\(B=x^8+2x^5-2x^4+x^2-2x-100+10x\left(x^4+x\right)+\left(5x-1\right)^2\)

\(=x^8+2x^5-2x^4+x^2-2x-100+10x^5+25x^2-10x+1\)

\(=x^8+12x^5-2x^4+36x^2-12x-99\)

\(=x^8+6x^5+9x^4+6x^5+36x^2+54x-11x^4-66x-99\)

\(=x^4\left(x^4+6x+9\right)+6x\left(x^4+6x+9\right)-11\left(x^4+6x+9\right)\)

\(=\left(x^4+6x+9\right)\left(x^4+6x-11\right)\)

a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)

\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)

\(< =>12-2+4x-2x^2=6x^2-13x+6\)

\(< =>10+4x-2x^2-6x^2+13x-6=0\)

\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)

b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)

\(< =>x-9=0< =>x=9\)

c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)

\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)

d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)

\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)

e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)

\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)

f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)

\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)

g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)

\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)

h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)

\(< =>x^2-16-6x+4=x^2-8x+16\)

\(< =>x^2-6x-12-x^2+8x-16=0\)

\(< =>2x-28=0< =>x=\frac{28}{2}=14\)

q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề

A. \(4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\)
\(\Leftrightarrow4x+8-14x+7+27x-36=30\)
\(\Leftrightarrow4x-14x+27x=30-8-7+36\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\) . Vậy \(S=\left\{3\right\}\)

B. \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=16-15-16+11\)
\(\Leftrightarrow10x=-4\)
\(\Leftrightarrow x=-\dfrac{2}{5}\) . Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)

Câu C) bạn xem lại đề nha mik tính ko đc

D. \(\left(5x-3\right)4x-2x\left(10x-3\right)=15\)
\(\Leftrightarrow20x^2-12x-20x^2+6x=15\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{5}{2}\) . Vậy \(S=\left\{-\dfrac{5}{2}\right\}\)

A = x⁸ + 2x⁵ - 2x⁴ + x² - 2x - 100 + 10x.(x⁴ + x) + (5x - 1)²

A = (x⁸ + 2x⁵ + x²) - (2x⁴ + 2x) + 10x.(x⁴ + x) + (5x - 1)² - 100

A = (x⁴ + x)² - 2(x⁴ + x) + 10x. (x⁴ + x) + (5x -1)² - 100

A = (x⁴ + x)² + (x⁴ + x).(10x - 2) + (5x - 1)² - 100

A = [(x⁴ + x)² + 2.(x⁴ + x).(5x - 1) + (5x - 1)² ] - 100

A = [x⁴ + x + 5x - 1]² - 10²

A = (x⁴ + 6x - 11).(x⁴ + 6x + 9)

Hok tốt ^_^

5: \(\Leftrightarrow9\left(x^2-5x-4\right)=36\left(x+1\right)+8\left(x^2-10x\right)\)

\(\Leftrightarrow9x^2-45x-36-36x-36-8x^2+80x=0\)

\(\Leftrightarrow x^2-x-72=0\)

=>(x-9)(x+8)=0

=>x=9 hoặc x=-8

6: \(\Leftrightarrow x^2-9=9x-x^2-9+x\)

\(\Leftrightarrow2x^2-10x=0\)

=>2x(x-5)=0

=>x=0 hoặc x=5

5, <=> 9x^2 - 45x - 36 = 36x + 36 + 8x^2 - 80x 

<=> x^2 - x - 72 = 0 <=> x = 9 ; x = -8 

6, <=> x^2 - 9 = 9x - x^2 - 9 + x = 10x - x^2 - 9 

<=> 2x^2 - 10x = 0 <=> x = 0 ; x = 5 

7, <=> (x-1)^2 = (3x+3)^2 

<=> (x-1-3x-3)(x-1+3x+3) = 0

<=> (-2x-4)(4x+2) = 0 <=> x = -2;x=-1/2

8, = (x^2-10x-15)(x^2-10x+25)