d: \(\dfrac{x^4-2x^3+2x-1}{x^2-1}\)

\(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)

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

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

sao làm có 1 ý vậy bạn ơi

\(a,=4x^2+3xy-y^2+4xy-4x^2=7xy-y^2\\ b,=x^2-9-x^3+3x+x^2-3=-x^3+2x^2+3x-12\\ c,=-2x^2+12x-18+5x^2+4x-1=3x^2+16x-19\\ d,=8x^3+1-3x^3+6x^2=5x^3+6x^2+1\\ e,=\left(3x^2+4x+15x+20\right):\left(3x+4\right)\\ =\left(3x+4\right)\left(x+5\right):\left(3x+4\right)\\ =x+5\\ f,=\left(x^3+4x^2-3x+3x^2+12x-9+3x+3\right):\left(x^2+4x-3\right)\\ =\left[\left(x^2+4x-3\right)\left(x+3\right)+3x+3\right]:\left(x^2+4x-3\right)\\ =x+3\left(dư.3x+3\right)\)

a.

\(\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-11x+28}+\dfrac{1}{x^2-19x+84}=\dfrac{1}{4}\\ \Rightarrow\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-7\right)}+\dfrac{1}{\left(x-7\right)\left(x-12\right)}=\dfrac{1}{4}\\ \Rightarrow\dfrac{1}{x-3}-\dfrac{1}{x-12}=\dfrac{1}{4}\\ \Rightarrow\dfrac{-9}{\left(x-3\right)\left(x-12\right)}=\dfrac{1}{4}\\ \Rightarrow x^2-15x+36=-36\\ \)

Tự giải tiếp

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

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

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

b) \(\left(x+2\right)\left(x-1\right)+2\left(3x-2\right)^2+4x-19x^2\)

\(=x^2+2x-x-2+2\left(9x^2-12x+4\right)+4x-19x^2\)

\(=x^2+2x-x-2+18x^2-24x+8+4x-19x^2\)

\(=-19x+6\)

c) \(2\left(3-x\right)\left(x-2\right)-\left(3x+1\right)^2+5x-11x^2\)

\(=6-2x\left(x-2\right)-\left(9x^2+6x+1\right)+5x-11x^2\)

\(=6-2x^3+4x-9x^2-6x-1+5x-11x^2\)

\(=-2x^3-20x^2+3x+5\)

a, \(3x+2\left(x-5\right)=6-\left(5x-1\right)\)

\(\Leftrightarrow3x+2x-10=6-5x+1\)

\(\Leftrightarrow-15\ne0\)Vậy phương trình vô nghiệm 

b, \(x^3-3x^2-x+3=0\)

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

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=3;\pm1\)

Vậy tập nghiệm của phương trình là S = { 1 ; -1 ; 3 }

c, \(\frac{1}{x-3}+\frac{x}{x+3}=\frac{2}{x^2-9}ĐK:x\ne\pm3\)

\(\Leftrightarrow\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)

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

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

Vậy ... 

a) \(\text{​​}/3x-5/-\frac{1}{7}=\frac{1}{3}\)                           b)\(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)

  \(/3x-5/=\frac{10}{21}\)                                           \([x.\left(\frac{3}{5}-\frac{2}{3}-1\right)]=\frac{-5}{21}.7\)

 \(\Rightarrow3x-5=\frac{10}{21}hay3x-5=\frac{-10}{21}\)         \(\left[x.\frac{-16}{15}\right]=\frac{-5}{3}\)

\(3x=\frac{115}{21}\)                \(3x=\frac{95}{21}\)                         \(x=\frac{25}{16}\)

\(x=\frac{115}{63}\)                  \(x=\frac{95}{63}\)                             Vậy x = \(\frac{25}{16}\)

                      Vậy x \(\in\left\{\frac{115}{63};\frac{95}{63}\right\}\)