\(A=2x^2-5xy-3y^2\)

\(A=2x^2-6xy+xy-3y^2\)

\(A=2x\left(x-3y\right)+y\left(x-3y\right)\)

\(A=\left(x-3y\right)\left(2x+y\right)\)

mk dùng p2 tách hạng tử ko bt đúng ko nha

A = x^2 - 4x -5

A = x^2 - 4x - x -4

A = x ( x - 1 ) - 4( x - 1)

A = ( x -1 ) . (x - 4)

câu b chờ mk suy nghĩ

4x-5 chứ có phải là 5x-4 dau ban

\(A=x^2-4x-5=x^2-4x+4-9=\left(x-2\right)^2-3^2\)

\(=\left(x-2-3\right)\left(x-2+3\right)=\left(x-5\right)\left(x+1\right)\)

\(B=4x^2-4x-15=4x^2-4x+1-16=\left(2x-1\right)^2-4^2\)

\(=\left(2x-1-4\right)\left(2x-1+4\right)=\left(2x-5\right)\left(2x+3\right)\)

Cách 2:

\(A=x^2-4x-5=x^2-5x+x-5=x\left(x-5\right)+\left(x-5\right)\)

\(=\left(x-5\right)\left(x+1\right)\)

\(B=4x^2-4x-15=4x^2-10x+6x-15=2x\left(2x-5\right)+3\left(2x-5\right)\)

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

\(=x^3-3x^2+6x^2-18x+8x-24\\ =\left(x-3\right)\left(x^2+6x+8\right)\\ =\left(x-3\right)\left(x^2+2x+4x+8\right)\\ =\left(x-3\right)\left(x+2\right)\left(x+4\right)\)

\(x^3+3x^2-10x-24=\left(x^3-3x^2\right)+\left(6x^2-18x\right)+\left(8x-24\right)=x^2\left(x-3\right)+6x\left(x-3\right)+8\left(x-3\right)=\left(x-3\right)\left(x^2+6x+8\right)=\left(x-3\right)\left[\left(x^2+2x\right)+\left(4x+8\right)\right]=\left(x-3\right)\left[x\left(x+2\right)+4\left(x+2\right)\right]=\left(x-3\right)\left(x+2\right)\left(x+4\right)\)

mình viết đề bị sai ạ -5z²

ko ghi lại đề nha: 

=(5x²-10x)+5y²-5z

=5(x-2)+5y²-5z

mk chỉ làm đc đến đây thôi

\(5x^2-10x+5y^2-5z^2\)

\(=5\left(x^2-2x+y^2-z^2\right)\)

jup mình với mọi người

júp mink với ace ơi

a) \(-10x^2-17x+6\)

\(=-10x^2-20x+3x+6\)

\(=-10x\left(x+2\right)+3\left(x+2\right)\)

\(=\left(x+2\right)\left(3-10x\right)\)

b)\(x^2-10x+9\)

\(=x^2-x-9x+9\)

\(=x\left(x-1\right)-9\left(x-1\right)\)

\(=\left(x-1\right)\left(x-9\right)\)

c) \(x^2-10x+24\)

\(=x^2-4x-6x+24\)

\(=x\left(x-4\right)-6\left(x-4\right)\)

\(=\left(x-4\right)\left(x-6\right)\)

a: \(4x^2-x-5=\left(4x-5\right)\left(x+1\right)\)

b: \(x^2-2x-15=\left(x-5\right)\left(x+3\right)\)

x^2 - 10x + 24

= x^2  - 4x - 6x + 24 

= x(x - 4) - 6(x - 4)

= (x - 6)(x - 4)

ko vt lại đề

x²-6x-4x+24

=(x²-6x)-(4x-24)

=x(x-6)-4(x-6)

=(x-6)(x-4)

\(x^2-10x+24\)

\(\Leftrightarrow x^2-4x-6x+24\)

\(\Leftrightarrow(x^2-4x)-\left(6x-24\right)\)

\(\Leftrightarrow x\left(x-4\right)-6\left(x-4\right)\)

\(\Leftrightarrow\left(x-4\right)\left(x-6\right)\)

a) \(-10x^3+2x^2=0\)

\(\Rightarrow-2x^2\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(5x\left(x-2016\right)-x+2016=0\)

\(\Rightarrow5x\left(x-2016\right)-\left(x-2016\right)=0\)

\(\Rightarrow\left(x-2016\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2016\\x=\dfrac{1}{5}\end{matrix}\right.\)

a: Ta có: \(-10x^3+2x^2=0\)

\(\Leftrightarrow-2x^2\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)