Bài 8: Phân tích đa thức thành nhân tử bằng phương pháp nhóm các hạng tử

=\(-\left(x^2-2xy+y^2\right)+16=16-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x-y\right)\)

\(-x^2+2xy-y^2+16\)

\(=-\left(x^2-2xy+y^2-16\right)\)

\(=-\left(x-y-4\right)\left(x-y+4\right)\)

a² - 2ab + b² - ab - b² + a² + ab = 2a² -2ab = 2a(a - b)

\(A=x\left(z-y\right)-2\left(y-z\right)=\left(z-y\right)\left(x+2\right)=\left(-2016,7-2,345\right)\left(-2+2\right)=0\)

\(6xy-12x^2y-18xy^2\)

\(=6xy\left(1-2x-3y\right)\)

Đề bài tính GTNN hay Phân tích đa thức thành nhân tử  hả bạn?

x²-3x=0

⇔x(x-3) = 0

⇔x=0 hoặc x-3=0⇒x=3

Vậy x∈0;3

a, x(x-3)=0

=>Th1:x=0

    Th2:x-3=0 => x=3

i, x³+3x²+9x+27

= x²(x+3)+9(x+3)

= (x+3)(x²+9)

a. x² - 3x = 0

<=> x(x - 3) = 0

<=> \(\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

b. (x² - 4x + 4) - x + 2 = 0

<=> (x - 2)² - (x - 2) = 0

<=> (x - 2 - 1)(x - 2) = 0

<=> (x - 3)(x - 2) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

c. 16x² - (9x² + 6x + 1) = 0

<=> 16x² - (3x + 1)² = 0

<=> (4x)² - (3x + 1)² = 0

<=> (4x - 3x - 1)(4x + 3x + 1) = 0

<=> (x - 1)(7x + 1) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\7x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{7}\end{matrix}\right.\)

d. 2x³ - 50x = 0

<=> 2x(x² - 25) = 0

<=> 2x(x - 5)(x + 5) = 0

<=> \(\left[{}\begin{matrix}2x=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

e. 7x(x - 2022) - x + 2022 = 0

<=> 7x(x - 2022) - (x - 2022) = 0

<=> (7x - 1)(x - 2022) = 0

<=> \(\left[{}\begin{matrix}7x-1=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=2022\end{matrix}\right.\)

f. 8x² - 4x = 0

<=> 4x(2x - 1) = 0

<=> \(\left[{}\begin{matrix}4x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)

g. (3x + 5)² = (2x - 1)²

<=> (3x + 5)² - (2x - 1)² = 0

<=> (3x + 5 - 2x + 1)(3x + 5 + 2x - 1) = 0

<=> (x + 6)(5x + 4) = 0

<=> \(\left[{}\begin{matrix}x+6=0\\5x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{-4}{5}\end{matrix}\right.\)

h. (x² + 2x + 1) - (4x² - 9) = 0

<=> x² + 2x + 1 - 4x² + 9

<=> -3x² + 2x + 10 = 0

<=> 3x² - 2x - 10 = 0

<=> \(3x^2-2.\sqrt{3}.x.\dfrac{\sqrt{3}}{3}+\dfrac{1}{3}-\dfrac{31}{3}=0\)

<=> \(\left(\sqrt{3}x-\dfrac{\sqrt{3}}{3}\right)^2=\dfrac{31}{3}\)

<=> \(\left[{}\begin{matrix}\sqrt{3}x-\dfrac{\sqrt{3}}{3}=\dfrac{\sqrt{93}}{3}\\\sqrt{3}x-\dfrac{\sqrt{3}}{3}=\dfrac{-\sqrt{93}}{3}\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}\sqrt{3}x=\dfrac{\sqrt{93}+\sqrt{3}}{3}\\\sqrt{3}x=\dfrac{-\sqrt{93}+\sqrt{3}}{3}\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{1+\sqrt{31}}{3}\\x=\dfrac{1-\sqrt{31}}{3}\end{matrix}\right.\)

i. x³ + 5x² + 4x + 20 = 0

<=> x²(x + 5) + 4(x + 5) = 0

<=> (x² + 4)(x + 5) = 0

<=> \(\left[{}\begin{matrix}x^2+4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=-4\left(Vlí\right)\\x=-5\end{matrix}\right.\)

dạng 2 ạ

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

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

x³ + 3x² + 3x + 9

= x²(x + 3) + 3(x + 3)

= (x² + 3)(x + 3)

\(x^3+3x^2+3x+9=x^2\left(x+3\right)+3\left(x+3\right)=\left(x^2+3\right)\left(x+3\right)\)