Question

[duong thi thanh thuy]

Bài 1:tìm x

a,(x+1)^3+(2-x)(4+2x+x^2)+3x(x+2)=17

b,(x+2)(x^2-2x+4)-x(x^2-2)=15

c,(x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2=15

d,x(x-5)(x+5)-(x+2)(x^2-2x+4)=3

Answer 1

[T.Thùy Ninh]

\(a,\left(x+1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)\(\Leftrightarrow x^3+3x^2+3x+1+8-x^3+3x^2+6x-17=0\)\(\Leftrightarrow6x^2+9x-8=0\)

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

\(\Leftrightarrow\left(x^2+\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{9}{16}-\dfrac{4}{3}=0\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right)^2=\dfrac{91}{48}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\sqrt{\dfrac{91}{48}}\\x+\dfrac{3}{4}=-\sqrt{\dfrac{91}{48}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{91}{48}}-\dfrac{3}{4}\\x=-\sqrt{\dfrac{91}{48}}-\dfrac{3}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{273}}{12}\\x=-\dfrac{9+\sqrt{273}}{12}\end{matrix}\right.\)

b, \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)

\(\Leftrightarrow x^3+8-x^3+2x-15=0\)

\(\Leftrightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)