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22 tháng 12 2019

\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)^3+9\left(x^2+1\right)=63\)

\(\Leftrightarrow9\left(x^2+1\right)=63\)

\(\Leftrightarrow x^2+1=7\)

\(\Leftrightarrow x^2=6\)

\(\Leftrightarrow x=\pm\sqrt{6}\)

22 tháng 12 2019

ta có: ( x –3 )³ – ( x – 3) . ( x² + 3x + 9) + 9 (x² + 1 ) = 63

\(\Leftrightarrow x^3-3.x^2.3+3.x.3^2-3^3+3.\left(x^2+3x+9\right)-x\left(x^2+3x+9\right)+9x^2+\)\(9=63\)

\(\Leftrightarrow\left(x^3-x^3\right)-\left(9x^2-9x^2\right)+27x-\left(27-27\right)-\left(9x-9x\right)-\left(3x^2-3x^2\right)\)\(+9=63\)

\(\Leftrightarrow27x+9=63\)

\(\Leftrightarrow3x+1=7\)

\(\Leftrightarrow3x=6\)

\(\Leftrightarrow x=2\)

vậy: x=2

3 tháng 9 2021

\(a,\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\\ \Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\\ \Leftrightarrow2\left(x-3\right)=0\\ \Leftrightarrow x=3\)

\(b,4x^2-9=0\\ \Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(c,x^2+6x+9=0\\ \Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)

3 tháng 9 2021

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

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

\(\Leftrightarrow2\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Bài 2:

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

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

Để 3/x+2 là số nguyên thì \(x+2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{-1;-3;1;-5\right\}\)

6 tháng 8 2018

\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)

\(=4x\left(x^2-9\right)-x^3+27\)

\(=4x^3-36x-x^3+27\)

\(=3x^3-36x+27\)

6 tháng 8 2018

\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)

\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)

\(=\left(x+6\right).0\)

\(=0\)

Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-9x^2+27x+9x^2+18x+9=15\)

\(\Leftrightarrow45x=6\)

hay \(x=\dfrac{2}{15}\)