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?

1)(x²-4x+16)(x+4)-x(x+1)(x+2)+3x²=0

\(\Rightarrow\)(x³+64)-x(x²+2x+x+2)+3x²=0

\(\Rightarrow\)x³+64-x³-2x²-x²-2x+3x²=0

\(\Rightarrow\)-2x+64=0

\(\Rightarrow\)-2x=-64

\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)

\(\Rightarrow x=32\)

2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50

\(\Rightarrow\)8x-24x²+2-6x+24x²-60x-4x+10=50

\(\Rightarrow\)-62x+12=50

\(\Rightarrow\)-62x=50-12

\(\Rightarrow\)-62x=38

\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)

3)(x²+2x+4)(2-x)+x(x-3)(x+4)-x²+24=0

\(\Rightarrow\)8-x³+x(x²+4x-3x-12)-x²+24=0

\(\Rightarrow\)8-x³+x³+4x²-3x²-12x-x²+21=0

\(\Rightarrow\)-12x+29=0

\(\Rightarrow\)-12x=-29

\(\Rightarrow\)x=\(\dfrac{-29}{-12}=\dfrac{29}{12}\)

\(f'\left(x\right)=4x^3\Rightarrow g\left(x\right)=4x^3-3x^2-6x+1\)

\(g'\left(x\right)=12x^2-6x-6=0\Rightarrow\left[{}\begin{matrix}x_2=-\dfrac{1}{2}\\x_1=1\end{matrix}\right.\)

\(\Rightarrow g\left(-\dfrac{1}{2}\right).g\left(1\right)=\dfrac{11}{4}.\left(-4\right)=-11\)

\(y'=\left(6x^5-6\right)f'\left(x^6-3x^2\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\f'\left(x^6-3x^2\right)=0\end{matrix}\right.\) trong đó \(x=1\) bội lẻ

\(f'\left(x\right)=0\) có các nghiệm \(x=-2;0;2;a;6\)

\(\Rightarrow f'\left(x^6-3x^2\right)=0\Leftrightarrow\) 5 trường hợp:

\(x^6-3x^2=-2\) \(\Leftrightarrow\left(x+1\right)^2\left(x-1\right)^2\left(x^2+2\right)=0\) có 2 nghiệm \(x=-1\) (bội chẵn) và \(x=1\) (bội chẵn)

.... làm tương tự

Riêng với \(x^6-3x^2=a\) thì dựa trên BBT của \(y=x^6-3x^2\) ta thấy pt này có 2 nghiệm đều bội lẻ khi \(4< a< 6\)

Đếm số nghiệm bội lẻ là được

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

\(\left(x-2\right)^2=0\)

=>x=2

b) \(2x^2-x=0\)

\(x\left(2x-1\right)=0\)

=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)

c) \(x^2-5x+6=0\)

\(x^2-2x-3x+6=0\)

\(\left(x-2\right)\left(x-3\right)=0\)

=> \(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

d) \(x^2+y^2=0\)

Vì \(x^2,y^2\ge0\forall x,y\)

=>x=y=0

e) \(x^2+6x+10=0\)

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

Vì \(\left(x+3\right)^2\ge0\forall x\)

=> VT>0 \(\forall x\)

=> phương trình vô nghiệm

a) \(x^2-4x+4=0\)

\(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

b) \(2x^2-x=0\)

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

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

c) \(x^2-5x+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\) \(\left(a+b+c=0\right)\)

d) \(x^2+y^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\y^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

e) \(x^2+6x+10=0\)

\(\Leftrightarrow x^2+6x+9+1=0\)

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

mà \(\left(x+3\right)^2+1\ge1>0,\forall x\in R\)

Nên phương trình (1) vô nghiệm

\(E=x^2+6x+11\)

\(=x^2+6x+9+2\)

\(=\left(x+3\right)^2+2>0\forall x\)

\(F=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

a) (x - 2)(x - 3).                        b) 3(x - 2)(x + 5).

c) (x - 2)(3x + 1).                     d) (x-2y)(x - 5y).

e) (x + l)(x + 2)(x - 3).             g) (x-1)(x + 3)( x 2  + 3).

h) (x + y - 3)(x - y + 1).

Ta có:  x 2  – 6x + 10 =  x 2  – 2.x.3 + 9 + 1 = x - 3 2  + 1

Vì  x - 3 2  ≥ 0 với mọi x nên  x - 3 2  + 1 > 0 mọi x

Vậy  x 2  – 6x + 10 > 0 với mọi x.(đpcm)

a) Thực hiện rút gọn VT = -2x – 64

Giải phương trình -2x – 64 = 0 thu được x = -32.

b) Thực hiện rút gọn VT = -62 x +12

Giải phương trình -62x + 12 = -50 thu được x = 1.

a) \(A=x^2+6x+10\)

\(=\left(x+3\right)^2+1=\left(-103+3\right)^2+1=100^2+1=10001\)

b) \(B=x^3+6x^2+12x+12\)

\(=\left(x+2\right)^3+4=\left(8+2\right)^3+4=1004\)

\(A=x^2+6x+10=\left(x^2+2.x.3+3^2\right)+1\\ =\left(x+3\right)^2+1=\left(-103+3\right)^2+1=10000+1=10001\\ b,B=x^3+6x^2+12x+12\\ =x^3+3x^2.2+3.x.2^2+2^3+4=\left(x+2\right)^3+4\\ =\left(8+2\right)^3+4=1000+4=1004\)