\(a,=3\left(x-5\right)-x\left(x-5\right)=\left(3-x\right)\left(x-5\right)\\ b,=7\left(x^2-2xy+y^2\right)=7\left(x-y\right)^2\\ c,=\left(x^2+y^2-2xy\right)\left(x^2+y^2+2xy\right)=\left(x-y\right)^2\left(x+y\right)^2\\ d,=\left(y^2-6y+9\right)-25x^2=\left(y-3\right)^2-25x^2=\left(y-5x-3\right)\left(y+5x-3\right)\)

7x2+15x-5=0

7*(x2+2.5x-5/7)=0

x2+2*x*1.25+1,5625-255/112=0

(x+1.25)2-255/112=0

+ )   x 3   +   7 x 2   +   12 x   +   4     =   x 3   +   6 x 2   +   x 2   +   12 x   +   8   –   4     =   ( x 3   +   6 x 2   +   12 x   +   8 )   +   ( x 2   –   4 )     =   ( x 3   +   3 . 2 . x 2   +   3 . 2 2 . x   +   2 3 )   +   ( x 2   –   4 )     =   ( x   +   2 ) 3   +   ( x   –   2 ) ( x   +   2 )     =   ( x   +   2 ) ( ( x   +   2 ) 2   +   x   –   2 )     =   ( x   +   2 ) ( x 2   +   4 x   +   x   –   2 )     =   ( x   +   2 ) ( x 2   +   5 x   +   2 )

Đáp án cần chọn là: A

a) Ta có: \(x^4-16x^2=0\)

\(\Leftrightarrow x^2\left(x^2-16\right)=0\)

\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

b) Ta có: \(x^8+36x^4=0\)

\(\Leftrightarrow x^4\left(x^4+36\right)=0\)

\(\Leftrightarrow x^4=0\)

hay x=0

c) Ta có: \(\left(x-5\right)^3-x+5=0\)

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

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)

d) Ta có: \(5\left(x-2\right)-x^2+4=0\)

\(\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

a:

=x(x²-y²-10x+25)

=x((x²-10x+25)-y²)

=x((x-5)²-y²)

=x(x-5-y)(x-5+y)

b

=>8x(x-5)-3(x-5)=0

=>(x-5)(8x-3)=0

x-5=0=>x=5 hoặc 8x-3=0=>x=3/8

a) ( x 2  – 4x + 1)( x 2  – 2x + 3).

b) ( x 2  + 5x – 1)( x 2  + x – 1).

\(5-7x^2=\left(\sqrt{5}\right)^2-\left(x\sqrt{7}\right)^2\)

\(=\left(\sqrt{5}-x\sqrt{7}\right)\left(\sqrt{5}+x\sqrt{7}\right)\)

\(3+4x=\left(\sqrt{3}\right)^2-\left(2\sqrt{x}\right)^2\) ( do x<0 )

\(=\left(\sqrt{3}-2\sqrt{x}\right)\left(3+2\sqrt{x}\right)\)

a) xy+3x-7y-21

=x(y+3)-7(x+3)

=(x-7)(y+3)

b)2xy-15-6x-5y

=2x(y-3)-5(-3+y)

=(2x-5)(y-3)

c)2x^2y+2xy^2-2x-2y

=2x(xy-1)+2y(xy-1)

=(2x+2y)(xy-1)

x(x+3)-5x(x-5)-5(x+3)

=(x-5)(x+3)-5x(x-5)

=(x-5)(x+3-5x)

Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn

a) xy+3x-7y-21=(xy+3x)-(7y+21)= x(y+3)-7(y+3)=(y+3)(x-7)

b)2xy-15-6x+5y=(2xy-6x)+(5y-15)=2x(y-3)+5(y-3)=(y-3)(2x+5)

c)2x^2y+2xy^2-2x-2y=2xy(x+y)-2(x+y)=2(x+y)(xy-1)

d) x(x+3)-5x(x-5)-5(x+3)=[x(x+3)-5(x+3)]-5x(x-5)=(x+3)(x-5)-5x(x-5)=(x-5)(x+3-5x)=(x-5)(3-4x)