a: x^3-7x-6

=x^3-x-6x-6

=x(x-1)(x+1)-6(x+1)

=(x+1)(x^2-x-6)

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

b: =2x^3+x^2-2x^2-x+6x+3

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

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

c: =2x^3-3x^2-2x^2+3x+2x-3

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

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

d: =2x^3+x^2+2x^2+x+2x+1

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

e: =3x^3+x^2-3x^2-x+6x+2

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

f: =27x^3-9x^2-18x^2+6x+12x-4

=(3x-1)(9x^2-6x+4)

a) \(x^3-7x-6\)

\(=x^3-x-6x-6\)

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

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

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

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

b) \(2x^3-x^2+5x+3\)

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

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

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

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

c) \(2x^3-5x^2+5x+1\)

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

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

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

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

d) \(2x^3+3x^2+3x+1\)

\(=2x^3+x^2+2x^2+x+2x+1\)

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

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

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

e) \(3x^3-2x^2+5x+2\)

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

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

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

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

f) \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=\left(27x^3-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)

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

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

a)  \(4\left(2x+7\right)^2=9\left(x+3\right)^2\)

\(\Leftrightarrow4\left(4x^2+28x+49\right)=9\left(x^2+6x+9\right)\)

\(\Leftrightarrow16x^2+112x+196=9x^2+54x+81\)

\(\Leftrightarrow7x^2+58x+115=0\)

\(\Leftrightarrow7x^2+35x+23x+115=0\)

\(\Leftrightarrow7x\left(x+5\right)+23\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\7x+23=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-\frac{23}{7}\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-5;-\frac{23}{7}\right\}\)

b) \(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow2x^3+2x^2+5x^2+5x+2x+2=0\)

\(\Leftrightarrow2x^2\left(x+1\right)+5x\left(x+1\right)+2\left(x+1\right)=0\)

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

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

\(\Leftrightarrow\left(x+1\right)\left[2x\left(x+2\right)+\left(x+2\right)\right]=0\)

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

\(\Leftrightarrow\)\(x+1=0\)

hoặc \(2x+1=0\)

hoặc  \(x+2=0\)

\(\Leftrightarrow\) \(x=-1\)

hoặc    \(x=-\frac{1}{2}\)

hoặc    \(x=-2\)

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

c) \(x^4+x^2+6x-8=0\)

\(\Leftrightarrow x^4-x^3+x^3-x^2+2x^2-2x+8x-8=0\)

\(\Leftrightarrow x^3\left(x-1\right)+x^2\left(x-1\right)+2x\left(x-1\right)+8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2-x^2-2x+4x+8\right)=0\)

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

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

\(\Leftrightarrow\)\(x-1=0\)

hoặc   \(x+2=0\)

hoặc   \(x^2-x+4=0\)

\(\Leftrightarrow\)\(x=1\)(tm)

hoặc   \(x=-2\)(tm)

hoặc  \(\left(x-\frac{1}{2}\right)^2+\frac{15}{4}=0\)(ktm)

Vậy tập nghiệm của phương trình là \(S=\left\{1;-2\right\}\)

d) \(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)

\(\Leftrightarrow x^3-3x^2+3x-1+8x^3+36x^2+54x+27=27x^3+8\)

\(\Leftrightarrow9x^3+33x^2+57x+26=27x^3+8\)

\(\Leftrightarrow18x^3-33x^2-57x-18=0\)

\(\Leftrightarrow18x^3-54x^2+21x^2-63x+6x-18=0\)

\(\Leftrightarrow18x^2\left(x-3\right)+21x\left(x-3\right)+6\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(18x^2+21x+6\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(18x^2+9x+12x+6\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[9x\left(2x+1\right)+6\left(2x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)\left(9x+6\right)=0\)

\(\Leftrightarrow\)\(x-3=0\)

hoặc  \(2x+1=0\)

hoặc  \(9x+6=0\)

\(\Leftrightarrow\)\(x=3\)

hoặc  \(x=-\frac{1}{2}\)

hoặc \(x=-\frac{2}{3}\)

Vậy tập nghiệm của phương trình là \(S=\left\{3;-\frac{1}{2};-\frac{2}{3}\right\}\)

\(2x^2+3x-27=2x^2-6x+9x-27=2x\left(x-3\right)+9\left(x-3\right)=\left(2x+9\right)\left(x-3\right)\)

\(x^3-7x+6=x^3-x-6x+6=x\left(x^2-1\right)-6\left(x-1\right)=x\left(x-1\right)\left(x+1\right)-6\left(x-1\right)=\left(x-1\right)\left(x^2+x-6\right)\)

\(x^3+5x^2+8x+4=x^3+x^2+4x^2+8x+4=x^2\left(x+1\right)+4\left(x^2+2x+1\right)=x^2\left(x+1\right)+4\left(x+1\right)^2\)

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

\(27x^3-27x^2+18x-4=27x^3-9x^2-18x^2+6x+12x-4\)

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

a) \(x^4+2x^3-3x^2-8x-4=0\)

\(\Leftrightarrow x^4-4x^2+2x^3-8x+x^2-4=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=4\\x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm2\\x=1\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{2;-2;1\right\}\)

b) \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)-72=0\)

Đặt \(t=x^2-4\), ta có :

\(t\left(t-6\right)-72=0\)

\(\Leftrightarrow t^2-6t-72=0\)

\(\Leftrightarrow\left(t-12\right)\left(t+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t-12=0\\t+6=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-16=0\left(tm\right)\\x^2+2=0\left(ktm\right)\end{cases}}\)

\(\Leftrightarrow x=\pm4\)

Vậy tập nghiệm của phương trình là \(S=\left\{4;-4\right\}\)

c) \(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow2x^3+2x^2+5x^2+5x+2x+2=0\)

\(\Leftrightarrow2x^2\left(x+1\right)+5x\left(x+1\right)+2\left(x+1\right)=0\)

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

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

\(\Leftrightarrow\)\(x+1=0\)

hoặc \(2x+1=0\)

hoặc \(x+2=0\)

\(\Leftrightarrow\)\(x=-1\)

hoặc \(x=-\frac{1}{2}\)

hoặc \(x=-2\)

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

a, \(x^4+2x^3-3x^2-8x-4=0\)

\(\Leftrightarrow\left(x^3+x^2-4x-4\right)\left(x+1\right)=0\)

TH1 : \(x+1=0\Leftrightarrow x=-1\)

TH2 : \(x^3+x^2-4x-4=0\Leftrightarrow\left(x+1\right)\left(x^2-4\right)=0\)

=> \(x=-1;x=\pm2\)

b, \(\left(x+2\right)\left(x-2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow x^4-14x^2+40=72\)

\(\Leftrightarrow x^4-14x^2-32=0\) Đặt \(x^2=t\left(t\ge0\right)\)

Ta có pt mới : \(t^2-14t-32=0\) Tự xử 

e, \(x^3+5x^2+8x+4=x^3+x^2+4x^2+4x+4x+4\)

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

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

d, \(27x^3-27x^2+18x-4=27x^3-9x^2-18x^2+6x+12x-4\)

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

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

\(a,\Rightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x+1\right)=0\\ \Rightarrow\left(2x+1\right)\left(2x-1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ c,\Rightarrow\left(x^2-8x+16\right)-10=0\\ \Rightarrow\left(x-4\right)^2-10=0\\ \Rightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\)

A= X²+5X+25/4-37/4 =(X+5/2)²-37/4 >= -37/4

A_min=-37/4

Đạt được khi : X=-5/2

B=-X²+7X+1=-(X²-7X-1)=-(X²+7X+49/4-53/4)=-(X+7/2)²+53/4<=53/4

B_Max=53/4

Đạt được khi:X=-7/2

C=2x²+6x=2x²+6x+9/4-9/4=2(x²+3x+9/4)-9/4=2(x+3/2)²-9/4>=-9/4

C_Min=-9/4

Đạt được khi:x=-3/2

1. Ta có: 

\(x^3-9x^2+27x-26=x^3-2x^2-7x^2+14x+13x-26\)

\(=x^2\left(x-2\right)-7x\left(x-2\right)+13\left(x-2\right)=\left(x-2\right)\left(x^2-7x+13\right)\)

Thay x = 23, ta có: \(C=\left(23-2\right)\left(23^2-7.23+13\right)=8001\)

2.

a) \(x^2+4y^2+6x-12y+18=0\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(4y^2-12y+9\right)=0\)

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

Mà \(\left(x-3\right)^2\ge0\) với mọi x, \(\left(2y-3\right)^2\ge0\) với mọi y

\(\Rightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)và \(\left(2y-3\right)^2=0\Leftrightarrow2y-3=0\Leftrightarrow y=\frac{3}{2}\)

Vậy \(\left(x,y\right)=\left(3;\frac{3}{2}\right)\)

b) \(2x^2+2y^2+2xy-10x-8y+41=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-10x+25\right)+\left(y^2-8y+16\right)=0\)

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

.....................................

Rồi giải tương tự như trên

i. (x²+y²+z²).(x+y+z)²+(xy+yz+zx)²

a. 6193 - [72 . (x - 104) + 98] = 5807

72 . (x - 104) + 98 = 6193 - 5807

72 . (x - 104) + 98 = 386

72 . (x - 104)         = 386 - 98

72 . (x - 104)         = 288

        x - 104          = 288 : 72 

        x - 104          = 4

        x                    = 4 + 104

        x                   = 108

e. (2x + 3)⁴ = 81

    (2x + 3)⁴ = 3⁴

 => TH1: 2x + 3 = 3

               2x       = 0

                 x       = \(\varnothing\)

TH2: 2x + 3 = -3

         2x       = -6

           x       = -3

Vậy .......

b. 5^{2x + 1} = 125

    5^{2x + 1} = 5³

    5^2x      = 5²

    5^x      = 5¹

=> x = 1

Vậy ......