\(A=3\left(x+\frac{4}{3}\right)^2+\frac{146}{3}\ge\frac{146}{3}\)

\(A_{min}=\frac{146}{3}\) khi \(x=-\frac{4}{3}\)

\(B=-\left(x-2\right)^2-5\le-5\)

\(B_{max}=-5\) khi \(x=2\)

\(a,\Leftrightarrow\left(x^2-8x+16\right)-10=0\\ \Leftrightarrow\left(x-4\right)^2-10=0\\ \Leftrightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\\ b,\Leftrightarrow10\left(2x-1\right)+6x=9x\\ \Leftrightarrow20x-10-3x=0\\ \Leftrightarrow17x=10\Leftrightarrow x=\dfrac{10}{17}\)

a)g(x)=0=>11x3+5x2+4x+10=0

=>(10x3+10)+(x3+x2)+(4x2+4x)=0

=>10(x3+1)+x2(x+1)+4x(x+1)=0

=>10(x+1)(x2−x+1)+x2(x+1)+4x(x+1)=0

=>(x+1)[(10(x2−x+1)+x2+4x]=0

=>(x+1)(11x2−6x+10)=0

=>(x+1)[(9x2−2.3x+1)+2x2+9]=0

=>(x+1)[(3x−1)2+2x2+9]=0

=>x+1=0

=>x=-1

Vậy x=-1

a) Thay đa thức này bằng 0, ta được: 

f(x) = x^3 - x^2 + x - 1 = 0

=> f(x) = x . x² - x . x + x - 1 = 0

=> f(x) = x. (x² - x + x) = 0 + 1 = 1

=> f(x) = x . x² = 1

=> x = 1   và    x² = 1

=> x = 1

Vậy nghiệm của đa thức là x = 1

x = 2 hoặc = 6

Cách làm:

x² - 8x + 12 = 0

x² - 6x - 2x + 12 = 0

( x² - 6x ) - ( 6x - 12 ) = 0

x . ( x - 2 ) - 6 . ( x - 2 ) = 0

( x - 2 ) . ( x - 6 ) = 0

\(\Rightarrow\hept{\begin{cases}x-2=0\\x-6=0\end{cases}}\hept{\begin{cases}x=2\\x=6\end{cases}}\)

x²-8x+12=0

<=> x²-6x-2x+12=0

<=> (x²-2x)-(6x-12)=0

<=> x.(x-2)-6.(x-2)=0

<=> (x-2)(x-6)=0

<=>\(\orbr{\begin{cases}x-2=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

\(=x^2-2x-6x+12\)

\(=x\left(x-2\right)-6\left(x-2\right)\)

\(=\left(x-2\right)\left(x-6\right)\)

\(a,\dfrac{1}{5}-\dfrac{1}{5}:x=\dfrac{3}{5}\\ \Rightarrow\dfrac{1}{5}:x=\dfrac{1}{5}-\dfrac{3}{5}\\ \Rightarrow\dfrac{1}{5}:x=\dfrac{-2}{5}\\ \Rightarrow x=\dfrac{1}{5}:\dfrac{-2}{5}\\ \Rightarrow x=\dfrac{-1}{2}\)

câu b thiếu đề

a: =>1/5:x=-2/5

hay x=-1/2