Question

1) Phân tích đa thức thành nhân tử

a) a³ + a²b -a²c - abc

b) x² + 2xy +  y² - xz + yz

c) 4 - x²  - 2xy - y²

d ) x² - 2xy + y² - z²

2) Tìm x, biết:

a) 2x (x - 3 ) - ( 3 - x ) = 0

b) 3x ( x + 5 ) - 6( x + 5 ) = 0

c) x ( x - 1 ) + 2x - 2 = 0

2x³ + 3x² + 2x + 3 = 0

Answer 1

a)  \(a^3+a^2b-a^2c-abc=a^2\left(a+b\right)-ac\left(a+b\right)=a\left(a+b\right)\left(a-c\right)\)

b) mk chỉnh lại đề

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

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

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

Answer 2

ồ cuk dễ nhỉ

Nếu các bn thích thì ...........

cứ cho NTN này nhé !

 

Answer 3

a)  \(2x\left(x-3\right)-\left(3-x\right)=0\)

\(\Leftrightarrow\)\(\left(x-3\right)\left(2x+1\right)=0\)

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

Vậy...

b)   \(3x\left(x+5\right)-6\left(x+5\right)=0\)

\(\Leftrightarrow\)\(3\left(x+5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

Vậy...

c)  \(x\left(x-1\right)+2x-2=0\)

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

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

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

Vậy...

d)  \(2x^3+3x^2+2x+3=0\)

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

\(\Leftrightarrow\)\(\left(2x+3\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\)\(2x+3=0\)  (do  x² + 1 > 0 )

\(\Leftrightarrow\)\(x=-\frac{3}{2}\)

Vậy...