TK

a) 

\(x^3+\left(x-5\right)\left(x+8\right)=2x^2-37\\ \Leftrightarrow x^3+x^2+3x-40=2x^2-37\\ \Leftrightarrow x^3-x^2+3x-3=0\\ \Leftrightarrow x^2\left(x-3\right)+3\left(x-3\right)=0\\ \Leftrightarrow\left(x^2+3\right)\left(x-3\right)=0\)

Vì \(x^2+3\ge3>0\Rightarrow x-3=0\\ \Leftrightarrow x=3\)

b)

\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\\ \Leftrightarrow\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]=24\\ \Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)

Đặt \(x^2+x=y\)

\(\Rightarrow y\left(y-2\right)=24\\ \Leftrightarrow y^2-2y+1=25\\ \Leftrightarrow\left(y-1\right)^2=25\\ \Leftrightarrow\left[{}\begin{matrix}y-1=5\\y-1=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}y=6\\y=-4\end{matrix}\right.\)

Nếu y = 6 

\(\Rightarrow x^2+x=6\\ \Leftrightarrow x^2+x-6=0\\ \Leftrightarrow x^2+2x-3x-6=0\\ \Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Nếu y = -4

\(\Rightarrow x^2+x=-4\\ \Leftrightarrow x^2+x+\dfrac{1}{4}=-4+\dfrac{1}{4}\\ \Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=-\dfrac{15}{4}\)

Mà \(\left(x+\dfrac{1}{.2}\right)^2\ge0>-\dfrac{15}{4}\)

`=> Loại`

c) Vế còn lại là bao nhiêu?

\(|x-6|=-5x+9\)

Xét \(x\ge6\)thì \(pt< =>x-6=-5x+9\)

\(< =>x-6+5x-9=0\)

\(< =>6x-15=0\)

\(< =>x=\frac{15}{6}\)(ktm)

Xét \(x< 6\)thì \(pt< =>x-6=5x-9\)

\(< =>4x-9+6=0\)

\(< =>4x-3=0< =>x=\frac{3}{4}\)(tm)

Vậy ...

\(|x+1|=x^2+x\)

Xét \(x\ge-1\)thì \(pt< =>x+1=x^2+x\)

\(< =>x^2+x-x-1=0\)

\(< =>\left(x-1\right)\left(x+1\right)=0\)

\(< =>\orbr{\begin{cases}x=1\\x=-1\end{cases}\left(tm\right)}\)

Xét \(x< -1\)thì \(pt< =>-x-1=x^2+x\)

\(< =>x^2+2x+1=0\)

\(< =>\left(x+1\right)^2=0\)

\(< =>x=-1\left(ktm\right)\)

Vậy ...

\(a,\left(2x-3\right)^2=\left(x+1\right)^2\\ \Leftrightarrow\left(2x-3\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-3+x+1\right)\left(2x-3-x-1\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-4\right)\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\x=4\end{matrix}\right. \\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{2}{3};4\right\}\)

\(b,x^2-6x+9=9\left(x-1\right)^2\\ \Leftrightarrow\left(x-3\right)^2=9\left(x-1\right)^2\\ \Leftrightarrow\left(x-3\right)^2-9\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-3\right)^2-3^2\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-3\right)^2-\left[3\left(x-1\right)\right]^2=0\\ \Leftrightarrow\left(x-3\right)^2-\left(3x-3\right)^2=0\\ \Leftrightarrow\left(x-3+3x-3\right)\left(x-3-3x+3\right)=0\\ \Leftrightarrow-2x\left(4x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-2x=0\\4x-6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\4x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{0;\dfrac{3}{2}\right\}\)

x 2   +   9   >   6 x   ⇔   ( x   -   3 ) 2  > 0 (đúng với mọi x)

Phương trình  ⇔ x 2 - 2 x + 1 = x 2 + 6 x - 3

⇔ - 8x + 4 = 0

⇔ x = 1/2

Vậy phương trình có nghiệm x = 1/2

Phương trình 9 x 2  +6x+1 =0 có hệ số a=9,b’=3,c=1

Ta có:  ∆ ’ =  b ' 2 – ac =  3 2  -9.1 = 9 - 9 = 0

Phương trình có nghiệm kép:

x 1  =  x 2  = -b'/a =-3/9 =-1/3