\(a,=12x^2-4x-6x-2-x-3=12x^2-11x-5\\ b,=12x^2-9x-12x^2-4x+5=5-13x\\ c,=12x^3-4x^2-12x^3-12x^2+7x-3=-16x^2+7x-3\\ d,=\left(x^2-4\right)\left(x^2+4\right)=x^4-16\)

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)

\(\left(x-3\right)\left(x-1\right)-3\left(x-3\right)\)

\(=\left(x-3\right)\left(x-1-3\right)\)

\(=\left(x-3\right)\left(x-4\right)\)

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

\(=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)\)

\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)

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

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

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

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

Còn lại bạn tự làm nhá

Em lớp 5 thôi nhưng em nghĩ bài này nhân vế bình thường thôi ạ 

Ví dụ câu a : \(\left(2x^2+x-1\right)\left(-3x^2-7x-5\right)\)

\(=-6x^4-17x^3-14x^2+2x+5\)

Tương tự chị nhé 

Câu b : 

\(\left(2x^2-3xy+y^2\right)\left(x+y\right)\)

\(=\)\(\left(2x^2-3xy+y^2\right).x+\left(2x^2-3xy+y^2\right).y\)

\(=\)\(2x^3-3x^2y+xy^2+2x^2y-3xy^2+y^3\) 

Câu b : 

\(\left(2x^2-3xy+y^2\right)\left(x+y\right)\)

\(=\left(2x^2-3xy+y^2\right).x+\left(2x^2-3xy+y^2\right).y\)

\(=2x^3-3x^2y+xy^2+2x^2y-3xy^2+y^3\)

\(=2x^3+y^3-x^2y-2xy^2\)

Câu c : 

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

\(=\left(-3x^3-3x\right)\left(-2-x\right)\left(3-2x\right)\)

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

\(=18x^3+9x^4+18x+9x^2-12x^4-6x^5-12x^2-6x^3\)

\(=12x^3-3x^4-3x^2+18x-6x^5\) 

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

Bài 1:

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

\(=12x^2-4x-6x-2-x-3\)

\(=12x^2-11x-5\)

b) \(=\left(-2x^2-1xy+2y^2\right)\left(-1x^2y\right)\)

\(=\left[\left(-1x^2y\right)\left(-2x^2\right)\right]-\left[\left(-1x^2y\right).1xy\right]+\left[\left(-1x^2y\right).2y^2\right]\)

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

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

c) \(4x\left(3x^2-x\right)-\left(2x+3\right)^2\left(6x^2-3x+1\right)\)

\(=\left(4x.3x^2\right)-\left(4x.x\right)-\left[\left(2x\right)^2+2.2x.3+3^2\right]\left(6x^2-3x+1\right)\)

\(=12x^3-4x^2-\left(4x^2+12x+9\right)\left(6x^2-3x+1\right)\)

\(=12x^3-4x^2-\left[4x^2\left(6x^2-3x+1\right)+12x\left(6x^2-3x+1\right)+9\left(6x^2-3x+1\right)\right]\)

\(=12x^3-4x^2-\left[\left(24x^4-12x^3+4x^2\right)+\left(72x^3-36x^2+12x\right)+\left(36x^2-27x+9\right)\right]\)

\(=12x^3-4x^2-24x^4+12x^3-4x^2-72x^3+36x^2-12x-36x^2+27x-9\)

\(=-48x^3-8x^2-24x^4+15x-9\)

Bài 1:

a) \(12x^2-11x-5\)

b,c,d tương tự.