sửa đề a, \(\left(x+3\right)^2+\left(4-x\right)^2+\left(x+2\right)\left(x-2\right)\)

\(=x^2+6x+9+x^2-8x+16+x^2-4=3x^2-2x+21\)

b, \(\left(x-5\right)^2-\left(x+1\right)^2+\left(x+3\right)\left(x-3\right)\)

\(=x^2-10x+25-x^2-2x-1+x^2-9=x^2-12x+15\)

c, \(\left(x+4\right)^2-\left(x-6\right)^2-\left(x-1\right)\left(x+1\right)\)

\(=x^2+8x+16-x^2+12x-36-x^2+1=-x^2+20x-19\)

A+4=mc2

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Ta có hằng đẳng thức: 

\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

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

do đó \(\left(x-1\right)^3+\left(x-2\right)^3+\left(3-2x\right)^3=3\left(x-1\right)\left(x-2\right)\left(3-2x\right)\)

suy ra \(\left(x-1\right)\left(x-2\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x_1=1\\x_2=2\\x_3=\frac{3}{2}\end{cases}}\)

\(S=\frac{29}{4}\).

trước 8h55 tối nay ạ

  bạn nên qanda nha :3 đảm bảo là có <3

\(a)\)

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

\(b)\)

\(15x^3y^5-20x^4y^4-25x^5y^3=5x^3y^3\left(3y^2-4xy-5x^2\right)\)

Trả lời:

+) Rút gọn:

( x + y )² + ( x - y )² 

= x² + 2xy + y² + x² - 2xy + y²

= 2x² + 2y² 

+) Phân tích đa thức thành nhân tử:

( x + y )² + ( x - y )² 

= x² + 2xy + y² + x² - 2xy + y²

= 2x² + 2y² 

= 2 ( x² + y² )

sai hay đúng?