Bài 1:

a) 2x^2 -3x + 1 = 2x^2 -2x -x +1 = 2x.(x-1) - (x-1) = (x-1).(2x-1)

b) 2x^3y - 2xy^3 - 4xy^2 - 2xy = 2xy.(x^2 - y^2 - 2y -1) = 2xy.[ x^2 - (y^2 + 2y+1)] = 2xy.[x^2 - (y+1)^2]

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

c) (x^2 + x+3).(x^2 + x +5) - 8 = (x^2+x+4-1).(x^2+x+4+1) - 8 = (x^2+x+4)^2 - 1 - 8 = (x^2+x+4)^2 - 3^2

= (x^2+x+4-3).(x^2+x+4+3) = (x^2+x+1).(x^2+x+7)

Bài 2:

a) (x+2).(x^2-2x+4) - (x^3+2x) = 0

x^3 + 8 - x^3 - 2x = 0

8 - 2x = 0

x = 4

b) x^2 - 2x - 8 = 0

x^2 +2x - 4x - 8 = 0

x.(x+2) - 4.(x+2) = 0

(x+2).(x-4) = 0

...

bn tự làm tiếp nha

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

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

\(3x^2-2x+1=3x^2+15x+12\)

\(\Rightarrow3x^2-2x+1-3x^2-15x-12=0\)

\(\Rightarrow-17x=11\)

\(\Rightarrow x=-\frac{11}{17}\)

\(b,M=x^2+12x+50\)

\(M=x^2+2.6.x+6^2+14\)

\(M=\left(x+6\right)^2+14\ge14>0\)

=> M luôn dương 

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

\(\Leftrightarrow x^2+2x+1+2x^2-4x=3.(x^2+x+4x+4)\)

\(\Leftrightarrow x^2-2x+2x^2+1=3x^2+15x+12\)

\(\left(x^2-3x^2+2x^2\right)=\left(15x+2x\right)+12-1\)

\(17x+11=0\)

\(\Leftrightarrow x=\frac{-11}{17}\)

\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)

\(x^2-2x+1+\left(\sqrt{3}y\right)^2+2.6.y+\left(2\sqrt{3}\right)^2+\left(\sqrt{2}z\right)^2+2.2.z+\left(\sqrt{2}\right)^2=0\)

\(\left(x-1\right)^2+\left(\sqrt{3}y+2\sqrt{3}\right)^2+\left(\sqrt{2}z+\sqrt{2}\right)^2=0\)

\(\Rightarrow x=1;y=-2;z=-1\)

<=>(x²-2x+1)+(3y²+12y+12)+(2z²+4z+2)=0

<=>(x-1)²+3(y+2)²+2(z+1)²=0

Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\3\left(y+2\right)^2\ge0\\2\left(z+1\right)^2\ge0\end{cases}\Rightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2\ge0}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-1=0\\y+2=0\\z+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\\z=-1\end{cases}}}\)

b đâu?

a) x2 - 7x + 5 = ( x2 - 2 . 7/2 . x + 49 / 4 ) + 5 - 49 / 4 
= (x - 7/2)^2 - 29/4
= (x - 7/2)^2 - (√ 29 / 2 )^2
= ( x - ( 7 + √ 29 / 2 )). ( x + ( 7 - √ 29 / 2 ))

giúp tui đi huhu TT 

Đặt \(f\left(x\right)=x^3-2x^2-6x+a\)

Gọi thương của \(f\left(x\right):\left(x-2\right)\)là \(P\left(x\right)\)

\(\Rightarrow f\left(x\right)=P\left(x\right).\left(x-2\right)\)

Thay \(x=2\)ta có: 

\(8-8-12+a=0\)

\(\Rightarrow a=12\)

Vậy \(a=2\)là giá trị cần tìm