=>a^2+b^2+c^2+3-2a-2b-2c=0

=>(a^2-2a+1)+(b^2-2b+1)+(c62-2c+1)=0

=>(3 hằng dẳng thức của a-1 b-1 c-1)

Suy ra (a-1)^2=0

và (b-1)^2=0

và(c-1)^2=0

thay vào A suy ra A=0

cố gắng trình bày lại nhé bạn!

Câu a :

\(x^2-2x-3=0\)

\(\Leftrightarrow x^2-x+3x-3=0\)

\(\Leftrightarrow x\left(x-1\right)+3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\x+3=0\Rightarrow x=-3\end{matrix}\right.\)

Câu b :

\(2x^2+3=-5x\)

\(\Leftrightarrow2x^2+3+5x=0\)

\(\Leftrightarrow2x^2+2x+3x+3=0\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\Rightarrow x=-1\\2x+3=0\Rightarrow x=-\dfrac{3}{2}\end{matrix}\right.\)

Mấy câu sau khó quá ko bt làm :)

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

\(=-3x.\left(x^2+2.x.2+2^2\right)+\left(x^2+x+3x-3\right).\left(x+1\right)-\left(2x\right)^2-2.2.x.\left(-3\right)+\left(-3\right)^2\)

\(=-3x.\left(x^2+4x+4\right)+\left(x^2+\left(x+3x\right)-3\right).\left(x+1\right)-4x+12x+9\)

\(=-3x.\left(x^2+4x+4\right)+\left(x^2+4x-3\right)\left(x+1\right)-4x+12x+9\)

\(=-3x^3-12x^2-12x+x^3+4x^2-3x+x^2+4x-3-4x+12x+9\)

\(=\left(-3x^3-x^3\right)+\left(-12x^2+4x^2+x^2\right)+\left(-12x-3x+4x-4x+12x\right)+\left(-3+9\right)\)

\(=-2x^3-7x^2-3x+6\)

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

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

\(=\left(x.x+x.3-3.x+\left(-3\right).3\right)\left(x+2\right)-\left(x.x^2+x.\left(-3\right)-1.x^2+\left(-1\right).\left(-3\right)\right)-5x.x+\left(-5x\right).4-x^2-2x5+5^2\)

\(=\left(x^2+3x-3x-9\right)\left(x+2\right)-x^3-3x-x^2+3-5x^2-20x-x^2-10x+25\)

\(=\left(x^2+\left(3x-3x\right)-9\right)\left(x+2\right)-x^3-3x-x^2+3-5x^2-20x-x^2-10x+25\)

\(=\left(x^2-9\right)\left(x+2\right)-x^3-3x-x^2+3-5x^2-20x-x^2-10x+25\)

\(=x^3+2x^2-9x-15-x^3-3x-x^2+3-5x^2-20x-x^2-10x+25\)

\(=\left(x^3-x^3\right)+\left(2x^2-x^2-5x^2-x^2\right)+\left(-9x-3x-20x-10x\right)+\left(-18+3+25\right)\)

\(=-5x^2-42x+10\)

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

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

B: ko hiểu đề@@

a) (3x-2)-(5x+3)=(x+4)-(x-1)

<=>3x-2-5x-3=x+4-x+1

<=>3x-5x-x+x=4+1+2+3

<=>-2x=10

<=>x=-5

Vậy S={-5}

b)2x²+9x=5

<=>2x²+9x-5=0

<=>2x²-x+10x-5=0

<=>x(2x-1)+5(2x-1)=0

<=>(2x-1)(x+5)=0

<=>2x-1=0 hoặc x+5=0

<=>x=0,5 hoặc x=-5

Vậy S={0,5;-5}

a) Cậu xem lại đề đi 

b) \(3x.\left(x-2\right)-5x.\left(1-x\right)-8.\left(x^2-3\right)=4\)\(\Leftrightarrow3x^2-6x-5x+5x^2-8x^2+24-4=0\Leftrightarrow-11x+20=0\Leftrightarrow-11x=-20\Leftrightarrow x=\frac{20}{11}\)

c) \(2x^2+3.\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\Leftrightarrow2x^2+3\left(x^2-1\right)-5x\left(x+1\right)=0\)

\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\Leftrightarrow-5x=3\Leftrightarrow x=-\frac{3}{5}\)

Trần Anh: Cảm ơn bạn nhiều nhé :)) Phần a đúng là có sai đề pạn ạ mik làm hoài mà cux ko ra hì hì !!~~ Dù sao mik cux cảm ơn pạn nhiều nhiều nhé :3