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

=(2xy)²-(x²+y²)²

=[2xy-(x²+y²)][2xy+(x²+y²)]

=(2xy-x²-y²)(2xy+x²+y²)

=-(-2xy+x²+y²)(x+y)²

=-(x-y)²(x+y)²

c) x^2+2x-3

=x²-x+3x-3

=x(x-1)+3(x-1)

=(x-1)(x+3)

d) 2a^3-54b^3

=2(a³-27b³)

=2[a³-(3b)³]

=2(a-3b)(a²+3ab+9b²)

\(a,\Rightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\\ \Rightarrow26x=26\Rightarrow x=1\\ b,\Rightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\\ \Rightarrow39x=-39\Rightarrow x=-1\)

a) \(\left(2x+3\right)\left(y-1\right)=54\) 

\(\Rightarrow2x+3,y-1\inƯ\left(54\right)\)

Ta có bảng sau: 

Vậy: ... 

câu 1:

a,x²+2x-4z²+1

=x²+2x.1+1²-(2z)²

=(x+1)²-(2z)²

=(x+1-2z)(x+1+2z)

bạn nên dùng hằng đẳng thức đã học

a) \(x^2+2x-4z^2+1=\left(x^2+2x+1\right)-4z^2\)

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

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

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

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

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

c)

a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2=\dfrac{x.y}{2.3}=\dfrac{54}{6}=9\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=36\\y^2=81\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm6\\y=\pm9\end{matrix}\right.\)

b) \(\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\left(\dfrac{x}{5}\right)^2=\left(\dfrac{y}{3}\right)^2=\dfrac{x^2-y^2}{5^2-3^2}=\dfrac{4}{16}=\dfrac{1}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{25}{4}\\y^2=\dfrac{9}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{5}{2}\\y=\pm\dfrac{3}{2}\end{matrix}\right.\)

c: Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\)

nên \(\dfrac{x}{10}=\dfrac{y}{15}\)

Ta có: \(\dfrac{y}{5}=\dfrac{z}{7}\)

nên \(\dfrac{y}{15}=\dfrac{z}{21}\)

mà \(\dfrac{x}{10}=\dfrac{y}{15}\)

nên \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{92}{46}=2\)

Do đó: x=20; y=30; z=42

a) $a^2-b^2-4a+4$

$\mathrm{=(a2-4a+4)-b2}$

$\mathrm{=(a-2)2-b2}$

$\mathrm{=(a-2-b)(a-2+b)}$

b) $x^2+2x-3$

$\mathrm{=x2-x+3x-3}$

$\mathrm{=x(x-1)+3(x-1)}$

$\mathrm{=(x+3)(x-1)}$

c) $4x^2{y}^2-{\left(x^2+y^2\right)}^2$

=(2xy-x²-y²)(2xy+x²+y²)

=-(x-y)²(x+y)²

d) $2a^3-54b^3$

=2(a³-27b³)

=2(a-3b)(a2+3ab+9b2)

Phân tích đa thức thành nhân tử

a, a²-b² -4a+4

\(=\left(a-2\right)^2-b^2\)

\(=\left(a-b-2\right)\left(a-b+2\right)\)

b, x²+2x-3

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

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

\(=\left(x+3\right)\left(x-1\right)\)

c,4x²y² - (x²+y²)²

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

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

d,2a³-54b³

^{\(=2\left(a^3-27b^3\right)=2\left(a-3b\right)\left(a^2+3ab+9b^2\right)\)}