\(A=x^2+4x+3=\left(x^2+4x+4\right)-1\)

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

Dấu "=" xảy ra <=> x = -2

Vậy ...

\(a,\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)

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

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

Với \(x=\frac{1}{2}\)

\(\Rightarrow\frac{2\left(x-2\right)}{x+2}=\frac{2\left(\frac{1}{2}-2\right)}{\frac{1}{2}+2}=\frac{2.-\frac{3}{2}}{\frac{5}{2}}=-3.\frac{2}{5}=\frac{-6}{5}\)

b,Do x = -5; y = 10=> y = -2x

Thay y = -2x vào biểu thức ta được

\(\frac{x^3-x^2\left(-2x\right)+x\left(-2x\right)^2}{x^3+\left(-2x\right)^3}\)

\(=\frac{x^3+2x^3+2x^2}{x^3-8x^3}\)

\(=\frac{3x^3+2x^2}{-7x^3}=\frac{3}{-7}+\frac{2}{-7x}\)

Thay x = -5 là đc

\(8x^3+12x^2+6x+1=0.\)

\(\Leftrightarrow8x^2\left(x+\frac{1}{2}\right)+8x\left(x+\frac{1}{2}\right)+2\left(x+\frac{1}{2}\right)=0\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)\left(8x^2+8x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\2\left(4x^2+4x+1\right)=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\2\left(2x+1\right)^2=0\Leftrightarrow x=-\frac{1}{2}\end{cases}}\)

Vậy pt có 1 No là...

\(2\left(x+5\right)-x^2-5x=0.\)

\(\Leftrightarrow2x+10-x^2-5x=0\)

\(\Leftrightarrow x^2+3x-10=0\)

\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)

\(\ge\forall\le\)Hình như cho Icons :>>

\(\forall\)là dấu : với mọi 

Ví dụ : \(\forall x\)thì \(x^2\ge0\)

Thế nhá

\(x^3+x^2-x-1\)

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

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

\(=\left(x+1\right)\cdot\left(x-1\right)\cdot\left(x+1\right)\)

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

A = x³ + x² - x - 1

    = x²( x + 1 ) - ( x + 1 )

    = ( x + 1 ) ( x² - 1 )

    = ( x + 1 ) ( x - 1 ) ( x + 1 )

    = ( x + 1 )² ( x - 1 )

Chúc bạn học tốt nha!!!!!

-(-219)+(-219)-401+12

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x^2-8x+16=0

(x+4)^2=0

x+4=0

x=-4

x²-8x=-16

<=>\(x^2-8x+16=0\)

<=> \(\left(x-4\right)^2=0\)

<=>\(x-4=0\)

<=>\(x=4\)