\(4x+6y=2.\left(2x+3y\right)\)

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

\(xy.\left(x-3\right)-2.\left(3-x\right)=-xy.\left(3-x\right)-2.\left(3-x\right)=\left(3-x\right).\left(-xy-2\right)\)

\(x^2-4x+4-y^2\)

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

\(=\left(x-2\right)^2-y^2\)

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

\(x^2+4x-2xy-4y+y^2\)

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

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

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

18𝑥2−24𝑥+8−9𝑥2+4=0

18x^{2}-24x+{\color{#c92786}{8}}-9x^{2}+{\color{#c92786}{4}}=018x2−24x+8−9x2+4=0

18𝑥2−24𝑥+12−9𝑥2=0

𝑥=2𝑥=2/3

2(3x - 2)² - 9x² + 4 = 0

<=> 2(9x² - 12x + 4) - 9x² + 4 = 0

<=> 9x² - 24x + 12 = 0

<=> 9x² - 18x - 6x + 12  = 0

<=> 9x(x - 2) - 6(x - 2) = 0

<=> (9x - 6)(x - 2) = 0 

<=> \(\orbr{\begin{cases}x-2=0\\9x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{2}{3}\end{cases}}\)

TL

x=2/3

x=2

2(3𝑥−2)2−9𝑥2+4=0

2\left(3x-2\right)^{2}-9x^{2}+4=02(3x−2)2−9x2+4=0

2(3𝑥−2)(3𝑥−2)−9𝑥2+4=0

2(3𝑥−2)(3𝑥−2)−9𝑥2+4=0

2{\color{#c92786}{(3x-2)(3x-2)}}-9x^{2}+4=02(3x−2)(3x−2)−9x2+4=0

2(3𝑥(3𝑥−2)−2(3𝑥−2))−9𝑥2+4=0

x = 2 

x = 2/3

ko biết

Giải

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

\(=\left(x-y\right)^2-z^2\)(Sử dụng hằng đẳng thức \(A^2-2AB+B^2=\left(A-B\right)^2\))

\(=\left(x-y+z\right)\left(x-y-z\right)\)(Sử dụng hằng đẳng thức \(A^2-B^2=\left(A+B\right)\left(A-B\right)\))
Vậy \(\left(x^2-2xy+y^2\right)-z^2=\left(x-y+z\right)\left(x-y-z\right)\)

-x²(3-2x)=-3x²+2x³