a) \(9y^2-4x^2+6x+9y\)

\(=\left(3y-2x\right)\left(3y+2x\right)+3\left(3y+2x\right)\)

\(=\left(3y+2x\right)\left(3y-2x+3\right)\)

b) \(x^3+4x^2-12x\)

\(=x\left(x^2+4x-12\right)\)

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

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

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

a)\(9y^2-4x^2+6x+9y=\left(9y^2-4x^2\right)+\left(6x+9y\right)=\left[\left(3y\right)^2-\left(2x\right)^2\right]+3\left(2x+3y\right)\)

\(=\left(3y-2x\right)\left(3y+2x\right)+3\left(3y+2x\right)=\left(3y+2x\right)\left[\left(3y-2x\right)+3\right]=\left(3y+2x\right)\left(3y-2x+3\right)\)

b)\(x^3+4x^2-12x=x^3-2x^2+6x^2-12x=\left(x^3-2x^2\right)+\left(6x^2-12x\right)\)

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

a) \(9x^2+6x-1=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}3x+1=\sqrt{2}\\3x+1=-\sqrt{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{2}-1}{3}\\\frac{-\sqrt{2}-1}{3}\end{cases}}\)

b) \(4x^3+4x^2+x=0\)

\(\Leftrightarrow x\left(4x^2+4x+1\right)=0\)

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

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

câu a bn thiếu vế phải kìa

Ta có: A = 2x² - 5x + 3 = 2(x² - 5/2x + 25/16) - 1/8 = 2(x - 5/4)² - 1/8 \(\le\)-1/8 \(\forall\)x

Dấu "=" xảy ra <=> x - 5/4 = 0 <=> x = 5/4

Vậy MinA = -1/8 <=> x = 5/4

\(A=2x^2-5x+3=2\left(x^2-\frac{5}{2}x+\frac{3}{2}\right)\)

\(=2\left(x^2-\frac{5}{2}x+\frac{25}{16}-\frac{1}{16}\right)\)

\(=2\left[\left(x-\frac{5}{4}\right)^2-\frac{1}{16}\right]\)

\(=2\left[\left(x-\frac{5}{4}\right)^2\right]-\frac{1}{8}\ge\frac{-1}{8}\)

i play free fire

a) Xét tứ giác AEMD có :

DÂE = 90° ; Góc ADM = 90° ; Góc AEM = 90°

\(\Rightarrow\)Tứ giác AEMD là hình chữ nhật ( theo định lí )

Bài 9:

Đặt f(x) = \(2x^3+ax+b\)

Vì f(x) = \(2x^3+ax+b\) chia cho x + 1 dư 6 và chia cho x - 2 dư 21 nên ta có:

\(\hept{\begin{cases}f\left(-1\right)=2\times\left(-1\right)^3-a+b=6\\f\left(2\right)=2\times2^3+2a+b=21\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-2-a+b=6\\16+2a+b=21\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-a+b=8\\2a+b=5\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-3a=3\\b=5-2a\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=-1\\b=7\end{cases}}\)

Vậy a = -1, b = 7