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a) \(\left(x+3\right)^3-x.\left(3x+1\right)^2+\left(2x+1\right).\left(4x^2-2x+1\right)-3x^2=54\)
\(\Leftrightarrow x^3+9x^2+27x+27-x.\left(9x^2+6x+1\right)+8x^3+1-3x^2=54\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\)
\(\Leftrightarrow26x+28=54\Leftrightarrow26x=54-28\Leftrightarrow26x=26\Leftrightarrow x=1\)
Vậy nghiệm của phương trình là x=1
b) \(\left(x-3\right)^3-\left(x-3\right).\left(x^2+3x+9\right)+6.\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+6.\left(x^2+2x+1\right)+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\)
\(\Leftrightarrow27x+12x+6=-33\Leftrightarrow39x=-33-6\Leftrightarrow39x=-39\Leftrightarrow x=-1\)
Vậy nghiệm của phương trình là x = -1
Trần Anh: Hí hí =)) ÀI LỚP DIU CHIU CHIU CHÍU :3 CẢM ƠN PẠN NHIỀU NHÁ ;) ;) ;)
\(\left(\dfrac{9}{x^2-9}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\) ( sửa đề \(x^3-9\) thành \(x^2-9\) )
\(=\left(\dfrac{9}{\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\dfrac{9+x-3}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{3\left(x-3\right)}{3x\left(x+3\right)}-\dfrac{x.x}{3x\left(x+3\right)}\right)\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}.\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)
\(=\dfrac{\left(x+6\right)3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(3x-9-x^2\right)}\)
\(=\dfrac{3x\left(x+6\right)}{\left(x-3\right)\left(3x-9-x^2\right)}\)
tự làm đi đừng ai giúp nhé lần này lại gặp mi nữa rồi
\(1,\left(2-x\right)^2-9=0\)
\(\Leftrightarrow\left(2-x-9\right)\left(2-x+9\right)=0\)
\(\Leftrightarrow\left(-7-x\right)\left(11-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\11-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=11\end{matrix}\right.\)
\(b,\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)=15-9\left(x+1\right)^2\)\(\Leftrightarrow x^3-9x^2+27x-27-x^3-27=15-9x^2-18x-9\)\(\Leftrightarrow x^3-9x^2+27x-x^3+9x^2+18x=15+27+27\)\(\Leftrightarrow45x=69\Rightarrow x=\dfrac{23}{15}\)
1. \(\left(2-x\right)^2-9=0\)
\(\left(2-x\right)^2=9\)
\(\left(2-x\right)^2=3^2\)
\(2-x=3\)
\(-x=-1\Rightarrow x=1\)
a) Cậu xem lại đề đi
b) \(3x.\left(x-2\right)-5x.\left(1-x\right)-8.\left(x^2-3\right)=4\)\(\Leftrightarrow3x^2-6x-5x+5x^2-8x^2+24-4=0\Leftrightarrow-11x+20=0\Leftrightarrow-11x=-20\Leftrightarrow x=\frac{20}{11}\)
c) \(2x^2+3.\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\Leftrightarrow2x^2+3\left(x^2-1\right)-5x\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\Leftrightarrow-5x=3\Leftrightarrow x=-\frac{3}{5}\)
Trần Anh: Cảm ơn bạn nhiều nhé :)) Phần a đúng là có sai đề pạn ạ mik làm hoài mà cux ko ra hì hì !!~~ Dù sao mik cux cảm ơn pạn nhiều nhiều nhé :3
Sửa đề: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9x^2+18x+9=15\)
=>45x+9=15
hay x=2/15