005=2031\)

(2x³-4x²+3x+12)=(x+2).(2x²-8x+21)-30

MK ko bk đúng ko nhé