a. \(7x-x^2-6=0\)

\(\Rightarrow-x^2+7x-6=0\)

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

\(\Rightarrow-x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(-x+6\right)=0\)

+) Nếu \(x-1=0\Rightarrow x=1\)

+) Nếu \(-x+6=0\Rightarrow x=6\)

Vậy x=1 hoặc x=6

b. \(8x^3-36x^2+57x-27=0\)

\(\Rightarrow\left(2x\right)^2-3.2^2.x^2.3+3.2x.3^2-3^3=0\)

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

\(\Rightarrow2x-3=0\Rightarrow x=\frac{3}{2}\)

Vậy...........

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

\(\Leftrightarrow2x+2-1=3-1+2x\)

\(\Leftrightarrow2x-2x=-2+1+3-1\)

\(\Leftrightarrow0x=1\)(vô lí)

Vậy tập nghiệm của phương trình trên bằng \(S=\varnothing\)

b)\(\left(5x-1\right)^2-x^2-8x-16=0\)

\(\Leftrightarrow\left(5x-1\right)^2-\left(x^2+8x+16\right)=0\)

\(\Leftrightarrow\left(5x-1\right)^2-\left(x+4\right)^2=0\)

\(\Leftrightarrow\left(5x-1-x-4\right)\left(5x-1+x+4\right)=0\)

\(\Leftrightarrow\left(4x-5\right)\left(6x+3\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}4x-5=0\\6x+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{4}\\x=-\frac{1}{2}\end{cases}}}\)

Vậy tập nghiệm của phương trình trên bằng\(S=\left\{\frac{5}{4};-\frac{1}{2}\right\}\)

 #hoktot<3# 

a) (2x - 1)(x^2 - 1 + 1) = 2x^3 - 3x^2 + 2

(2x - 1).x^2 = 2x^3 - 3x^2 + 2

2x^3 - x^2 = 2x^3 - 3x^2 + 2

-x^2 = -3x^2 + 2

2x^2 = 2

x^2 = 1

=> x = 1; -1

b) (x + 2)(x + 2) - (x - 2)(x - 2) = 8x

(x + 2)^2 - (x - 2)^2 = 8x

x^2 + 4x + 4 - x^2 + 4x - 4 = 8x

8x = 8x

=> x thuộc N*

c) (x + 1)(x + 2)(x + 5) - x^3 - 8x^2 = 27

x^3 + 5x^2 + 2x^3 + 10x + x^2 + 5x + 2x + 10x - x^3 - x^2 = 27

17x + 10 = 27

17x = 27 - 10

17x = 17

=> x = 1

d) (x + 1)(x^2 + 2x + 4) - x^3 - 3x^2 + 16 = 0

x^3 + 2x^2 + 4x + x^2 + 2x + 4 - x^3 - 3x^2 + 16 = 0

6x + 20 = 0

6x = -20

x = -20/6

=> x = -10/3

a.\(x^3-6x^2+12x-8=0\Rightarrow\)\(\left(x-2\right)^3=0\Rightarrow x=2\)

b.\(x^3+9x^2+27x+27=0\Rightarrow\left(x+3\right)^3=0\)\(\Rightarrow x=-3\)

c. \(8x^3-12x^2+6x-1=0\)

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

\(\Rightarrow x=\frac{1}{2}\)

Bài 1 :

\(x^2\left(x-3\right)-4x+12=0\)

\(x^2\left(x-3\right)-4\left(x-3\right)=0\)

\(\left(x-3\right)\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\left\{\pm2\right\}\end{cases}}}\)

Bài 2 :

\(x-1-x^2\)

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

\(=-\left[x^2-2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\right]\)

\(=-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\)

Vì \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge0\forall x\)

\(\Rightarrow-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\le0\forall x\left(đpcm\right)\)

a) 2x²+3x-5=0

=> 2x²+5x-2x-5=0

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

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

=> 2x-5=0,   x-1=0

=> x=5/2; 1

 \(2x^2+3x-5=0< =>2x^2-2+3x-3=0\)

\(< =>2\left(x+1\right)\left(x-1\right)-3\left(x-1\right)=0\)

\(< =>\left(x-1\right)\left(2x-1\right)=0< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)