1

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

\(\left(x+5y\right)\left(x-5y\right)=x^2-25y\)

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

\(\left(x-5\right)\left(x^2+5x+25\right)=x^3-125\)

Bài 3:

a: \(\Leftrightarrow x^2+8x+16-x^2+1=16\)

=>8x+1=0

=>x=-1/8

b: \(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

=>2x+255=0

=>x=-255/2

c: \(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6x^2+12x+6=49\)

=>24x+62=49

=>24x=-13

=>x=-13/24

d: =>x^3+8-x^3-2x=15

=>-2x=15-8=7

=>x=-7/2

gt : \(x^2-4x+1=0\Leftrightarrow x^2+1=4x\)(1)

\(\Leftrightarrow\left(x^2+1\right)^2=16x^2\Leftrightarrow x^4+2x^2+1=16x^2\Rightarrow x^4+1=14x^2\)(2)

\(\Leftrightarrow\left(x^2+1\right)^3=64x^3\Leftrightarrow x^6+3x^4+3x^2+1=64x^3\)

\(\Leftrightarrow x^6+3x^2\left(x^2+1\right)+1=64x^3\Leftrightarrow x^6+12x^3+1=64x^3\)

\(\Rightarrow x^6+1=52x^3\)(3)

Thay (1);(2);(3) vào T ta dược :

\(T=\left(\frac{x^2+1}{x}\right)^2+\left(\frac{x^4+1}{x^2}\right)^2+\left(\frac{x^6+1}{x^3}\right)^2\)

\(=\left(\frac{4x}{x}\right)^2+\left(\frac{14x^2}{x^2}\right)^2+\left(\frac{52x^3}{x^3}\right)^2=4^2+14^2+52^2=2916\)

Bài 4:

1: \(\left(x-1\right)\left(x^2+x+1\right)-x^3-6x=11\)

=>\(x^3-1-x^3-6x=11\)

=>-6x-1=11

=>-6x=11+1=12

=>\(x=\dfrac{12}{-6}=-2\)

2: \(16x^2-\left(3x-4\right)^2=0\)

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

=>\(\left(4x-3x+4\right)\left(4x+3x-4\right)=0\)

=>(x+4)(7x-4)=0

=>\(\left[{}\begin{matrix}x+4=0\\7x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{4}{7}\end{matrix}\right.\)

3: \(x^3-x^2-3x+3=0\)

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

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

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

=>\(\left[{}\begin{matrix}x-1=0\\x^2-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

4: \(\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}\)(ĐKXĐ: \(x\notin\left\{-2;-1\right\}\))

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

=>\(x^2+4x+4=x^2-1\)

=>4x+4=-1

=>4x=-5

=>\(x=-\dfrac{5}{4}\left(nhận\right)\)

5: ĐKXĐ: \(x\notin\left\{0;-1\right\}\)

\(\dfrac{1}{x}+\dfrac{2}{x+1}=0\)

=>\(\dfrac{x+1+2x}{x\left(x+1\right)}=0\)

=>3x+1=0

=>3x=-1

=>\(x=-\dfrac{1}{3}\left(nhận\right)\)

6: ĐKXĐ: \(x\notin\left\{0;3\right\}\)

\(\dfrac{9-x^2}{x}:\left(x-3\right)=1\)

=>\(\dfrac{-\left(x^2-9\right)}{x\left(x-3\right)}=1\)

=>\(\dfrac{-\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}=1\)

=>\(\dfrac{-x-3}{x}=1\)

=>-x-3=x

=>-2x=3

=>\(x=-\dfrac{3}{2}\left(nhận\right)\)

1) \(a,\left(x-3\right)\left(x+7\right)-\left(x+5\right)\left(x-1\right)=x^2+4x-21-x^2-4x+5=-16\)

\(b,\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2=\left(x+8-x+2\right)^2=\left(10\right)^2=100\)

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

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

hình như sai r thui ko làm nữa

1.

a,\(\left(x-3\right)\left(x+7\right)-\left(x+5\right)\left(x-1\right)\)

\(=(x^2+7x-3x-21)-\left(x^2-x+5x-5\right)\)

\(=x^2+7x-3x-21-x^2+x-5x+5\)

\(=-16\)

b,\(\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2\)

\(=\left[\left(x+8\right)-\left(x-2\right)\right]^2\)

\(=\left(x+8-x+2\right)^2\)

\(=10^2=100\)

c,\(\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)

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

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

\(=2\)

2.

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

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

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

\(=6x^2+2\)

(Bạn xem lại đề câu này nhé,hoặc là mình làm sai)

b,\(\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)

\(=\left(x^3+6x^2+12x+8+x^3-6x^2+12x-8\right)-2x^3-24x\)

\(=x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x^3-24x\)

\(=x^3+x^3-2x^3+6x^2-6x^2+12x+12x-24x+8-8\)

\(=0\)

Vậy giá trị của biểu thức trên không phụ thuộc bởi x

(Câu này cũng không chắc lắm,bạn cũng xem lại xem mình đúng chưa nhé!!!)

\(\)

\(\)

a: =>(3/2-2x):2/3=1/6

=>3/2-2x=1/6x2/3=2/18=1/9

=>2x=25/18

hay x=25/36

b: \(\Leftrightarrow2x-2x+\dfrac{5}{2}-2=x-\dfrac{1}{4}\)

=>x-1/4=1/2

=>x=3/4

c: \(\Leftrightarrow2x-\dfrac{2}{3}-\dfrac{1}{3}x+\dfrac{1}{4}x=0\)

=>23/12x=2/3

=>x=8/23

Bài 2: 

a: \(\Leftrightarrow x^3-27-x\left(x^2-4\right)=1\)

\(\Leftrightarrow x^3-27-x^3+4x=1\)

=>4x-27=1

hay x=7

b: \(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x-1\right)^2+10=0\)

\(\Leftrightarrow6x^2+12-6x^2+12x-6=0\)

=>12x+6=0

hay x=-1/2