\(A=\left(x+3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=-3\\ B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{29}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{29}{4}\ge-\dfrac{29}{4}\\ B_{min}=-\dfrac{29}{4}\Leftrightarrow x=-\dfrac{3}{2}\\ C=\left(9x^2-12x+4\right)+2017=\left(3x-2\right)^2+2017\ge2017\\ C_{min}=2017\Leftrightarrow x=\dfrac{2}{3}\)

\(A=9x^2-6x+2=\left(9x^2-6x+1\right)+1\) 

\(=\left(3x-1\right)^2+1\) 

Với mọi giá trị của x , ta có:

\(\left(3x-1\right)^2\ge1\Rightarrow\left(3x-1\right)^2+1\ge1\) 

Vậy \(Min_A=1\) 

Để A = 1 thì \(3x-1=0\Rightarrow3x=1\Rightarrow x=\frac{1}{3}\) 

\(B=x^2-7x+11=\left(x^2-7x+\frac{49}{4}\right)-\frac{5}{4}\) 

\(=\left(x-\frac{7}{2}\right)^2-\frac{5}{4}\) 

Với moị giá trị của x , ta có:

\(\left(x-\frac{7}{2}\right)^2\ge0\Rightarrow\left(x-\frac{7}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\) 

Vậy \(Min_B=-\frac{5}{4}\)

Để B = \(-\frac{5}{4}\) thì \(x-\frac{7}{2}=0\Rightarrow x=\frac{7}{2}\) 

\(C=x^2+x+5=\left(x^2+x+\frac{1}{4}\right)+\frac{19}{4}\) 

\(=\left(x+\frac{1}{2}\right)^2+\frac{19}{4}\) 

Với mọi giá trị của x thì :

\(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\) 

Vậy : \(Min_C=\frac{19}{4}\) 

Để \(C=\frac{19}{4}\) thì \(x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\) 

\(D=\left(x-1\right)\left(x+2\right)+1=x^2+x-2+1\) 

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

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

Với mọi giá trị của x . ta có:

\(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\) 

Vậy \(Min_D=-\frac{5}{4}\) 

Để \(D=-\frac{5}{4}\) thì \(x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)

\(x^3-9x+7x^2-63=0\)

\(\Rightarrow\left(x^3+7x^2\right)-9x-63=0\)

\(\Rightarrow x^2\left(x+7\right)-9\left(x+7\right)=0\)

\(\Rightarrow\left(x^2-9\right)\left(x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2-9=0\\x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=9\\x=-7\end{cases}\Rightarrow}\orbr{\begin{cases}x=\pm3\\x=-7\end{cases}}}\)

Vậy ...

x3−9x+7x2−63=0x3−9x+7x2−63=0

⇒(x3+7x2)−9x−63=0⇒(x3+7x2)−9x−63=0

⇒x2(x+7)−9(x+7)=0⇒x2(x+7)−9(x+7)=0

⇒(x2−9)(x+7)=0⇒(x2−9)(x+7)=0

⇒{x2−9=0x+7=0⇒{x2=9x=−7⇒{x=±3x=−7⇒{x2−9=0x+7=0⇒{x2=9x=−7⇒{x=±3x=−7

Vậy ...

a: A=x^2-6x+9+2=(x-3)^2+2>=2

Dấu = xảy ra khi x=3

b: B=x^2-20x+100+1=(x-10)^2+1>=1

Dấu = xảy ra khi x=10

d: C=x^2-16x+8+3

=(x-4)^2+3>=3

Dấu = xảy ra khi x=4

a) \(9x^2-6x+3=0\)

\(\Leftrightarrow\left(3x\right)^2-2.3x.1+1^2+2=0\)

\(\Leftrightarrow\left(3x-1\right)^2=-2\) ( vô lí )

b) \(x^2-7x+12=0\)

\(\Leftrightarrow x^2-2.x.\frac{7}{2}+\left(\frac{7}{2}\right)^2-\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2=\frac{1}{4}=\left(-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{7}{2}=\frac{1}{2}\\x-\frac{7}{2}=-\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}\)

Vậy : \(x\in\left\{3,4\right\}\)

c) \(x^2-8x+6=0\)

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

\(\Leftrightarrow\left(x-4\right)^2=10\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=\sqrt{10}\\x-4=-\sqrt{10}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{10}+4\\x=-\sqrt{10}+4\end{cases}}\)

\(a.\frac{x^3-6x^2+12x-8+x^2-4x+4}{x-2}\)\(=\frac{\left(x-2\right)^3+\left(x-2\right)^2}{x-2}\)\(=2\left(x-2\right)^2\)

a)

A(x)= 5x^4 - 3 + 2x^2 - 6x + 7x^2 - x^4

A(x)= 4x^4 + 9x^2 - 6x - 3.

Bậc: 4.

B= -9x^2 + x - 3 - 4x^4 + 5x^3

B(x)= -4x^4 + 5x^3 - 9x^2 + x - 3

b)

N(x) = A(x) + B(x)= ( 4x^4 + 9x^2 - 6x - 3 ) + (-4x^4 + 5x^3 - 9x^2 + x - 3)

N(x)= 5x^3 - 5x - 6

M(x) = A(x) - B(x)= ( 4x^4 + 9x^2 - 6x - 3 ) - 

(-4x^4 + 5x^3 - 9x^2 + x - 3)

M(x)= 8x^4 - 5x^3 + 18x^2 - 7x.

\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)

\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)

Mina giúp Shino đây nè:3(lần lượt nhá)

Ta có:\(4x^2-4x+1=0\)

\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)

\(\Leftrightarrow\left(2x-1\right)^2=0\)

\(\Leftrightarrow2x=1\)

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

1/ f(x) = 4x² - 4x + 1

4x² - 4x + 1 = 0

=> 4x² + 2x + 2x + 1 = 0

=> 2x(2x + 1) + (2x + 1) = 0

=> (2x + 1)(2x + 1) = 0

=> (2x + 1)² = 0

=> 2x + 1 = 0

=> 2x = -1

=> x = -1/2

Vậy nghiệm của đa thức f(x) là x = -1/2

\(9x^2+6x+1=0\)

\(\Leftrightarrow\left(3x\right)^2+2\cdot3x\cdot1+1^2=0\)

\(\Leftrightarrow\left(3x+1\right)^2=0\)

\(\Leftrightarrow x=-\frac{1}{3}\)

Áp dụng 2 hằng đẳng thức:

\(a^2+2ab+b^2=\left(a+b\right)^2\)

\(a^2-2ab+b^2=\left(a-b\right)^2\)