\(a,\left(8-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\\ b,2x\left(x+81\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)

a)\(\left(8-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\)
b)\(2x\left(x+81\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)

a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)

b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)

c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Xem lại đề câu d 

a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

hay \(x=-\dfrac{1}{4}\)

c) Ta có: \(8x^3-50x=0\)

\(\Leftrightarrow2x\left(4x^2-25\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

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

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

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

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

b) 9x^2-(x-4)^2+0

\(a,\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)-9=0\\ \Leftrightarrow4x^2-4x^2+4x-16x+1-16-9=0\\ \Leftrightarrow-12x=24\\ \Leftrightarrow x=\dfrac{24}{-12}=-2\\ b,\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\\ \Leftrightarrow x^2+6x+9-\left(x^2+4x-32\right)=1\\ \Leftrightarrow x^2-x^2+6x-4x=1-9-32\\ \Leftrightarrow2x=-40\\ \Leftrightarrow x=-20\\ c,3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\\ \Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-4x+1\right)-7\left(x^2-9\right)=36\\ \Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\\ \Leftrightarrow3x^2+4x^2-7x^2+12x-4x=36-12-1-63\\ \Leftrightarrow8x=-40\\ \Leftrightarrow x=\dfrac{-40}{8}=-5\)

a) \(18-\left(2x+5\right)=9\)

\(2x+5=18-9\)

\(2x+5=9\)

\(2x=9-5\)

\(2x=4\)

\(x=2\)

a) \(18-\left(2x+5\right)=9\)

\(\Rightarrow2x+5=18-9=9\)

\(\Rightarrow2x=9-5=4\Rightarrow x=4:2=2\)

b) \(23x-4=32\Rightarrow23x=32+4=36\Rightarrow x=\dfrac{36}{23}\)

c) \(\left(3x+2\right)^2=64\)

\(\Rightarrow\left[{}\begin{matrix}3x+2=8\\3x+2=-8\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{10}{3}\end{matrix}\right.\)

d) \(x\left(2x-12\right)=0\Rightarrow6x\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

2x+5=18-9

2x+5=9

2x=9-5

2x=4

  x=4:2

  x=2

c)3x+2=64:2

   3x+2=32

    3x=32-2

    3x=30

     x=30:3

     x=10

B3  a) x=4        b) x=-7         c) x=5          d) x=4

B2  a) -3+ -2+ -1+0+1+2+3+4=4

      b) -6+ -5+ -4+ -3+ -2+ -1+0+1+2+3+4=-11

 c) -18+-17+-16+-15+-14+-13+-12+-11+-10+-9+-8+-7+-6+-5+-4+3+-2+-1+0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19=19

a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Rightarrow\left[2x+1-2\left(x+2\right)\right]\left[2x+1+2\left(x+2\right)\right]=9\\ \Rightarrow\left(2x+1-2x-4\right)\left(2x+1+2x+4\right)=9\\ \Rightarrow-3\left(4x+5\right)=9\\ \Rightarrow4x+5=-3\\ \Rightarrow4x=-8\\ \Rightarrow x=-2\)

b c d nữa bạn

\(b) (x+3)^2-(x-4)(x+8)=1 <=>x^2+6x=9-(x^2+8x-4x-32)=0 \)

\(.<=> X^2+6x+9-x^2-8x+4x+32-1=0\)

\(<=>2x=-40<=>x=-20\)

=> ptrình có tập nghiêm S={-20}

c) 3(x + 2)^2 + (2x - 1)^2 - 7(x + 3)(x - 3) = 36

\(<=>3(x^2+4x+4)+(4x^2-4x+1)-7(x^2-9)=36\)

\(<=>3x^2+12x+12+4x^2-4x+1-7x^2+49=0\)

\(<=>8x=-62<=>x=7,75\)

=> ptrình có tập nghiệm S={7,75}

d)d)(x - 3)(x^2 + 3x + 9) + x(x + 2)(2 - x) = 1

\(<=> x^3+3x^2+9x-3x^2-9x-27-x(x^2-4)=1\)

\(<=>x^3+3x^2+9x-3x^2-9x-27-x^3+4x-1=0\)

\(<=> 4x=28<=> x=7\)

=> ptrình có tập nghiệm S={7}