d) \(\left(2x+1\right)^2-2\left(x-3\right)\left(2x+1\right)+\left(x-3\right)^2=\left[\left(2x+1\right)-\left(x-3\right)\right]^2=\left(x+4\right)^2\)

e) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{128}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{128}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{128}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)...\left(2^{128}+1\right)\)

\(=...=\left(2^{128}-1\right)\left(2^{128}+1\right)=2^{256}-1\)

\(Q=-x^2+4x-10\)

\(=-x\left(x+4\right)-10\)

Ta có : 

\(-x\left(x+4\right)< 0\text{ với mọi x }\)

\(-10< 0\)

\(\Rightarrow-x\left(x+4\right)-10< 0\text{ với mọi x }\)

\(\Rightarrow Q< 0\text{ với mọi x }\)

=> Q luôn âm với mọi x 

\(Q=-x^2+4x-10=-\left(x^2-4x+4\right)-6\)

\(=-\left(x-2\right)^2-6< 0\)

Vậy Q luôn âm

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=2x+2\\x-2=-2x-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=0\end{cases}}\)

Vậy x = 0 và x = -4

a/6x² - 9xy

= 3x . 2x - 3x . 3y

= 3x(2x-3y)

\(b,y^2+10y+25\)

\(=y^2+5y+5y+25\)

\(=y\left(y+5\right)+5\left(y+5\right)\)

\(=\left(y+5\right)\left(y+5\right)\)

\(=\left(y+5\right)^2\)

a) Ta có: M+(5x2−2xy)=6x2+9xy−y2M+(5x2−2xy)=6x2+9xy−y2

⇔M=6x2+9xy−y2−5x2+2xy⇔M=6x2+9xy−y2−5x2+2xy

⇔M=x2+11xy−y2⇔M=x2+11xy−y2

Vậy: M=x2+11xy−y2

b) Ta có: (3xy−4y2)−N=x2−7xy+8y2(3xy−4y2)−N=x2−7xy+8y2

⇔N=3xy−4y2−x2+7xy−8y2⇔N=3xy−4y2−x2+7xy−8y2

⇔N=−x2+10xy−12y2⇔N=−x2+10xy−12y2

Vậy: N=−x2+10xy−12y2

TL

TL

a) ( 6x²-9x ) y                                                                         b) y= -5

HT

Vào đây xem nha 

https://h7.net/hoi-dap/toan-8/phan-h-da-thuc-4x-2-2x-3y-9y-2-thanh-nhan-tu-faq289539.html

hoctot

\(x^2-4x=-4\)

\(x^2-4x+4=0\)

\(x^2+4-2x-2x=0\)

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

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

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

\(x-2=0\)

\(x=2\)