\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

Bài 1: 

a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)

\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)

\(\Rightarrow16x-5=x-2\)

\(\Rightarrow16x-x=5-2\)

\(\Rightarrow15x=3\)

\(\Rightarrow x=\dfrac{15}{3}=5\)

b) \(12x^2-4x\left(3x+5\right)=10x-17\)

\(\Rightarrow12x^2-12x^2-20x=10x-17\)

\(\Rightarrow-20x=10x-17\)

\(\Rightarrow-20x-10x=-17\)

\(\Rightarrow-30x=-17\)

\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)

c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)

\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)

\(\Rightarrow-8x=12\)

\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)

Bài 2: 

a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)

\(=x^2-7x+5x-35-7x^2+21x\)

\(=-6x^2+19x-35\)

b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)

\(=x^3-x^2-2x-x^2+x-5x-5\)

\(=x^3-2x^2-6x-5\)

c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)

\(=x^2-7x-5x+35-x^2-3x+4x-12\)

\(=11x+23\)

d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)

\(=x^2-2x-x+2-x^2+2x+5x+10\)

\(=4x+12\)

a) 5.(x^2-3x+1)+x.(1-5x)=x-2

\(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)

\(\Leftrightarrow-14x-x=-2-5\)

\(\Leftrightarrow-15x=-7\)

\(\Leftrightarrow x=\frac{7}{15}\)

b\(,3x.\left(\frac{4}{3}+1\right)-4x\left(x-2\right)=10\)

\(\Leftrightarrow4x+3x-4x^2+8x-10=0\)

\(\Leftrightarrow-4x^2+15x-10=0\)

Đề sai???

\(c,12x^2-4x\left(3x-5\right)=10x-17\)

\(\Leftrightarrow12x^2-12x^2+20x-10x=-17\)

\(\Leftrightarrow10x=-17\)

\(\Leftrightarrow x=-\frac{17}{10}\)

\(d,4x\left(x-5\right)-7x\left(x-4\right)+3x^2=12\)

\(\Leftrightarrow4x^2-20x-7x^2+28x+3x^2=12\)

\(\Leftrightarrow8x=12\)

\(\Leftrightarrow x=\frac{3}{2}\)

a: =-x^2+6x-4

=-(x^2-6x+4)

=-(x^2-6x+9-5)

=-(x-3)^2+5<=5

Dấu = xảy ra khi x=3

b: =3(x^2-5/3x+7/3)

=3(x^2-2*x*5/6+25/36+59/36)

=3(x-5/6)^2+59/12>=59/12

Dấu = xảy ra khi x=5/6

c: \(=-\left(x-3\right)^2+2\left|x-3\right|\)

\(=-\left[\left(\left|x-3\right|\right)^2-2\left|x-3\right|+1-1\right]\)

\(=-\left(\left|x-3\right|-1\right)^2+1< =1\)

Dấu = xảy ra khi x=4 hoặc x=2

1.

\(G=\dfrac{2}{x^2+8}\le\dfrac{2}{8}=\dfrac{1}{4}\)

\(G_{max}=\dfrac{1}{4}\) khi \(x=0\)

\(H=\dfrac{-3}{x^2-5x+1}\) biểu thức này ko có min max

2.

\(D=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{3}{2}\)

\(D_{min}=\dfrac{3}{2}\) khi \(x=4\)

\(E=\dfrac{4x^4-x^2-1}{\left(x^2+1\right)^2}=\dfrac{-\left(x^4+2x^2+1\right)+5x^4+x^2}{\left(x^2+1\right)^2}=-1+\dfrac{5x^4+x^2}{\left(x^2+1\right)^2}\ge-1\)

\(E_{min}=-1\) khi \(x=0\)

\(G=\dfrac{3\left(x^2-4x+5\right)-5}{x^2-4x+5}=3-\dfrac{5}{\left(x-2\right)^2+1}\ge3-\dfrac{5}{1}=-2\)

\(G_{min}=-2\) khi \(x=2\)