\((x-2)(x^2+2x+5+2x+4-5)=0<=> (x-2)(x^2+4x+4)=0 <=> (x-2)(x+2)^2 = 0 <=> x = 2 ; x = -2\)

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

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

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

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

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

2:

a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8

=>x^2-x-12-x^2+4x+5=8

=>3x-7=8

=>3x=15

=>x=5

b: =>3x^2+3x-2x-2-3x^2-21x=13

=>-20x=15

=>x=-3/4

c: =>x^2-25-x^2-2x=9

=>-2x=25+9=34

=>x=-17

d: =>x^3-1-x^3+3x=1

=>3x-1=1

=>3x=2

=>x=2/3

bài 1:

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

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

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

=-x^2-3

a) 5^x-1 = 5^2 + 100

=> 5^x-1 = 125 = 5³

=> x - 1 = 3

=> x = 4.

b) x¹⁰ = x

=> x = {-1;0;1}

2.5².3² + {[2.5³ - (5x + 4).5 : (2³.3.5)]} = 455

=> 450 + {250 - (5x + 4).5 : 120} = 455

=> 250 - (5x + 4).5 : 120 = 5

=> (5x + 4).5 : 120 = 245

=> (5x + 4).5 = 245.120 = 29400

=> 5x + 4 = 29400 : 5 = 5880

=> 5x = 5876

=> x = 1175,2

a)    \(\left(3^2-2\right).\left(x-12+35\right)\)\(=\)\(5^2+279:5\)

\(7.\left(x-12+35\right)=80,8\)

\(x-12+35=80,8:7\)

\(x-12+35=\frac{404}{35}\)

\(x-12=\frac{404}{35}-35\)

\(x-12=\frac{-821}{35}\)

\(x=\frac{-821}{35}+12\)

\(x=\frac{-401}{35}\)

b)    \(260:\left(x+4\right)\)\(=\)\(5\left(2^3+5\right)-3\left(3^2+2^2\right)\)

     \(260:\left(x+4\right)=26\)

     \(x+4=260:26\)

    \(x+4=10\)

   \(x=10-4\)

   \(x=6\)

c)   \(7^{x-3}=343\)

    \(7^x:7^3=343\)

    \(7^x=343.7^3\)

    \(7^x=117649\)

vì \(117649=7^6\Rightarrow x=6\)

d)    \(\left(x-3\right)^{2017}=\left(x-3\right)^{2016}\)

\(\Rightarrow\hept{\begin{cases}x-3=1\\x-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=4\\x=3\end{cases}}}\)

e)     \(\left(2^3+3\right).\left(x-5\right)+14\)\(=\)\(5^2+124:2^2\)

      \(\left(2^3+3\right).\left(x-5\right)+14=56\)

       \(\left(2^3+3\right).\left(x-5\right)=56-14\)

       \(\left(2^3+3\right).\left(x-5\right)=42\)

        \(x-5=42:\left(2^3+3\right)\)

        \(x-5=\frac{42}{11}\)

      \(x=\frac{42}{11}+5\)

     \(x=\frac{97}{11}\)

a: \(\left(x+5\right)^2-\left(x-5\right)^2-2x+1=0\)

=>\(x^2+10x+25-\left(x^2-10x+25\right)-2x+1=0\)

=>\(x^2+8x+26-x^2+10x-25=0\)

=>18x+1=0

=>\(x=-\dfrac{1}{18}\)

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

=>\(4x^2-28x+49-\left(x^2+6x+9\right)-3x^2-6=0\)

=>\(x^2-28x+43-x^2-6x-9=0\)

=>34-34x=0

=>34x=34

=>x=1

c: \(\left(3x+2\right)^2-9\left(x-5\right)\left(x+5\right)=225-5x\)

=>\(9x^2+12x+4-9\left(x^2-25\right)-225+5x=0\)

=>\(9x^2+17x+4-225-9x^2+225=0\)

=>17x+4=0

=>x=-4/17

a: \(\Leftrightarrow\left(x+2\right)\left(x+2-2x+10\right)=0\)

\(\Leftrightarrow x\in\left\{-2;12\right\}\)

`a,x(x-1)-(x+2)^2=1`

`<=>x^2-x-x^2-4x-4=1`

`<=>-5x=5`

`<=>x=-1`

`b,(x+5)(x-3)-(x-2)^2=-1`

`<=>x^2+2x-15-x^2+4x-4+1=0`

`<=>6x-18=0`

`<=>x-3=0`

`<=>x=3`

`c,x(2x-4)-(x-2)(2x+3)=0`

`<=>2x(x-2)-(x-2)(2x+3)=0`

`<=>(x-2)(2x-2x-3)=0`

`<=>-3(x-2)=0`

`<=>x-2=0`

`<=>x=2`

`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`

`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`

`<=>4x+26=-12`

`<=>4x=-38`

`<=>x=-19/2`

Bài 1: 

a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)

\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)

\(\Rightarrow16x-5=x-2\)

\(\Rightarrow16x-x=5-2\)

\(\Rightarrow15x=3\)

\(\Rightarrow x=\dfrac{15}{3}=5\)

b) \(12x^2-4x\left(3x+5\right)=10x-17\)

\(\Rightarrow12x^2-12x^2-20x=10x-17\)

\(\Rightarrow-20x=10x-17\)

\(\Rightarrow-20x-10x=-17\)

\(\Rightarrow-30x=-17\)

\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)

c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)

\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)

\(\Rightarrow-8x=12\)

\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)

Bài 2: 

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

\(=x^2-7x+5x-35-7x^2+21x\)

\(=-6x^2+19x-35\)

b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)

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

\(=x^3-2x^2-6x-5\)

c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)

\(=x^2-7x-5x+35-x^2-3x+4x-12\)

\(=11x+23\)

d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)

\(=x^2-2x-x+2-x^2+2x+5x+10\)

\(=4x+12\)