a) với a = -2 ta được phương trình:

3.[(-2) - 2].x + 2.(-2).(x - 1) = 4.(-2) + 3

<=> 3.(-4x) - 4.(x - 1) = (-8) + 3

<=> -12x - 4(x - 1) = -5

<=> -12x - 4x + 4 = -5

<=> -16x + 4 = -5

<=> -16x = -5 - 4

<=> -16x = -9

<=> x = 9/16

b) để x = 1, ta có:

3.(a - 2).1 + 2a(1 - 1) = 4a + 3

<=> 3(a - 2) + 0 = 4a + 3

<=> 3a - 6 = 4a + 3

<=> 3a - 6 - 4a = 3

<=> -a - 6 = 3

<=> -a = 3 + 6

<=> a = -9

sin³x + cos³x = sin2x + 1 + sinx + cosx

⇔ (sinx + cosx)(sin²x + cos²x - sinx.cosx) = 2sinxcosx + sin²x + cos²x + sinx + cosx 

⇔ (sinx + cosx)(1 - sinx . cosx) = (sinx + cosx)² + (sinx + cosx)

⇔ (sinx + cosx)(1 - sinx.cosx - sinx - cosx - 1) = 0

⇔ (sinx + cosx)(sinx + cosx + sinx.cosx) = 0

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\sinx+cosx+sinx.cosx=0\left(2\right)\end{matrix}\right.\)

(1) ⇔ \(\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=0\)

⇔ \(x+\dfrac{\pi}{4}=k\pi\)

⇔ \(x=-\dfrac{\pi}{4}+k\pi\)

(2) ⇔ \(\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+\dfrac{1}{2}sin2x=0\)

⇔ \(\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)-\dfrac{1}{2}cos\left(2x+\dfrac{\pi}{2}\right)=0\)

⇔ \(\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)-\dfrac{1}{2}.\left[1-2sin^2\left(x+\dfrac{\pi}{4}\right)\right]=0\)

⇔ \(\dfrac{1}{4}sin^2\left(x+\dfrac{\pi}{4}\right)+\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)-\dfrac{1}{2}=0\)

⇔ \(\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{10}-2\sqrt{2}\\sin\left(x+\dfrac{\pi}{4}\right)=-\sqrt{10}-2\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{10}-2\sqrt{2}\)

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

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

Đặt \(x^2-2x=t\)

\(\Rightarrow t\left(t-3\right)=4\)

\(\Leftrightarrow t^2-3t-4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-2x=-1\\x^2-2x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+1=0\\x^2-2x-4=0\end{matrix}\right.\) (bấm máy)

Ta có: \(\Delta'=32>0\)

\(\Rightarrow\) Phương trình có 2 nghiệm phân biệt

Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=12\\x_1x_2=4\end{matrix}\right.\)

Mặt khác: \(T=\dfrac{x_1^2+x^2_2}{\sqrt{x_1}+\sqrt{x_2}}\)

\(\Rightarrow T^2=\dfrac{x_1^4+x^4_2+2x_1^2x_2^2}{x_1+x_2+2\sqrt{x_1x_2}}=\dfrac{\left(x_1^2+x_1^2\right)^2}{x_1+x_2+2\sqrt{x_1x_2}}\) \(=\dfrac{\left[\left(x_1+x_2\right)^2-2x_1x_2\right]^2}{x_1+x_2+2\sqrt{x_1x_2}}=\dfrac{\left(12^2-2\cdot4\right)^2}{12+2\sqrt{4}}=1156\)

Mà ta thấy \(T>0\) \(\Rightarrow T=\sqrt{1156}=34\) 

\(x\left(a^2+a-3a-3\right)+a^2-9=0\)0

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

nếu \(a=3\)thì phương trình nghiệm đúng với mọi x

nếu \(a\ne3\)thì phương trình có nghiệm a+1