Question

a) (x+2)\(^2\)+2(x-4)=(x-4)(x-2)

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

c) (x+3)\(^2\)-(x-3)\(^2\)=6x+18

d) (x-1)\(^3\)-x(x+1)\(^2\)=5x(2-x)-11(x+2)

 

Answer 1

a) Ta có: \(\left(x+2\right)^2+2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)

\(\Leftrightarrow x^2+4x+4+2x-8=x^2-6x+8\)

\(\Leftrightarrow x^2+6x-4-x^2+6x-8=0\)

\(\Leftrightarrow12x-12=0\)

\(\Leftrightarrow12x=12\)

hay x=1

Vậy: S={1}

b) Ta có: \(\left(x+1\right)\left(2x-3\right)-3\left(x-2\right)=2\left(x-1\right)\)

\(\Leftrightarrow2x^2-3x+2x-3-3x+6=2x-2\)

\(\Leftrightarrow2x^2-4x+3-2x+2=0\)

\(\Leftrightarrow2x^2-6x+5=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{5}{2}\right)=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{4}=0\)(Vô lý)

Vậy: \(S=\varnothing\)

c) Ta có: \(\left(x+3\right)^2-\left(x-3\right)^2=6x+18\)

\(\Leftrightarrow x^2+6x+9-\left(x^2-6x+9\right)-6x-18=0\)

\(\Leftrightarrow x^2-9-x^2+6x-9=0\)

\(\Leftrightarrow6x-18=0\)

\(\Leftrightarrow6x=18\)

hay x=3

Vậy: S={3}

d) Ta có: \(\left(x-1\right)^3-x\left(x+1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)

\(\Leftrightarrow x^3-3x^2+3x-1-x\left(x^2+2x+1\right)=5x-5x^2-11x-22\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=-5x^2-6x-22\)

\(\Leftrightarrow-5x^2+2x-1+5x^2+6x+22=0\)

\(\Leftrightarrow8x+21=0\)

\(\Leftrightarrow8x=-21\)

hay \(x=-\dfrac{21}{8}\)

Vậy: \(S=\left\{-\dfrac{21}{8}\right\}\)