1.a. \(\left(x-7\right)^2-x\left(x+25\right)=x^2-14x+49-x^2-25x\)

\(=-39x+49\)

b. \(\left(2x+5\right)^2-2x\left(2x-13\right)=4x^2+20x+25-4x^2+26x\)

\(=46x+25\)

c.\(\left(x+3\right)^2-\left(x+2\right)^2-3\left(x+1\right)\left(x-1\right)\)

\(=x^2+6x+9-x^2-4x-4-3x^2+3\)

\(=-3x^2+2x+8\)

 mơn nha!!!

không có j nè, câu 2 để minh2 suy nghỉ làm cho đúng cách đã ^.^

bn ơi chỗ khúc giữa của câu là trừ à hay là cộng(+) vậy mình thắc mắc á

lần lượt là +;+; -; +;+ bạn ah

Ta có:

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

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

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

\(=y^2+2yz+z^2-4xy-4xz+4x^2\) \(+z^2+2xz+x^2-4xy-4yz+4y^2+x^2\)

\(+2xy+y^2-4xz-4zy+4z^2\)

\(=6\left(x^2+y^2+z^2\right)-6\left(xy+yz+zx\right)\)

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

\(\Rightarrow\left(y+z-2x\right)^2+\left(z+x-2y\right)^2+\left(x+y-2z\right)^2\)\(-\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Rightarrow6\left(x^2+y^2+z^2\right)-6\left(xy+zx+yz\right)\)\(-\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

\(\Rightarrow6x^2+6y^2+6z^2-6xy-6zx-6yz\)\(-\left(x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2\right)\) =0

\(6x^2+6y^2+6z^2-6xy-6zx+6yz-\)\(\left(2x^2+2y^2+2z^2-2yz-2xy-2xz\right)=0\)

\(\Rightarrow4x^2+4y^2+4z^2-4xy-4zx-4yz=0\)

\(\Rightarrow2\left(x^2-2xy+y^2\right)+2\left(x^2-2xz+z^2\right)\)\(+2\left(y^2-2yz+z^2\right)=0\)

\(\Rightarrow2\left(x-y\right)^2+2\left(x-z\right)^2+2\left(y-z\right)^2=0\)

\(\Rightarrow2\left[\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2\right]=0\)

\(\Rightarrow\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x-y=0\\x-z=0\\y-z=0\end{cases}\Rightarrow\hept{\begin{cases}x=y\\x=z\\y=z\end{cases}}}\)

=> x = y = z

=> (đpcm)

thanks Bùi Minh Anh