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

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

a, 7x - 14

= 7(x-2)

b, 2x - 2y + \(x^2\)- xy 

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

= 2(x-y) + x(x-y)

= (x-y)(2+x)

c, 6x + 12

= 6(x+2)

\(a,=7\left(x-2\right)\\ b,=2\left(x-y\right)+x\left(x-y\right)=\left(x+2\right)\left(x-y\right)\\ c,=6\left(x+2\right)\\ d,\text{Sai đề}\)

Hướng dẫn giải:

Lời giải của các bạn đều thỏa mãn yêu cầu đề bài là phân tích đa thức thành nhân tử

a; Đặt x+7=0

=>x=-7

b: Đặt x-4=0

=>x=4

c: Đặt -8x+20=0

=>-8x=-20

=>\(x=\frac{20}{8}=\frac52\)

d: Đặt \(x^2-100=0\)

=>(x-10)(x+10)=0

=>\(\left[\begin{array}{l}x-10=0\\ x+10=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=10\\ x=-10\end{array}\right.\)

e: Đặt \(4x^2-81=0\)

=>(2x-9)(2x+9)=0

=>\(\left[\begin{array}{l}2x-9=0\\ 2x+9=0\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=9\\ 2x=-9\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac92\\ x=-\frac92\end{array}\right.\)

f: Đặt \(x^2-7=0\)

=>\(x^2=7\)

=>\(x=\pm\sqrt7\)

g: Đặt \(x^2-9x=0\)

=>x(x-9)=0

=>\(\left[\begin{array}{l}x=9\\ x=0\end{array}\right.\)

h: Đặt \(x^3+3x=0\)

=>\(x\left(x^2+3\right)=0\)

mà \(x^2+3\ge3>0\forall x\)

nên x=0

\(3x^4-48\)

\(=\left(3x^4-6x^3\right)+\left(6x^3-12x^2\right)+\left(12x^2-24x\right)+\left(24x-48\right)\)

\(=3x^3\left(x-2\right)+6x^2\left(x-2\right)+12x\left(x-2\right)+24\left(x-2\right)\)

\(=\left(x-2\right)\left[\left(3x^3+6x^2\right)+\left(12x+24\right)\right]\)

\(=\left(x-2\right)\left[3x^2\left(x+2\right)+12\left(x+2\right)\right]\)

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

\(x^4-8x\)

\(=x\left(x^3-8\right)\)

\(=x\left[\left(x^3-2x^2\right)+\left(2x^2-4x\right)+\left(4x-8\right)\right]\)

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

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

b: 1/2x-4=0

=>1/2x=4

hay x=8

a: x+7=0

=>x=-7

e: 4x²-81=0

=>(2x-9)(2x+9)=0

=>x=9/2 hoặc x=-9/2

g: x²-9x=0

=>x(x-9)=0

=>x=0 hoặc x=9

a)\(x+7=0=>x=-7\)

b)\(\dfrac{1}{2}x-4=0=>\dfrac{1}{2}x=4=>x=8\)

c)\(-8x+20=0=>-8x=-20=>x=\dfrac{5}{2}\)

d)\(x^2-100=0=>x^2=100=>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)

a: x+7=0

nên x=-7

b: x-4=0

nên x=4

c: -8x+20=0

=>-8x=-20

hay x=5/2

d: x²-100=0

=>(x-10)(x+10)=0

=>x=10 hoặc x=-10

a) x +7 =0

=>x = -7

b) x - 4 =0=>x = 4

c) -8x + 20 = 0 =>-8x =-20 =>\(x=-\dfrac{20}{-8}=\dfrac{5}{2}\)

d)\(x^2-100=0=>x^2=100>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)